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Unit 5

Arithmetic Progression

Exercise 5.1

Question 1:

In which of the following situations, does the list of numbers involved make as arithmetic progression and why?

  1. The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.
  2. The amount of air present in a cylinder when a vacuum pump removes of the air remaining in the cylinder at a time.
  3. The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.
  4. The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.

Answer:

  1. It can be observed that Taxi fare for 1st km = 15

     Taxi fare for first 2 km = 15 + 8 = 23

     Taxi fare for first 3 km = 23 + 8 = 31

     Taxi fare for first 4 km = 31 + 8 = 39

     Clearly 15, 23, 31, 39 … forms an A.P. because every term is 8 more than the       preceding term.

  1. Let the initial volume of air in a cylinder be V lit. In each stroke, the vacuum pump removes of air remaining in the cylinder at a time. In other words, after every stroke, only part of air will remain.

VoiZNurbndMZZAIDsww4xq2F5XegEH30rcDZsilep3FqQr 7p IkZoGGYyMBWwHlnNa6PoaFXdXB6e j27oSnDG7rSRK3CDWzSSkPwtaPuyQgUrYP0SNVicAnKKNCQyWfJu02gM

bZzg4s7zzAa6dkG OpYis VZeq7OOvo 7rylk8m73cEtpTgRu9KHO0B51aot7HvLZ0SbyWVlIlyQwXT

Therefore, volumes  will be

Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

  1. Cost of digging for first metre = 150

Cost of digging for first 2 metres = 150 + 50 = 200

Cost of digging for first 3 metres = 200 + 50 = 250

Cost of digging for first 4 metres = 250 + 50 = 300

Clearly, 150, 200, 250, 300 … forms an A.P. because every term is 50 more than the preceding term.

  1. We know that if Rs P is deposited at r% compound interest per annum for n years, our money will be after n years.

Therefore, after every year, our money will be

VViU2eF eEb4Gv9IAVY4mSbcL9miATvVZeUQAyIbsv1ezhAQVSP7KpZAzCOmKG6TCumNLMJQSJPgQ jNBWh4zJaOJLAMMFkzuBu3UB 9ERfsqLzUV1URQfQF3ghkHR JtPtgcNg

Clearly, adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

Question 2:

Write first four terms of the A.P. when the first term a and the common difference d are given as follows

(i) a = 10, d = 10

  1. a = − 2, d = 0
  2. a = 4, d = − 3
  3. a = − 1 d =

(v) a = − 1.25, d = − 0.25

Answer:

(i) a = 10, d = 10

Let the series be a1, a2, a3, a4, a5 … a1 = a = 10

a2 = a1 + d = 10 + 10 = 20 a3 = a2 + d = 20 + 10 = 30 a4 = a3 + d = 30 + 10 = 40

a5 = a4 + d = 40 + 10 = 50

Therefore, the series will be 10, 20, 30, 40, 50 …

First four terms of this A.P. will be 10, 20, 30, and 40.

  1. a = −2, d = 0

Let the series be a1, a2, a3, a4 … a1 = a = −2

a2 = a1 + d = − 2 + 0 = −2 a3 = a2 + d = − 2 + 0 = −2

a4 = a3 + d = − 2 + 0 = −2

Therefore, the series will be −2, −2, −2, −2 …

First four terms of this A.P. will be −2, −2, −2 and −2.

  1. a = 4, d = −3

Let the series be a1, a2, a3, a4 … a1 = a = 4

a2 = a1 + d = 4 − 3 = 1 a3 = a2 + d = 1 − 3 = −2

a4 = a3 + d = − 2 − 3 = −5

Therefore, the series will be 4, 1, −2 −5 …

First four terms of this A.P. will be 4, 1, −2 and −5.

  1. a = −1, d =

Let the series be a1, a2, a3, a4

AIVcmlSm4 mfJ HaZ4GPQQ9NiU3N3CzC8ryGdSUWh8P5LiECBUuo4GSwHnh83x3 1pSBhoglLVdun4Q9mC0jChF uQX Y7z8P7 2zOXdCYS3AMCSRJ8ITCS6rmHA0eYtdn51Fkg

Clearly, the series will be

upMwxIbmdY05sql74xcUQ6SVFGRYFKVPNwZapbKACxbVuwF dx1r45OG  ………….

First four terms of this A.P. will be and .

(v) a = −1.25, d = −0.25

Let the series be a1, a2, a3, a4

a1 = a = −1.25

a2 = a1 + d = − 1.25 − 0.25 = −1.50 

a3 = a2 + d = − 1.50 − 0.25 = −1.75 

a4 = a3 + d = − 1.75 − 0.25 = −2.00

Clearly, the series will be 1.25, −1.50, −1.75, −2.00 ……..

First four terms of this A.P. will be −1.25, −1.50, −1.75 and −2.00.

Question 3:

For the following A.P.s, write the first term and the common difference. (i) 3, 1, − 1, − 3 …

(ii) − 5, − 1, 3, 7 …

(iii) zcsi Md56EMX6DJRhqBHPe4laeQZoeyUtCJbtJ3Ib99Rbf1H9FCY6MD89eZc0YNif1wW iwEIgzeFfHYe6OHCrmG5sx6

(iv) 0.6, 1.7, 2.8, 3.9 …

Answer:

(i) 3, 1, −1, −3 …

Here, first term, a = 3

Common difference, d = Second term − First term

= 1 − 3 = −2

(ii) −5, −1, 3, 7 …

Here, first term, a = −5

Common difference, d = Second term − First term

= (−1) − (−5) = − 1 + 5 = 4
zcsi Md56EMX6DJRhqBHPe4laeQZoeyUtCJbtJ3Ib99Rbf1H9FCY6MD89eZc0YNif1wW iwEIgzeFfHYe6OHCrmG5sx6

(iii)

Here, first term,

Common difference, d = Second term − First term

V

(iv) 0.6, 1.7, 2.8, 3.9 …

Here, first term, a = 0.6

Common difference, d = Second term − First term

= 1.7 − 0.6

= 1.1

Question 4:

Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.

(i) 2, 4, 8, 16 …

(ii)

(iii) −1.2, − 3.2, − 5.2, − 7.2 …

(iv) −10, − 6, − 2, 2 …

  1. TaAUFlW1Eraz8k4gTc1T7LTuUTtLHr6tBcEUcv4bqvB FLYXqRIYy5lo8xWgJI31S9ML1mOAl0 D6GXlCMayA j7EVuFWT (vi) 0.2, 0.22, 0.222, 0.2222 ….

(vii) 0, − 4, − 8, − 12 …

(viii)      (ix)     1, 3, 9, 27 …

(x) a, 2a, 3a, 4a …

(xi) a, a2, a3, a4 … (xii) efzuvbrgs4BKT6G4KJWiFIpd47yTj T24JNZgUoM4XTv9WO2B9d

(xiii) pSlv cbVA5xTBqo9XSlT5Scq7pJOBq3NkkjT95UuVEa7M5bl9st7UjKp8IL9w64PTcwxxgHO7

(xiv) 12, 32, 52, 72

(xv) 12, 52, 72, 73 …

Answer:

(i) 2, 4, 8, 16 …

It can be observed that a2 − a1 = 4 − 2 = 2

a3 − a2 = 8 − 4 = 4

a4 − a3 = 16 − 8 = 8

i.e., ak+1− ak is not the same every time. Therefore, the given numbers are not forming an A.P.

(ii) 

It can be observed that

t17fAgAK11 9mZPnErbxSj0kf0fkbCnZ3UHizE7x8FrbnyFP2y97jHtJNMr1 2MipZ9RVXVmyZHP7twkC6seJJ x8CEoVWhrg ceTo068SVIM68cpBVItMUvCw0Zs3kCv5 F0Lc

i.e., ak+1− ak is same every time.

Therefore, and the given numbers are in A.P.

Three more terms are

(iii) −1.2, −3.2, −5.2, −7.2 …

It can be observed that

a2 − a1 = (−3.2) − (−1.2) = −2

a3 − a2 = (−5.2) − (−3.2) = −2

a4 − a3 = (−7.2) − (−5.2) = −2

i.e., ak+1− akis same every time. Therefore, d = −2 The given numbers are in A.P.

Three more terms are a5 = − 7.2 − 2 = −9.2

a6 = − 9.2 − 2 = −11.2

a7 = − 11.2 − 2 = −13.2

(iv) −10, −6, −2, 2 …

It can be observed that a2 − a1 = (−6) − (−10) = 4

a3 − a2 = (−2) − (−6) = 4

a4 − a3 = (2) − (−2) = 4

i.e., ak+1 − akis same every time. Therefore, d = 4 The given numbers are in A.P.

Three more terms are a5 = 2 + 4 = 6

a6 = 6 + 4 = 10

a7 = 10 + 4 = 14

(v) OyV34AcZaWNvCP1iMcFFKIwJSVn40hoa c96djEyicii89d8vRtJO h8cZTuU7kfj2XeGNaAkZaFuMmf

It can be observed that

UtqeifteFNcaiXnewc 5B NTzG5xg85zB0v 3WNJmVcPJNoc1jJ3h aJBCcdImROr5ZWVIAl1imgbo21Fsg1xDcK7TJ Czq9LJ7FeIt frvv6QgjbdgOIZwGsoIyGeIQMOVBHqE

i.e., ak+1 − ak is same every time. Therefore,

The given numbers are in A.P. Three more terms are

(vi) 0.2, 0.22, 0.222, 0.2222 ….

It can be observed that

a2 − a1 = 0.22 − 0.2 = 0.02

M3yyTDCRmYisNAK6w1CnpniPNzQrXhqSuLGEPxNR7oe01UnxfF 16v 8RmqItGBD6rSwFEjfATTioB1 1fDmciM2ZIfYQj1kr8BBym

a3 − a2 = 0.222 − 0.22 = 0.002

a4 − a3 = 0.2222 − 0.222 = 0.0002

i.e., ak+1 − ak is not the same every time. Therefore, the given numbers are not in A.P. 

(vii) 0, −4, −8, −12 …

It can be observed that a2 − a1 = (−4) − 0 = −4

a3 − a2 = (−8) − (−4) = −4

a4 − a3 = (−12) − (−8) = −4

i.e., ak+1 − ak is same every time. Therefore, d = −4 The given numbers are in A.P.

Three more terms are a5 = − 12 − 4 = −16

a6 = − 16 − 4 = −20

a7 = − 20 − 4 = −24

(viii)

It can be observed that

xfVwShfX6l g68 yxczz8JMiVYKCl1R8txOz5XYt8R7OvPIsUPUSL9MHMuPWsz4nVMx2BklPw3ZQwX96znj7tP6nu O4YYOwjA3VuPdMmHyVy3e3YLFAHRbMWwZXyNvi749YDEk

i.e., ak+1 − ak is same every time. Therefore, d = 0 The given numbers are in A.P.

Three more terms are

G7mXF0eCab9dOY5gbr7FrQaKi

(ix) 1, 3, 9, 27 …

It can be observed that a2 − a1 = 3 − 1 = 2

a3 − a2 = 9 − 3 = 6

a4 − a3 = 27 − 9 = 18

i.e., ak+1 − ak is not the same every time. Therefore, the given numbers are not in A.P.

(x) a, 2a, 3a, 4a …

It can be observed that

a2 − a1 = 2a − a = a a3 − a2 = 3a − 2a = a a4 − a3 = 4a − 3a = a

i.e., ak+1 − ak is same every time. Therefore, d = a The given numbers are in A.P.

Three more terms are a5 = 4a + a = 5a

a6 = 5a + a = 6a

a7 = 6a + a = 7a

(xi) a, a2, a3, a4

It can be observed that a2 − a1 = a2 − a = a (a − 1)

a3 − a2 = a3 − a2 = a2 (a − 1)

a4 − a3 = a4 − a3 = a3 (a − 1)

i.e., ak+1 − ak is not the same every time. Therefore, the given numbers are not in A.P.

(xii)

It can be observed that

IxZIPl1YYNS1cFU0EhIwbC9q2YnzRVvDO9dAKw8ESequlSzeeTv8M4HIokxoLPJ votIPcIR j9PR7d PQkLdh52PF7Tl61tW0txV V6U9UAz7OndyVFlj

i.e., ak+1 − ak is same every time. Therefore, the given numbers are in A.P.

And,

Three more terms are

(xiii)pSlv cbVA5xTBqo9XSlT5Scq7pJOBq3NkkjT95UuVEa7M5bl9st7UjKp8IL9w64PTcwxxgHO7

It can be observed that

suq48FfjWL4lwzyTZgFYm9oxQDTzsfylJI11K4TxnRNs5k86 OdjSJFHZOUVR678T76EUcrEt6RAhZLwOCETbxAv9XawuYQAFGeTLsWGnKBQrVo dizpuzYC9bu9ddB NlwLhKY

i.e., ak+1 − akis not the same every time. Therefore, the given numbers are not in A.P.

(xiv) 12, 32, 52, 72

Or, 1, 9, 25, 49 …..

It can be observed that a2 − a1 = 9 − 1 = 8

a3 − a2 = 25 − 9 = 16

a4 − a3 = 49 − 25 = 24

i.e., ak+1 − ak is not the same every time. Therefore, the given numbers are not in A.P.

(xv) 12, 52, 72, 73 …

Or 1, 25, 49, 73 …

It can be observed that a2 − a1 = 25 − 1 = 24

a3 − a2 = 49 − 25 = 24

a4 − a3 = 73 − 49 = 24

i.e., ak+1 − ak is same every time. Therefore, the given numbers are in A.P. And, d = 24

Three more terms are a5 = 73+ 24 = 97

a6 = 97 + 24 = 121

a7 = 121 + 24 = 145

Exercise 5.2

Question 1:

Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the A.P.

adnan
I738……
II− 18…..100
III…..− 318− 5
IV− 18.92.5…..3.6
V3.50105…..

Answer:

I. a = 7, d = 3, n = 8, an = ?

We know that,

For an A.P.

= 7 + (8 − 1) 3

= 7 + (7) 3

= 7 + 21 = 28

Hence, an = 28

  1. Given that

a = −18, n = 10, an = 0, d = ?

We know that,

x1G uq0Ze3VYe28oI4WvgwPusUqMS59yDxFdFBifMsKpZtPp qHRdF G1jM4b cL5cO8ugCgFriH0Rp5ZZigQlq1Nun JAQnR nGpHgM8rMUc6dplag1gV3lBKnR9yGswBb 7Vw
y8o2R7RgvoMc5gOFoJxXVds6rteomI5JdIpqSVqiukAHa5xKznSHFKMnvstGbslaxXJCzSyFtY50EmR

18 = 9d
GVbSSbQyIRUci8lYyCJfwAkkDGkqIxK2SWcLkyE0LJSmapAUex8r4CuVQ9KeheXnTyryc0fe9p8D GS1E9D60mcEt0ul 42E m4b34dEnBI7F8 lm5GXmaWADC 7HZlGmgDdHYA

Hence, common difference, d = 2

  1. Given that

d = −3, n = 18, an = −5

We know that,

MW40Q2u3KQjgOPp0rYp7rwFZyEyXDkX jhgfZuqYqXnupCSPx07Ab2Yi1rGg Oizmzx6ye5DEVpWv1Hnn rdfYzBbUjKU8zNyHORyuCmS 4yw8S7CqT7BMRl31OGIwAhI1g1K Q
UYnzM7YYGK 8AoHCAWjeS1DPRhavTBqjpEuIy vbA2aSvQr yB 8ImNMife KfXItOIN0 8BpaAH38jVkP48ErjGP17XFktmoV e9emTcpnOPdR8f0kf87LX5kiuK1UoiY7J7XA
a BtvGur08i2TaPA696m2RHEqovHkpWrqmpUxOmBchagP5V44wpJmcOrxn A49 Aa4nGw2S2gRUVyLQj1tvUI8P5WDAZX9JyETAbx 8VT2aOLO9y4TQD3YnTzUUhp4KC4oA7K U

Hence, a = 46

IV. a = −18.9, d = 2.5, an = 3.6, n = ?

We know that,

gODQs0278Wn8vG2sR79MA0bzcAYyxMBisCokRudpTR7MQHxpTHTOR9TLuGLBGZQxQiY9bVjhVEvcVTVuGbQ7NY 5ZqdfaZe4ldZWMoD VHt6LJcArFrbtNuS9ZFPAheiiuYz7vY
kNVJhDLNJt90hge40JQnLuF65Y14o7CgC9ITqwDQKT6QIHLRdPBjaQ3w4zoKb80lvb5SKY 9GJc94Z FGdv3U5NyATfAeFwjtOFXYBKbrbP86PGZ
u 8lRyx rcvNO8mXnxOlzZWOOLUYGDZ8JJNaw6CBJjoBbI03TIKQ PLPXUyzgyDhRns50h4JqgCPWseqEL7RcDnlJgDbJrWTnfwmXL69u7hjqtSFdlg8RSkNp0thUpgIK atFk
qUF6ksbQK5C41NI
idq2xvlJoJDkCJCbdLbmI6ddw4mEfWHJA5AedzrWy95IT0X1u5bGiedpKn dTbEHfXq1nfhPKXi37xSEoCaO MxOmOprhwxZyvEgLHKjF9iHZakvNnjIpI7xFI13rAO9Qkw G84

Hence, n = 10

V. a = 3.5, d = 0, n = 105, an = ?

We know that, an = a + (n − 1) d

qdys3mrIm8azpzUgwi54Ms KzceVLFE4Oiz5CrBc 2 cB6H2MbSI Ssw1twuBOVuBC9Hq0DGlBaZzKGkd0VZmMQKGEtmyHT2ixV riXU00 Rg CRYFHvmbrqbbUe9 PKIaGfTS4
sMXfqhoKkQ7n6H3i0g9PtO8 LWPCjrF2r7yOmYVM
WcFKw rGGrQrnMwhvCleViIeYtmEItunwEFNaWOpR2W6PVNkdmujN9J1X SBayXxWVBAIqJkk0I3KIEepUhTSUW0IkXExc51e7gDt5cCHxaevJQrL a0etOGyFnzS2Ylrd0 FvE

Hence, an = 3.5

Question 2:

Choose the correct choice in the following and justify

I. 30th term of the A.P: 10, 7, 4,…, is 

A. 97 B. 77 C. − 77 D. − 87

II 11th term of the A.P. is

A. 28 B. 22 C. − 38 D.

Answer:

I. Given that

A.P. 10, 7, 4, …

First term, a = 10

Common difference, d = a2 − a1 = 7 − 10

= −3

We know that,

a30 = 10 + (29) (−3)

a30 = 10 − 87 = −77

Hence, the correct answer is C.

II. Given that, A.P.

First term a = −3

Common difference, d = a2 − a1

We know that,

4Oy2DQ UcdXNMJRkSyqc6fknhN5k7xhWoMfs5iDvzfPFEXzFvebemV1yB4qlXMes YjZYgTTnb6rLHdrE0gJgUl 2 UbvUW25nhSsFc756yf5FitWTnkTtFvq1TdBh9BUXc4s s

Hence, the answer is B.

Question 3:

In the following APs find the missing term in the boxes

I.01AAP8YUYp1iADTV3rMkA5qXItgHAZCXGTSI1SAWqxkLYCk1v

II.

1Ivj4XtJ T2LVGeTuUW3iA3T2vUG

III.

ONCnXlxHi2mnqMrC5YDQWpQ04NjExyzoYjUARIKrXciYzd7BcAwSRW4bIBB0i nzeoGa7XrSdNGD6 paI3fg7o suaHWQ6oOI46OJs HPsfJCIm4vx9J qju9qs

IV.

npDP3WAbsZhpPXxsDpPaQfiJaDT5jp cdqN3fVGWCMO7N0Ri8dHd96d zoRWLiJtira8j6

V.

Answer:
01AAP8YUYp1iADTV3rMkA5qXItgHAZCXGTSI1SAWqxkLYCk1v

I.

For this A.P., a = 2

a3 = 26

We know that,

Vs0qlwVmC10k2jKlq9Hm6f1wdEeETOHePA7S4h0f7rcYuhmfdM1p0brjY8ClLwmeIBrFGZohunhbdMvKGqvVbn0buZmCOiP4HIW9tDEL46D IISoRysR6R5ZAlTpv3ZFpPFdmgY
2P9GB5W2ncctp2wlIL TTkcLPQrcVwFdnUNfMrw04ydLO2 kl6QtULZcyAJs7p45QgzyxNDwtvzr3vLK5fpWM2Ovtzf2KrGauTc7431d HZpnAjjqiQfTJFkXdWJkSnKMi6UqFE

d = 12

a2 = 2 + (2 − 1) 12

= 14

Therefore, 14 is the missing term.

II.

For this A.P., a2 = 13 and a4 = 3

We know that, an = a + (n − 1) d

a2 = a + (2 − 1) d 13 = a + d (I)

a4 = a + (4 − 1) d 3 = a + 3d (II)

On subtracting (I) from (II), we obtain

−10 = 2d

d = −5

From equation (I), we obtain

13 = a + (−5)

a = 18

a3 = 18 + (3 − 1) (−5)

= 18 + 2 (−5) = 18 − 10 = 8

Therefore, the missing terms are 18 and 8 respectively.

1Ivj4XtJ T2LVGeTuUW3iA3T2vUG

III.

For this A.P.,
SyrvmLWHlY4l3D8AH3XmRdiIorF0K6EfyXDiWHNnJe7s4PGOT4h3CTbXGDUo7xi1NzBhKV9Xvb 66Gftwb RWW3CAEVO2Ix5EzYxmkrlwqW5tIcjlGNS uMA1sLobE7XB6B2FD0

We know that,
O NHi1mVUv6RHuxlZxE7g94fjFPnFMgle

Therefore, the missing terms are and 8 respectively.BUuhXY 28tTiEami4Otm3apDY12QBPkbaitIb1MRGTQAJPhoX72GpZhyhWGf8Yqjh2Hmxs1q94LjyPyrCA7dal2YNhrSzly5J3D8olS35dHKzGmPa9XWjfSwdv2FF OQeP73rpg
ONCnXlxHi2mnqMrC5YDQWpQ04NjExyzoYjUARIKrXciYzd7BcAwSRW4bIBB0i nzeoGa7XrSdNGD6 paI3fg7o suaHWQ6oOI46OJs HPsfJCIm4vx9J qju9qs

IV.

For this A.P., a = −4 and

a6 = 6

We know that, an = a + (n − 1) d

8sh xC34WLYA4PEpH

10 = 5d

d = 2

a2 = a + d = − 4 + 2 = −2

crYjJazj1VBh9XgHosQw2yXJs05Kb4eDcTvHNUL5nkplcse7 F6qLCN6TxVjJTw97UcQ 2AfYqgYfIW3xrobm0y7W4xYyjpNgebdCY6O rSkAey6qqy3Yrb
GXWoKFlyxA0nfcTZyzXul1BbxN5RAq9E0OQF48RcoU9TQcQYm4G4fBLwLm20JYUWTTRZbbq1qzuIQSseOglX2ODif1BgpZsY5vIYWKSGrpWBhdm Te4c31QQDMb K5eWbtDn GE
wBd dSSI yPYYH6SlTmab11TpWpXm68cnqQH4KnyLLtd8nYKO alZBjFxRs4P wVfE5X7rBtL6DfYzKcQNjH9YRMmUAgwbIwPWY diYoJqddWokA8OonHImJbwBOg8V5fuIyOhM

Therefore, the missing terms are −2, 0, 2, and 4 respectively.

npDP3WAbsZhpPXxsDpPaQfiJaDT5jp cdqN3fVGWCMO7N0Ri8dHd96d zoRWLiJtira8j6

V.

For this A.P., a2 = 38

a6 = −22

We know that

an = a + (n − 1) d

85XaJpj3mOCWXcr7UnHepdkRqM1o

a6 = a + (6 − 1) d

−22 = a + 5d (2)

On subtracting equation (1) from (2), we obtain

− 22 − 38 = 4d

−60 = 4d

d = −15

Pow6BkKE 8jux2dnM XvA1fAWbZp9SQsIhMTSTbVBsvRUeZBTlQpUlEGQW sDf24VAiMAkLNyBZ kLkaJ02vy43JT4z1WQm0YLoVt8TK4rjeWlO Dt
IQcf26udWQ2s33jx2ArxtN o7RhdHsk5z4qn5AYtTSFj8maZnx5MB5shBNNOstbbDLlimtszhssoGPG7RNa1DV5hp2JtR1MIB9agBHCa5QwSEWipwUnU3LB8cHxf3IEwJDnJL k
yXeqP12k8xGBn1PyUj8QXOFF9UXHYbia0oIdSJ VstyEtP6x sQWMOY3MziT KBHYsfcpGPlmIlMs6NRXAqcL4l1xCFjLug5SRh 4p0iqR3Ol0zjpnbpbinBRPgwEfM6pyqJy9c

Therefore, the missing terms are 53, 23, 8, and −7 respectively.

Question 4:

Which term of the A.P. 3, 8, 13, 18, … is 78?

Answer:

3, 8, 13, 18, …

For this A.P.,

a = 3

d = a2 − a1 = 8 − 3 = 5

Let nth term of this A.P. be 78.

78 = 3 + (n − 1) 5

75 = (n − 1) 5

(n − 1) = 15

n = 16

Hence, 16th term of this A.P. is 78.

Question 5:

Find the number of terms in each of the following A.P.

I. 7, 13, 19, …, 205
mgHegQKTx jRfE0TVvy8uqjKV XDYUhJM75CsnxZVjPgKsJKJJKQlWJDwXyP3XIafDcULTuRiG0IXBwk14FzzL4pzzc4DBAM0baMWHL19o3Y3nZ 7r 2Gqk5ekGmqj3doSm8iA

II.

Answer:

I. 7, 13, 19, …, 205

For this A.P., a = 7

d = a2 − a1 = 13 − 7 = 6

Let there are n terms in this A.P.

an = 205

We know that

an = a + (n − 1) d

Therefore,

198 = (n − 1) 6

33 = (n − 1)

n = 34

Therefore, this given series has 34 terms in it.

S1xskQCdSi3aw9zz7E0GslYzgKHU3bbl8 sdVyzoaIY DwRaNdXHO4zovfQWNNdBim49epDMRyf HJt0veYOKcLYPdYICwF Kw1eJZne3A9d3ZoPr TRK0iv RA DsFXILIHaQ

II.

For this A.P.,
Z4mBSKfs62FnzBdzLdHm r915b3QqootaVNIEGKtmL919H1uV98narI52H5u1pj95 RpevSHSWHVJJU8ae91v6H1Gcrxg84cR9eBH7 two tcfdN4uez7oIiDde2mfrNfTDvqTA

Let there are n terms in this A.P. Therefore, an = −47 and we know that,

dU7GLPdxdWPUbDa2ZC od17b8zRiY4i6ii7 sAtoXyIWB40eR4

Therefore, this given A.P. has 27 terms in it.

Question 6:

Check whether − 150 is a term of the A.P. 11, 8, 5, 2, …

Answer:

For this A.P., a = 11

d = a2 − a1 = 8 − 11 = −3

Let −150 be the nth term of this A.P. We know that,ngmdj85oOh WKNO1hVeNl1ebZRm9V5rUiRd VaF9f813iz Ge0GmYBKXC WW2l7Ri hX70A QaoS e1tpVSJPyA4pPCswdCzqjAvyLB68VIZjiHBwgOhZ6LGgMmgp9MGTT9etaM

Clearly, n is not an integer.

Therefore, −150 is not a term of this A.P.

Question 7:

Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73

Answer:

Given that, a11 = 38

a16 = 73

We know that,

AmkHPiJhplhUVATQ0luTTs2EaG TzM5G3lkI1vzfQ4gwQdw23x5en8zlQB39T9YUQmjQAxlWD2B4FTJV3ACdQptFL5XB99Uts24yi86sAjqCRT66YiJs0 mfHFmM0a0w1 ahIFU
XAeyMAzsalmQkc73cvxPN ZOwHFn2YHsPDQJ0idCSl2VR h2eLAeawqUNUw6ZDdiJKriD0FWmFFST000ZdEaFe4XMqHm64uUKZbKipdCYD 5NFXflctre8q5Hl11MMCFuhX7QVw

Similarly,

GWEqI1imjvoK9a7 k9y5fLUO rRWyYWhlcIGbfrkC3j0R9B kwAoDYemqz4CSkaGiPi9O xIOIlUDFVJac0kiRV1Ckyb4ZXHy t0HpO2YTF85FIlc7PChLwZn V99wPE4zMUNls

On subtracting (1) from (2), we obtain 35 = 5d

d = 7

From equation (1),

image

38 − 70 = a

a = −32

a31 = a + (31 − 1) d

= − 32 + 30 (7)

= − 32 + 210

= 178

Hence, 31st term is 178.

Question 8:

An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term

Answer:

Given that, a3 = 12

a50 = 106

We know that,

JedeU3H6QLf61UTj cP73ujazeY9db 9Wk8B1cBpVZEiur3zyI VJeu2o qkAUe7rng3J5is3SWiLmBHlW3HkLgC SsnvQ1PfJkRbIoiYV4W71OMaHtDhYbUqIgJzhfZmNSUSpo

12 = a + 2d (I)

Similarly, (II)

On subtracting (I) from (II), we obtain 94 = 47d

d = 2

From equation (I), we obtain 12 = a + 2 (2)

a = 12 − 4 = 8

a29 = a + (29 − 1) d a29 = 8 + (28)2

a29 = 8 + 56 = 64

Therefore, 29th term is 64.

Question 9:

If the 3rd and the 9th terms of an A.P. are 4 and − 8 respectively. Which term of this A.P. is zero.

Answer:

Given that, a3 = 4

a9 = −8

We know that, an = a + (n − 1) d

OMG3RqPue5McfqRBNn1MFVO8n0t5IZbqg1pDEbyeNHKaSYhajNLQtMRRlq9H0oAsKiqpP5r1BYJAysl15b oFiQTR7vRdMU1FSUepE8Mga WDEj4VGlNDNbgRnwBbXL

(I)

x9 kG9On9Q Tbn

-8 = a + 8d (II)

On subtracting equation (I) from (II), we obtain

−12 = 6d

d = −2

From equation (I), we obtain 4 = a + 2 (−2)

4 = a − 4

a = 8

Let nth term of this A.P. be zero. an = a + (n − 1) d

IO6ZWwYMWNmL7fmxWA W8n2rxv7PYeIat9rSUnPJyakdEFrEqpp6ust58L8EfvTHiwbTppCYCaimqV22PuYvr8xg HD0ZbqmG0DyERkDSSYveqA3aAUQ0nEe4gh0KytBN8jgpAA

2n = 10

n = 5

Hence, 5th term of this A.P. is 0.

Question 10:

If 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.

Answer:

We know that,

For an A.P.,

v 4VGq
6uiyIORfzJ Hilyi Cqjluuh3lIP3VBtjIyTutwQXIwWvNO7U caoMHWu3ZxF eHMK6oXUu0upG7slevYqn jd69b5yrdyEqpCWyZikI76mqJLS1DLu6sNFp2h4RBAkihccvdOE

Similarly, a10 = a + 9d It is given that

a17 − a10 = 7

(a + 16d) − (a + 9d) = 7

 7d = 7

d = 1

Therefore, the common difference is 1.

Question 11:

Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54th term?

Answer:

Given A.P. is 3, 15, 27, 39, …

a = 3

d = a2 − a1 = 15 − 3 = 12 a54 = a + (54 − 1) d

= 3 + (53) (12)

= 3 + 636 = 639

132 + 639 = 771

We have to find the term of this A.P. which is 771.

Let nth term be 771.

an = a + (n − 1) d 771 = 3 + (n − 1) 12

768 = (n − 1) 12

(n − 1) = 64

n = 65

Therefore, 65th term was 132 more than 54th term. Alternatively,

Let nth term be 132 more than 54th term.

TgzqNQYotIMxpNPyqgc EuI8PW7 Rq6jZH9NXey4QwWpl50VXn U1W Qhg7zVxnAN PTuuijKo8uOQw4l4Jj0W oBYIOgdvw28uQKT3C6 Z1Dpolm6 flYLNlg8sVWeDqgnRoak

Question 12:

Two APs have the same common difference. The difference between their 100th term is 100, what is the difference between their 1000th terms?

Answer:

Let the first term of these A.P.s be a1 and a2 respectively and the common difference of these A.P.s be d.

For first A.P.,

dZUWpvRqXuZWgbiGzWhIeSYlp4QQHTYx543Y Zuuckv8XuNOdlgTyyBiY1whYBwYlO35CRmeMDDKZkQ9aIBoee3COza3Fqzm9lfmpq9eF2 IRO3zJW3fFd lPHJZha4SLDpHuUs
rcjN0
BNJ6ozYQfZ8US6H06XQt91O7RJlDNgl7kdckSH5qQfgeBzGKiipj0dqVY5Ir N2w6CzeJCHIP bIIZdhxiyOsa w1ssXeV1QLzrTL6 Cm0YFG204jzoF6QPCAkYUIOiy7KpQmIM
3mb5Q99pijLvZ62YO AGDGDvS EwBB1x2N6Jv qDZHklitARNq93MK7fzLg2foFFGciQldtrfUu5TCkdKbbSNLO 58J yhz4AQCb6UapvmjfVnuajm3Cm R JvVvgWj9iCJYA8k

For second A.P.,

OvWprsBZ6FwHlIROsj7CzaNi7Hkb rr8KKiFxXrlMXgLkBUbMAIMdr3YSnHLERQxnl84Y7dYTt9LrASe3kND7 2zlih3tQ6ZZ1J0tdvm9ojhLLMghZe4n7GSxNVx0 FsdP7apn4
7g0Aahvicc7hDhOA1fBaptGqgHdOM6odxwJFkA27SAMhWyLd8UhRJFNAtsWA 55RGF71XJY5 J QL5j035OgALtTplmrkQBR3fS4RflIVc02sNmHbRLSUY1Y3L KkHOOiBc3gfE
RMEC7X4lwR68DjmHr9tWNp4H3TVcjXjA YkRqUjdtZzAJ6dUwI9eWyifQ17ml7jol9cupheaSuAdLMj1CHWJjFarIOWB25fDdLHZU7 Gbf677T4HhCJHusjinLlug67Kzt qlAU
B9rIrxxZAb013WzNZkadnmUJfpJBveugbR8CKCUQGsh7SBk3MKmErjTzEJSWYks40PM7ipfl16UQjux

Given that, difference between 100th term of these A.P.s = 100

Therefore, (a1 + 99d) − (a2 + 99d) = 100 a1 − a2 = 100 (1)

Difference between 1000th terms of these A.P.s

(a1 + 999d) − (a2 + 999d) = a1 − a2

From equation (1),

This difference, a1 − a2 = 100

Hence, the difference between 1000th terms of these A.P. will be 100.

Question 13:

How many three digit numbers are divisible by 7

Answer:

First three-digit number that is divisible by 7 = 105 Next number = 105 + 7 = 112

Therefore, 105, 112, 119, …

All are three digit numbers which are divisible by 7 and thus, all these are terms of an

A.P. having first term as 105 and common difference as 7.

The maximum possible three-digit number is 999. When we divide it by 7, the remainder will be 5. Clearly, 999 − 5 = 994 is the maximum possible three-digit number that is divisible by 7.

The series is as follows. 105, 112, 119, …, 994

Let 994 be the nth term of this A.P. a = 105

d = 7

GEEY6Ig44yMKzKrAu6h30Y juKdtlZcIvoJSgu42zqEn9YXOnSHszPjcxGSL0lvHecuVZtAs8DBugasqrg7sfeBxU195vL7ajLKiDGYFs E3p1X54OZ66n3eoGfN5Z45an36eqk
XqbFI9XMol3GrU GSRKr9N6 0ddVuRIxnsoPV6CbO qXV9Cml1B3iOZBVEzfyu67ocsDFe0gRRUqQbDVUYqOsCcChCTJTj5mFbi
UixIsxTE2urerZQyHusczorvKOrsgrXq1pT5 7JpZJwHPQ0ukGWdERMJhIE

Therefore, 128 three-digit numbers are divisible by 7.

Question 14:

How many multiples of 4 lie between 10 and 250?

Answer:

First multiple of 4 that is greater than 10 is 12. Next will be 16. Therefore, 12, 16, 20, 24, …

All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.

When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4.

The series is as follows. 12, 16, 20, 24, …, 248

Let 248 be the nth term of this A.P.
Fg6l7Vgtu Fb XlaYnpOI BBNS

Therefore, there are 60 multiples of 4 between 10 and 250.

Question 15:

For what value of n, are the nth terms of two APs 63, 65, 67, and 3, 10, 17, … equal

Answer:

63, 65, 67, …

a = 63

d = a2 − a1 = 65 − 63 = 2

nth term of this A.P. = an = a + (n − 1) d 

4v0RXkYdHxN1JGnuw6IVzhNJOoTu4sXSn30A84dNWqsdgZx79A4FX7bXrj6UWZSGoCbD hgT70fE46jnP7oWDEottSrMvGGtoeMtp6QZ5D9CscYBkktgGcWkg9im2f5ra r9KCM

an = 61 + 2n (1)

3, 10, 17, …

a = 3

d = a2 − a1 = 10 − 3 = 7

nth term of this A.P. = 3 + (n − 1) 7 an = 3 + 7n − 7

an = 7n − 4 (2)

It is given that, nth term of these A.P.s are equal to each other. Equating both these equations, we obtain

61 + 2n = 7n − 4

61 + 4 = 5n

5n = 65

n = 13

Therefore, 13th terms of both these A.P.s are equal to each other.

*Question 16:

Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.

Answer:

=a3 = 16

a + (3 − 1) d = 16

a + 2d = 16 (1) a7 − a5 = 12

Ue 1CuMjYA0lIjtKc1BoIS eyF s5TDDPlW4gQ8kkY9wmZZRyvAhmqSS5DZjG4of
hCuTSQhR mWmHbE9BonKzAEu3OJ48s pIV DGRnlvwT4 0tWy F XPvw25W2jlE02H8ECDv7m28J1Fv5uyQBArUVMSVyJYmlXYNS91W YA8JZOGPLtRZBtFarFMyHLlYX8eBFIU

d = 6

From equation (1), we obtain a + 2 (6) = 16

a + 12 = 16

a = 4

Therefore, A.P. will be 4, 10, 16, 22, …

*Question 17:

Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253

Answer:

Given A.P.  is 3, 8, 13, …, 253

Common difference for this A.P. is 5.

Therefore, this A.P. can be written in reverse order as

253, 248, 243, …, 13, 8, 3

For this A.P., a = 253

d = 248 − 253 = −5

n = 20

Therefore, 20th term from the last term is 158.

*Question 18:

The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is

44. Find the first three terms of the A.P.

Answer:

We know that,

nyO Ru0w3tioUH6 4fSrU9V5HOyQP5RLYbeQEF9Qh5O2wi9coAycw60PIecT4iAYn VD7 WfWowkjmpqKGAtMCSpCzgg2WFR1CIvU3GCLyRmjlWKNVj25qIYcjg73IAL5rF5Lz4
Qb V QgQ9bEwBBFOGQ Z u3ZUvfxk QOt RC0kye3iDu 2pqUjibk93TCuug9gf00WpPwYl5AGonmnD pJ9df3EdZUh Mlm3fnJvLsK47NSWYiIlPwef0ewbFzcMuIjFOerZCo

Similarly, a8 = a + 7d

a6 = a + 5d 

a10 = a + 9d

Given that, a4 + a8 = 24 

a + 3d + a + 7d = 24

2a + 10d = 24 a + 5d = 12 (1)

a6 + a10 = 44

lGpTIeF1jkAbxy NxVXs4TlL0OuG8se8UdeqFewZxrCSYGfpKR6Hz8VySYJ2tp0er3aJMeGFolEQ0Rrwczb7UlCebUKx0fOW7UWZdMbORN26P7uiJPaKZi0hP4EAAv fJL1Vhwc

On subtracting equation (1) from (2), we obtain 2d = 22 − 12

2d = 10

d = 5

From equation (1), we obtain a + 5d = 12

a + 5 (5) = 12

a + 25 = 12

a = −13

a2 = a + d = − 13 + 5 = −8

a3 = a2 + d = − 8 + 5 = −3

Therefore, the first three terms of this A.P. are −13, −8, and −3.

*Question 19:

Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?

Answer:

It can be observed that the incomes that Subba Rao obtained in various years are in A.P. as every year, his salary is increased by Rs 200.

Therefore, the salaries of each year after 1995 are 5000, 5200, 5400, …

Here, a = 5000 d = 200

Let after nth year, his salary be Rs 7000.

Therefore,

200(n − 1) = 2000

(n − 1) = 10

n = 11

Therefore, in 11th year, his salary will be Rs 7000.

*Question 20:

Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the nth week, her week, her weekly savings become Rs 20.75, find n.

Answer:

Given that, a = 5

d = 1.75

an = 20.75

n = ?

RHDu3j1ObbN d0 MrP4H M7GDPlEecZVP2VGVapDjMISTEaJdxIenx69oUl0gh3RsovPuswYVMTsu7qEmce9Wly2g0FcsWmcmYC
668B QaJuIk2xaOqkmBC6JxMf41OifeakFbLmZFhz5bNANFyEg swd0deYbiizAKin3OvRA43yleYVu3e9Kl61IFkYPgOwtRxQR2pc2fz7jc8ih6F5hp5InyKPSYID23JJ

n − 1 = 9

n = 10

Hence, n is 10.

Exercise 5.3

Question 1:

Find the sum of the following APs. 

(i) 2, 7, 12, …., to 10 terms.

(ii) −37, − 33, − 29, …, to 12 terms

(iii)     0.6, 1.7, 2.8, …., to 100 terms

kM1O3h1PzLGfp8vNpmGumXp0PVw7 oedFiu6z76V

(iv) ,… , to 11 terms

Answer:

(i) 2, 7, 12, …, to 10 terms For this A.P.,

a = 2

d = a2 − a1 = 7 − 2 = 5 n = 10

We know that,
6XKbAQfVKiMdCdh5G0eUE6ZT elnLkWt9ygEMMNPmjsAkQdP8XANEjE04poI7zsTSGPgVWNdn95lwQsZWBr1rZNjz0o5AcK9hAtIG14rAmfu2o4FG80vVEa8gtBVVKSYqDDsHiU

(ii) −37, −33, −29 ,…, to 12 terms For this A.P.,

a = −37

d = a2 − a1 = (−33) − (−37)

= − 33 + 37 = 4

n = 12

We know that,

9dwvU5HckyaGdb9EeOgu8uY7b0EAm35gO VPqboIli5AUHGqSr5ajuuAudDMNCyAb3pIDZcaBWraJXmKFwJGudNMmmimJoA6DF3p1aoFghEnOU4k6NSVE2yv7UjC1 fiVGbB98

(iii) 0.6, 1.7, 2.8 ,…, to 100 terms For this A.P.,

a = 0.6

d = a2 − a1 = 1.7 − 0.6 = 1.1 n = 100

We know that,9gicRkT QunGBb3nSUupI5iUqDwSSFvBPrkwN60sKDyqcYOpZQkl n2T4zes3SxGilAcl76j9fEHIcuhB NZT2WxSxFC4rIEbYTxqge1hxKgm1hhznIJov PJNlltd pgr6IejE

Ngefjv lsbFodO WZ2jhMBJJS2vixRArwvkNStmvJywauEYjCSAe8NoVZYkcc8PIIHMdTI6NwbnPl9S m0EqWfq20r UK4r3uqZppC qAd4EBT5QiKFo AVAvYC31CAmz2yrVwE

(iv) , to 11 terms

For this A.P.,
WZNGF93dJCnw5zZ6pkJA1MtKq0Chg4efRn2lqu0jCONDU3WIY94 IPlEbZYtMihwXPFdGa6ziugg9PD UBKhNpWqYSmlJS8yT1gGM1ZSHYoYAH36DvX5QMMBI1F9JhM0 dqrKQg

n = 11N EuXNOtZyFw7gpxyEWMMe1iAFjXNXGKY5t92OarkvjCpCfzv nferufctHiZdD1ooE5PJDnZenlwLdyMT4xmcnMxQOMT6wSvyyld3EcE QvjG2QLcdewA3BVesI CPtfnE9bHM

We know that,7jqUAqvp h6Zmk GVycA4kqni2 e6RyoWyhfHwCeCxu4x 5P4HyfgVcIjYK4Jwv8DLKamd4sZN54wMzNVpLQm1AicduyLpaNsUX9zll9hFq2tBqlVWj7whq xsaTn fmUpdp2TU

Question 2:

Find the sums given below

(i)

(ii) 34 + 32 + 30 + 10

(iii) − 5 + (− 8) + (− 11) + (− 230)

Answer:

(i)

For this A.P., a = 7

l= 84v xbhdw5xg0gdeU5Vy6UQn5QRWyNLuVz5suKeSUAxGmyJa31MQ6m7OoEuDZml3EcrTErt IZw4j0g9UCHw iq0pgC5LXBUGUKJ6S7lwjs1QsaWy45ODZNiPVyQPgVSNMWsqGrwk

Let 84 be the nth term of this A.P. l = a + (n − 1)dBLN13EqbBpP3p5ZKWKchmgVNgeFoBGktxpBDqSRaA7 vVL2mWzRQYDriTA SCntiq1pFNK3tG7XAt YtiIvK8Fu2NiRZflRlNJTkfLmIypm75o9v70nWSN pf7uiFsnwzFObMCc

22 = n − 1TNz3UIBUe8kx4MHuv fM5JzPqG vUSC61C9ZtUFTxskSAdkTVWQAUZQKpZqy5 oXa6ShdvBy1AVL4ecPCcYfIzoWA7TIoaSUQ8vTHlt U3s5272YqcWVZwC8s2fDhtPjloXqi1s

n = 23

We know that,
8EdVdUwW4f736WCnZKOWXHaEKCF9Qf75VVTshus8E8VpsMOs6S dB8uER1blV9szwrd T eizhU58YgJk if5VMjPpEKAuYJ6aZnczzAZgjQjoqnBDZ2OjwuTIOQ8N V7c9tN1g

(ii) 34 + 32 + 30 + 10

For this A.P., a = 34

fVFo1q5q7gAJqQLBU11PiJh3cR0liakwdDoyskC wGOeU3fmMcfvwXB 210k mCC5jmljaIgezcI 3g Zv3Zd8Ilkh dvUju kA2ctlr0va fnUsj3UEOlv0gveOQ0gqduES1Ro

Let 10 be the nth term of this A.P. l = a + (n − 1) d

TbSe4VB IJ0HOw11KsHOgTRTMs5Kh tpMdVeo5xTz 9QZGso j XgzedUZGpKO J7OnvM71spyI
4nxEjQcheB0bwPnEoEvE2MoBZPHyrPY0t aj5FG4tANmnj gbf0bU

n = 13KcKZdsjrLjBtBi00D1mQPzd6Cz7VFZRGVAtOtrmtARJil0a78n7TTd57ZDAfkJnZqsQ z5deW9139iPylREa2q1uuf9PSey3sUQOdyrjkvn5qUEdFz570ML5MK5Op0h05 OrI8v3bfOoxzAwuXwMrMpq3WS4A8m aX1eChBHMkY08CyhwMFDbeSdbxuXMU8ATVtQXAxRiNy0vBgUUNdKP9y HWpMYZzkde03B9w0D0nz A zmszupfs

(iii) (−5) + (−8) + (−11) + (−230)

For this A.P., a = −5

l = −230

Jn fiibvi5xNFSfqzSJ6poR7 RTR8 MtrqmNifnWjBPClkRiV5 9aaFGW77xE92SUzWLIIFIl23P VxQiKR npvfB 9aTiN8fyqQyLbaCThTWe2RqWWDD2tNu0ze7MQX8dAz PA

Let −230 be the nth term of this A.P.

rngW0V6i3Zi0SJN iTwU2ZYD1qcBELqkl2WogeD kQTkJ2j3l zlXlWFrFfSw33panFjP o7OXzxymMwrgQZjUaFzuwsJSWemEk6qW38 7tH KH41DDMGTu9spNmYJVd8XHQqvI
OuIZL96HqtN9UwrKYb17rdVu32UZ72hMjOzcM0mm1a jiXGeSRywiROZ5bC0CvybxcNtRwCqshOWZoDCY53iDVSSjroE6R aoaRrVg9Lt

(n − 1) = 75

n = 76
Fz570ML5MK5Op0h05 OrI8v3bfOoxzAwuXwMrMpq3WS4A8m aX1eChBHMkY08CyhwMFDbeSdbxuXMU8ATVtQXAxRiNy0vBgUUNdKP9y HWpMYZzkde03B9w0D0nz A zmszupfs

And,

iYlRgI sbBr0YJIxMS58s8l jdDxcZz746kQNfg561h J0VbpF43DOyyRSNuY4dx2MA4INgLsSDRo2mIfwEsJGwg0WXG2K8Z2W BcyFsT4ZaDS4imcCNqDCcOd0cmlvRqNA9GCk

Question 3:

In an AP

  1. Given a = 5, d = 3, an = 50, find n and Sn.
  2. Given a = 7, a13 = 35, find d and S13.
  3. Given a12 = 37, d = 3, find a and S12.
  4. Given a3 = 15, S10 = 125, find d and a10.
  5. Given d = 5, S9 = 75, find a and a9.
  6. Given a = 2, d = 8, Sn = 90, find n and an.
  7. Given a = 8, an = 62, Sn = 210, find n and d.
  8. Given an = 4, d = 2, Sn = − 14, find n and a.
  9. Given a = 3, n = 8, S = 192, find d.
  10. Given l = 28, S = 144 and there are total 9 terms. Find a.

Answer:

  1. Given that, a = 5, d = 3, an = 50

As an = a + (n − 1)d,

∴ 50 = 5 + (n − 1)3

45 = (n − 1)3

15 = n − 1

n = 16

Rq7ccYCKluFBIoVVUw9tKfnqxK4o 0hOSMOmhq94oq7sj7O9xQcVf28EZR2nbfl3MKjcIL7gH7j wtU29FFFEbUsnfp1OuZ5j8maC8VLfPo1NcIp8d6Yjhfojr1QkKrV0qsq00
  1. Given that, a = 7, a13 = 35 As an = a + (n − 1) d,

∴ a13 = a + (13 − 1) d 35 = 7 + 12 d

35 − 7 = 12d

28 = 12d

n79DbDXwhHRjZqZeB6ccaMylnufdvlKCL8kNbjHP1fWFvQHtJcWIftdSFvPIIY4EYYvcsFvOi kDYeGJyeeCazjeIMKwS4WWoXzhz5HGkyev1qGUL8HynVG5cFOq0FTzmp9MCLCf7nZ8Bmf0Y dPtTdG5Y0lH8lHHjGqZXyowdtlzXU1SsOaHDAfSfYwn QqXIo

  1. Given that, a12 = 37, d = 3

As

a12 = a + (12 − 1)3 

37 = a + 33

a = 4
hM2wjupE3jdqWutsp2DUcVhQgo3NBMdE8LNZUhN7d gcXPd1v sToSiAX5uP wmfPakyKz8c2Z5LSApG0iJLf3f8xG3pkjuwS23m4 kuDw kKhVzfIxuefVSRYSA5YnTeLMf nU

  1. Given that, a3 = 15, S10 = 125 As an = a + (n − 1)d,

a3 = a + (3 − 1)d 15 = a + 2d (i)

gw3VXjj ZE9jxqtBJd8F4Jx8gJ8QFKIvGLC5VFMHog 5WY6K2FFZoOLtQtVdPn1wMlln fpkIAsrgUlV8p3PBUsGwEtW 6 G57jZNwuP9TD e4cSnx6uT1CtAJeU7CDvie dIA

On multiplying equation (1) by 2, we obtain 30 = 2a + 4d (iii)

On subtracting equation (iii) from (ii), we obtain

−5 = 5d

d = −1

From equation (i), 15 = a + 2(−1)

15 = a − 2

a = 17

a10 = a + (10 − 1)d

a10 = 17 + (9) (−1)

a10 = 17 − 9 = 8

  1. Given that, d = 5, S9 = 75

VQVyJjbz6yBKKr65tizi3rOF86uJOoxRnaz4zJpzxLC0BaECRnGBPwXUwVErsmgNz7I9qCJdkfJT6LGhRDPBCMFZXpAdsXj RjbnCmJ7jjsw5Tzxl7 ID2WfqjA4uORRGkJwOyg

,n5cR4dZcKud1zRXZ65SawpnnBwUDCE0ZtFKiBTc4iQMUsL10T wxqWJ7FXfjE6UFeZE37VC3EK RF

25 = 3(a + 20)

25 = 3a + 60

3a = 25 − 60
sTdhk84l9ThntBDs2nSteYrJyXGQSbJBokrTqcCqdLyjjru4cK5WOUJtn8AYxvKXmYnBqN05LvsSZmZvybypu1FVgnuSG2LJWOayH 66rqu

an = a + (n − 1)d

a9 = a + (9 − 1) (5)

smIK TPs
  1. Given that, a = 2, d = 8, Sn = 90

VQVyJjbz6yBKKr65tizi3rOF86uJOoxRnaz4zJpzxLC0BaECRnGBPwXUwVErsmgNz7I9qCJdkfJT6LGhRDPBCMFZXpAdsXj RjbnCmJ7jjsw5Tzxl7 ID2WfqjA4uORRGkJwOyg

       As ,

lnazz ferrKYgVZxvTxnaouDU3ScwNERLqqv74Js1 qA1iNJJ3ef Lp1GLCWn0MLzS4N8Lj40Kt8MLNBd8RJ Z1yXfVH39xI0tazkn9fAW WLUqadS6 tFcpPwu8p3ez09LubWA

90 = n [2 + (n − 1)4]

90 = n [2 + 4n − 4]

90 = n (4n − 2) = 4n2 − 2n

4n2 − 2n − 90 = 0

4n2 − 20n + 18n − 90 = 0

4n (n − 5) + 18 (n − 5) = 0

(n − 5) (4n + 18) = 0

Either n − 5 = 0 or 4n + 18 = 0

n = 5 or

However, n can neither be negative nor fractional. Therefore, n = 5

an = a + (n − 1)d

a5 = 2 + (5 − 1)8

= 2 + (4) (8)

= 2 + 32 = 34

  1. Given that, a = 8, an = 62, Sn = 210U7ZnutIsP1GDFc96npxtLeiWJ3s2hIv4GUQGooi 9FQdGKNvmOlV 6m3fxvwu32TyD0 ea qoh5ljQ23EZheKQ74RtUZeuS9Q0aA8H 3rMaREhVaY7ORQkvrCrjm VAPtNLD 1w

n = 6

an = a + (n − 1)d

62 = 8 + (6 − 1)d

62 − 8 = 5d

54 = 5dd2F7MlrpTBCLb5jBo ykeItNIyqfhjIjfAP nFTU8fz4jnWJm 2S62vaTsfO3MuHfk9T qLIcnLm4QSYjkS KRJMCaHsX vx4nvIjxErnVql31o CUm tuMT0l3hY5WGLLOivM

  1. Given that, an = 4, d = 2, Sn = −14

an = a + (n − 1)d 4 = a + (n − 1)2

4 = a + 2n − 2

a + 2n = 6

a = 6 − 2n (i)

1RiLeDDh4Ym98bdbGps6MtJLj96ejyx gGJbJsqL41iWvmMwecZAQqVn4JJMPvrqJn2pnqp2iLhUeDxZF6pxafTs AY o5tQKHu3mkB59hUa40JRL0hZFhU7QobI1YBo 9dGtts

−28 = n (a + 4)

−28 = n (6 − 2n + 4) {From equation (i)}

−28 = n (− 2n + 10)

−28 = − 2n2 + 10n 2n2 − 10n − 28 = 0 

n2 − 5n −14 = 0

n2 − 7n + 2n − 14 = 0 

n (n − 7) + 2(n − 7) = 0 

(n − 7) (n + 2) = 0

Either n − 7 = 0 or n + 2 = 0 n = 7 or n = −2

However, n can neither be negative nor fractional. Therefore, n = 7

From equation (i), we obtain a = 6 − 2n

a = 6 − 2(7)

= 6 − 14

= −8

  1. Given that, a = 3, n = 8, S = 192
1dLhiC BPNlnxlNfQnqol36l96dVHYpn9lYMER3hKzqo5V5PIxj6Mdg 54Bmf0zkKjFZLKbjdJJB7XJeYtB5GtFqR0Cyqh C5agm2QAMZehDFAhD4NMLPdujyYELGVvaHCYP0U

192 = 4 [6 + 7d]

48 = 6 + 7d

42 = 7d

d = 6

  1. Given that, l = 28, S = 144 and there are total of 9 terms.9dfGz nktEObunH

(16) × (2) = a + 28

32 = a + 28

a = 4

Question 4:

How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?

Answer:

Let there be n terms of this A.P.

For this A.P., a = 9

d = a2 − a1 = 17 − 9 = 8

636 = n [9 + 4n − 4]

636 = n (4n + 5)

4n2 + 5n − 636 = 0

4n2 + 53n − 48n − 636 = 0

n (4n + 53) − 12 (4n + 53) = 0

(4n + 53) (n − 12) = 0

Either 4n + 53 = 0 or n − 12 = 0J 9DpbONFChZsKCskIPQP2YQq4oBNPJjzxcbjQ

or n = 12

n cannot be . As the number of terms can neither be negative nor fractional, therefore, n = 12 only.

Question 5:

The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Answer:

Given that, a = 5

l = 45

Sn = 400

X1Joo8mtdCy69Ua7uqST4aYv1Zhcmsuf7Baw2MfJAg3z2MTybAqqI rg5AeYPbuP0fBsw3f TphZ4E TwrR51bqwOGEAie3GBBqdUva BhqckrrVv0px5wYOiaFTorvdXnRl9 Q

n = 16

l = a + (n − 1) d

45 = 5 + (16 − 1) d

40 = 15d

GfUeR72UJ4tP3Tn8AWyvFbbQtZ10JIeOpScURaddHgKR UT0mHvsRJBwZrOUfkL

Question 6:

The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Answer:

Given that, a = 17

l = 350

d = 9

Let there be n terms in the A.P. l = a + (n − 1) d

350 = 17 + (n − 1)9

333 = (n − 1)9

(n − 1) = 37

n = 38
48 UIS1wzqaLsec9t V

Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.

Question 7:

Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

Answer:

d = 7

a22 = 149, S22 = ?

an = a + (n − 1)d

a22 = a + (22 − 1)d 149 = a + 21 × 7

149 = a + 147

a = 2

Question 8:

Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Answer:

Given that, a2 = 14

a3 = 18

d = a3 − a2 = 18 − 14 = 4

a2 = a + d 

14 = a + 4

a = 10RB00UP5pgGaUR EXb8Gq3dksVq1d0kYnLzoJh6gXoSsS1nr0hCIa 8mBFnqy sF0QuC2EpNjQkkmQ8Xtgs8dyD9BCqVwMDH2HF3WdiE5 rZX4a2 NDYOnkGBouvhm05WPRv3z8k

= 5610

Question 9:

If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Answer:

Given that, S7 = 49

S17 = 289

7 = (a + 3d)

a + 3d = 7 (i)

Similarly,
Guw45M5XDB59sxXCgIeCOEU7evZHO9V56FcWFhD 3JEAUyYT5Ef7XIES

17 = (a + 8d)

a + 8d = 17 (ii)

Subtracting equation (i) from equation (ii), 5d = 10

d = 2

From equation (i), a + 3(2) = 7

a + 6 = 7

a = 1
o7FVS1wB03RrR1TdCIhWIbldqYC

= n2

Question 10:

Show that a1, a2 … , an , … form an AP where an is defined as below

  1. an = 3 + 4n
  2. an = 9 − 5n

Also find the sum of the first 15 terms in each case.

Answer:

  1. an = 3 + 4n a1 = 3 + 4(1) = 7

a2 = 3 + 4(2) = 3 + 8 = 11

a3 = 3 + 4(3) = 3 + 12 = 15

a4 = 3 + 4(4) = 3 + 16 = 19

It can be observed that

a2 − a1 = 11 − 7 = 4

a3 − a2 = 15 − 11 = 4

a4 − a3 = 19 − 15 = 4

i.e., ak + 1 − ak is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.

JzsjoXiiz9TRCTJj3PiPFYsyaJxPWxqcUZ2mAF67NlC3CfCeolTUxn144lbNae nGvkEFGB2FIlfW6CYu2tDewxwFsbHObnmvXw

= 15 × 35

= 525

  1. an = 9 − 5n

a1 = 9 − 5 × 1 = 9 − 5 = 4

a2 = 9 − 5 × 2 = 9 − 10 = −1

a3 = 9 − 5 × 3 = 9 − 15 = −6

a4 = 9 − 5 × 4 = 9 − 20 = −11

It can be observed that a2 − a1 = − 1 − 4 = −5

a3 − a2 = − 6 − (−1) = −5

a4 − a3 = − 11 − (−6) = −5

i.e., ak + 1 − akis same every time. Therefore, this is an A.P. with common difference as

−5 and first term as 4.

CeF2cxv9Ia7WypWM963u06o0lQH 7 yDGZuU0V5XoU8lShpybmbEf9LNGUcfI5gckzYHCbORMX6VygFuUB0jK3yUmF4G8uNeArNRivc8YWQn51YeSmEOXDA5o4ZlfW0Cv5HkpQ4

= −465

Question 11:

If the sum of the first n terms of an AP is 4n − n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly find the 3rd, the10th and the nth terms.

Answer:

Given that, Sn = 4n − n2

First term, a = S1 = 4(1) − (1)2 = 4 − 1 = 3

Sum of first two terms = S2

= 4(2) − (2)2 = 8 − 4 = 4

Second term, a2 = S2 − S1 = 4 − 3 = 1

d = a2 − a = 1 − 3 = −2

an = a + (n − 1)d

= 3 + (n − 1) (−2)

= 3 − 2n + 2

= 5 − 2n

Therefore, a3 = 5 − 2(3) = 5 − 6 = −1

a10 = 5 − 2(10) = 5 − 20 = −15

Hence, the sum of first two terms is 4. The second term is 1. 3rd, 10th, and nth terms are

−1, −15, and 5 − 2n respectively.

Question 12:

Find the sum of first 40 positive integers divisible by 6.

Answer:

The positive integers that are divisible by 6 are 6, 12, 18, 24 …

It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.

a = 6

d = 6 S40 =?

= 20[12 + (39) (6)]

= 20(12 + 234)

= 20 × 246

= 4920

Question 13:

Find the sum of first 15 multiples of 8.

Answer:

The multiples of 8 are 8, 16, 24, 32…

These are in an A.P., having first term as 8 and common difference as 8.

Therefore, a = 8 d = 8

S15 =?Wk1vaZ6IUZ7qCPCrKUDtfDD7KwguBINip1f UZFfMPOZ7X1CmAySq5ZnoHxWy8uc8Cn06uEE4b83QXKWGm

= 960

Question 14:

Find the sum of the odd numbers between 0 and 50.

Answer:

The odd numbers between 0 and 50 are 1, 3, 5, 7, 9 … 49

Therefore, it can be observed that these odd numbers are in an A.P. a = 1

d = 2

l = 49

l = a + (n − 1) d 49 = 1 + (n − 1)2

48 = 2(n − 1)

n − 1 = 24

n = 25
Fz570ML5MK5Op0h05 OrI8v3bfOoxzAwuXwMrMpq3WS4A8m aX1eChBHMkY08CyhwMFDbeSdbxuXMU8ATVtQXAxRiNy0vBgUUNdKP9y HWpMYZzkde03B9w0D0nz A zmszupfs6IwZsjTtocgqchwYZmp2iekiB82IGMs AFLyV1zJT7ZX0xt1ZkLPA0mKXVZMPhFT91z3JXLEU Zfq Ene7tCwMzc

= 625

Question 15:

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days.

Answer:

It can be observed that these penalties are in an A.P. having first term as 200 and common difference as 50.

a = 200

d = 50

Penalty that has to be paid if he has delayed the work by 30 days = S30

z4n MXSdVsVy2QNkgkIasNnhpjdgxpc3ZnZiW8HJ5oQtq18jKFnhWo33sWbYNfsxmiCghSMYS0oEBfPAvtw6a4cLSD5aYmaYF0ZiFE5zOvu5GVu8OqSCD tb2udh pYJfQ43ZUs

= 15 [400 + 1450]

= 15 (1850)

= 27750

Therefore, the contractor has to pay Rs 27750 as penalty.

Question 16:

A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.

Answer:

Let the cost of 1st prize be P. Cost of 2nd prize = P − 20 And cost of 3rd prize = P − 40

It can be observed that the cost of these prizes are in an A.P. having common difference as −20 and first term as P.

a = P

d = −20

Given that, S7 = 700

E9KekAFeb8DCrnioh8ZK0rJm8RsXCl 1cokLTCHPJ5d1y7SWXAJT9j6devsElyq3 cbnExUNfo

a + 3(−20) = 100

a − 60 = 100

a = 160

Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.

*Question 17:

In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?

Answer:

It can be observed that the number of trees planted by the students is in an AP. 1, 2, 3, 4, 5… 12

First term, a = 1

Common difference, d = 2 − 1 = 1

uyjFzsZ7G m7QEE1 1exFxE1pa33wsmuCW2nZYlm6fgLNoDn u8MfCcnUd97ZkFtisQaa5Z uNxOIhYwqSjlq4qxZLxJNVx0xLYobVwbCGmCmEbeSXhG5eFaWA2VmOl kFAWFTU

= 6 (2 + 11)

= 6 (13)

= 78

Therefore, number of trees planted by 1 section of the classes = 78 Number of trees planted by 3 sections of the classes = 3 × 78 = 234 Therefore, 234 trees will be planted by the students.

*Question 18:

A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, ……… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles?SSapQ01ttCbqooA8yWUbzeboKjVFMMo9UVkoL8bH n4amJtb58yAnaK3mf 6Gg

Answer:

Semi-perimeter of circle

rYTg5aUogKjDG8t0OQXd3w3qM2 NHrSLIOjBd7iGFnHtyXBNrmK4mnOv IxoF7UJGMeitl6id W5 hv5
o1AU0WnX8GAR9YP 0ARNc2jjMH7ueEFG67 eoKa85 c0qVRAhqGQMAga wNWYoU9Y87HB3KCwggg sKkBaLYRWC7XXMfrXVLRDcld910Dfo7EJjIG

Therefore, I1, I2, I3 ,i.e. the lengths of the semi-circles are in an A.P.,
SE6X0dyM 8n95XNog4aavoVY q2ZMT0QyKXbXOP5NR0C9LAveCLo3dPMLW5ZOblNEbpK6T4lYceWJa1zNINLbMtavL5O5vlzyBdzukr3XcvT5Hx388613s7lpF6081qMFEqHq 4

jSnS8WuyHLWCfaZB7hzlyzwfQZS9DODOidutCx ehb7LwUdv3K86RASX0 xrVL902 4z6VxxOQPoHbu2xjDTYaCMDvI7L04ScRWX8x0d8HKd2h36aKeTc38EHKKXZuUt 6rk0pw

S13 =?

We know that the sum of n terms of an a A.P. is given byh82XXDyCTZ5ByYHOFFh5svABda7OyzizkYjNMP kj3zK gKtdEIf sxta4d2ZpIhztwX8ZZ5UDT C8IgODyyvRt6HorT12lrqPgXRzsT0Raa1vF6oRXTML38sHreST8yeoCwPDw

= 143

Therefore, the length of such spiral of thirteen consecutive semi-circles will be 143 cm.

*Question 19:

200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?HgQqUq7gDoEXtoPTci9oD GVFIAOhZEKkAJxE9kxkBs2c2qIwhkeHCytgsCBeGcJC hV4dP XjeeYLE5aGlvh3Wuj9ZN3OhHmQd8B Ptuv63klfS5 oOjB20dEHXub7c cqBgc

Answer:

It can be observed that the numbers of logs in rows are in an A.P. 20, 19, 18…

For this A.P., a = 20

d = a2 − a1 = 19 − 20 = −1

Let a total of 200 logs be placed in n rows. Sn = 200
qQ3F5ZFkFDOaBTYvz7NKX IJz g6j5ddeHzMc40QjwPURFbm40pJmcWBKUdqbUy5351mdPZH5apfqV1joSeT5D76SYw3Ci0rXwBgmeGc9HjFSxOOlhx1 jEoGQMYRJkKpTjT4C8

400 = n (40 − n + 1)

400 = n (41 − n)

400 = 41n − n2

jyu wPoJW4StHII4Cy x59lKARE5NzI RtzpBOCwJdXcAP UocfdOOIJNWSSW6HGIcHy XUGC5zseqFnDGR4IBAYTa8EV 0kFFEsQ6grqEtBTPtYFE S72duZ p HKooMFazfoo

n2 − 16n − 25n + 400 = 0

n (n − 16) −25 (n − 16) = 0

(n − 16) (n − 25) = 0

Either (n − 16) = 0 or n − 25 = 0 n = 16 or n = 25

an = a + (n − 1)d

a16 = 20 + (16 − 1) (−1)

a16 = 20 − 15

a16 = 5

Similarly,

a25 = 20 + (25 − 1) (−1)

a25 = 20 − 24

= −4

Clearly, the number of logs in 16th row is 5. However, the number of logs in 25th row is negative, which is not possible.

Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16th row is 5.

*Question 20:

In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?yDV iEJG618Ba5Fj 09qiGGsmDpmsH1Whz0w9VQfn7RQqL0oXCsvYjAhfv99bqu01Kk5tJP19Anmo1UamR0ruBVxNAvpc yaHqS4ahVuq2SwYGow03w4Chq 6 PI7N M2Aumr8

[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 ×(5 + 3)]

Answer:
u KMWDf6F9LVnn28mje4RqcUp6SaZdkN07iNLTZFD1MpVVZC Omj62afK3DBS9e j5ZzHJXY0piapya3 604B Kph0RDqUGYEt23DocbG

The distances of potatoes are as follows. 5, 8, 11, 14…

It can be observed that these distances are in A.P. a = 5

d = 8 − 5 = 3j4 POwEchI96YplZL9yPFtRB4RyMEdN43uj0vRYkAmNf0gAqZQOfxBcgIzLQl3kQICKAe mbQzBl KYOHwz80QSHuBN4d93VXSkKnlaJMYTyK9 qM6L

= 5[10 + 9 × 3]

= 5(10 + 27) = 5(37)

= 185

As every time she has to run back to the bucket, therefore, the total distance that the competitor has to run will be two times of it.

Therefore, total distance that the competitor will run = 2 × 185

= 370 m

Alternatively,

The distances of potatoes from the bucket are 5, 8, 11, 14…

Distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept. Therefore, distances to be run are

10, 16, 22, 28, 34,……….

a = 10

d = 16 − 10 = 6 S10 =?
3dTB2 bTHp755 ZxEk fDnx1BuC5S2ZvHrQLCThvDAJ5 gXxG5jc7Mcz O0Y3XqmUERaNiPNKoavAb0ayrwlS0kA6Ata2

= 5[20 + 54]

= 5 (74)

= 370

Therefore, the competitor will run a total distance of 370 m.

Exercise 5.4

*Question 1:

Which term of the A.P. 121, 117, 113 … is its first negative term?

[Hint: Find n for an < 0]

Answer:

Given A.P. is 121, 117, 113 …

a = 121

d = 117 − 121 = −4

an = a + (n − 1) d

= 121 + (n − 1) (−4)

= 121 − 4n + 4

= 125 − 4n

We have to find the first negative term of this A.P.

IwLGMC0nrRzpVJ03K2Tjjr1bkQTTsUGevLG4aCumlNSFe4ALsk1kHl9cnKurg2qrKuqMWJg

Therefore, 32nd term will be the first negative term of this A.P.

Question 2:

The sum of the third and the seventh terms of an A.P is 6 and their product is 8. Find the sum of first sixteen terms of the A.P.

Answer:

We know that, an = a + (n − 1) d a3 = a + (3 − 1) d

a3 = a + 2d

Similarly, a7 = a + 6d

Given that, a3 + a7 = 6 (a + 2d) + (a + 6d) = 6 2a + 8d = 6

a + 4d = 3

a = 3 − 4d (i)

Also, it is given that (a3) × (a7) = 8 (a + 2d) × (a + 6d) = 8

From equation (i),

bl3QtbHUuTe3tq0vhCkM8

From equation (i)

Where d is

yzRa kvV mHqPJcGzYmhaHSYuqP9p2G6poihjthjIo2XERVEb7pbJbs
8SQnEl1ocmeA5AM6nH XC3APpkZzwurL247OB8K RdyUizzAzOBifjBPMZwyPZ8HdOwrTVi9bvz4DdC mbiKvUFch1mjmEuhqS8wL5BibmRWULqwQthMjLke DXfQsfzYsU3fms
wYqrzWrVDc7l GZRYBPwkvBXtlccZUiWefRz66P5iy9ki8vaIdAIN7hck

When d is

db oNkrD316Kz poQHUxF3pcP4uuCGACia1ZOg5nUOrYoTt z9xfVIJAIvh9 EUZeVbkeW29e8KcbbWpwlC6ytErKWr V7bm5xHGCT5smSdq TMABIz3s8namR079zlqC wFgTo
2kWTXuAGQW3qR6o2NRoWaLG9kafQpG0xgmOnSaO5rojOCM41dDfL86FWybCDr3bJyT CQzrWH ia3A0AOuzwhl0j2e6aA4i 1E 7CA kRi2MEDUhPJbBwllpTZktDr61yJ 4Zs
aMftBnpeFfrvNaFSlEZNK2e9Li3YAj4LGxgeh3algsTU0o WNad4NCq0zxQkI7YgJvemHLuQEGkjW4Y13vqGcP BJF0DyiDo5RU7qLcsD0uA2iR1okKltW lEGaw iQ1l2Saabk

When a is 1 and d is

t10YQ58eNOUsLklK6kmoHIaOaY5htPoAfXcBTIPYt7Lgyt8pPWZAmCo8xCP8pU93iZLWWa9ghm4Bvo OxFq
6uHH0sFYasV87XsTJ sq yf8x1nfVmg3XsfJL4bZhvg tmmwJDI8Mm FNDuba4S70H1hkCaOYauq4Wi9XblbNetGHXy0ClM6KRPXVrzuqybKJdBk

= 76

When and d is

msH4o3p8W5NGP31aLJ83lxNpZvyICnqq0JePCtr3YrAjIP psusGjR xawzARhsuuwPiI0w1Nsat5KLnDOlG0vr71AbWa11GJrWVuEvnkMvqRtDYJaEDKgZltjzfBzje v6NW00
banUKWMzHe2l0UCxPzlL3JAy3VA0KNvZoo4dYgjdqKqJ8VtPqU U70OhLcX5DZ98i v HLoFQrUvSFzLd5Ct1CftL4XgM91PP8c9L6y1aWJ7ciaXaQYuFsrr9nfI34foy0F98A
wjYAHrYPbNlrmyyusATqnCA40tnEj3FDlX0g9XoFz SbrGRnIe4YCFbHTJT4Hwbrbffl662HixrnVL 6pGEV
yb7HqLI2IHh SDceUIaBROWfOkHmpx9JzCiR82fu8DIY6tPFI00K2SvduAUf03TeTY9MIqoEOTsdf

*Question 3:

A ladder has rungs 25 cm apart. (See figure). The rungs decrease uniformly in length

from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs m are apart, what is the length of the wood required for the rungs?

[Hint: number of rungs ]MUqUyBGzbwtWfdna1LO54lxi7DXSMUACocSXo1gqA irAewYFVr2v1F2jcNxQHHw9YvCRxep5Lzn7YUQgMyWDQo optP0cuXm6GgCN7YwathmsfiE Ovy2348hLBuLA9Ntq FDM

Answer:

It is given that the rungs are 25 cm apart and the top and bottom m rungs are apart.

∴ Total number of rungs

Now, as the lengths of the rungs decrease uniformly, they will be in an A.P.

The length of the wood required for the rungs equals the sum of all the terms of this A.P. First term, a = 45

Last term, l = 25 n = 11
mSkgTdDkHltIfRs2q011g eOYEvc4OWwMY6U2CEgq iVRZ38b5Pz8C6GlelIZMsqWlEG1GwwZG

Therefore, the length of the wood required for the rungs is 385 cm.

*Question 4:

The houses of a row are number consecutively from 1 to 49. Show that there is a value of x such that the sum of numbers of the houses preceding the house numbered x is equal to the sum of the number of houses following it.

Find this value of x.

[Hint Sx − 1 = S49 − Sx]

Answer:

The number of houses was 1, 2, 3 … 49

It can be observed that the number of houses are in an A.P. having a as 1 and d also as 1.

Let us assume that the number of xth house was like this. We know that,

Sum of n terms in an A.P.

Sum of number of houses preceding xth house = Sx − 1
2io3TC88R4mEbr5cRSXVjXMZ3gzY NwlPrxxijKfFzHpd9BLNmwTUqUyvKpXzmvN5Z76StL KFmjV16u7 3AIOWXhR p9bQnyy7sgflhalkvX3q68 Zw4k0SpY0iukNnwAz4b Y

O1osxl48nqDcqpea3ueKY9Y1NGWmm 7iKxsQ0Ux6SL iUq1i3B9I47XCzFH6NKcJENEMAxjyiPzB2 1Qtl s1klTzT wlcK9KVc8p1W9Z0psH2m fsLAYY9X9Q8LOyZeKcfbdlA

Sum of number of houses following xth house = S49 − Sx

It is given that these sums are equal to each other.UpLmf2nO8S 3gXFH4zL4 nP JNNxTsvKsMrJGpYn2OwcM4sjn0nAlm7z DsUUdQFRDi9C1kzJhTSJRvnyOkKalGLjHQJEXoHwyChU
SStZx6RhELm QOzFyaiIn5i6p2fH15WtsL4VRsKo8AWeALlOXJdmVUh2DmeUUJW qTkr8Qpg9Q5Szhidtz8CAPF0 Kc1hGD8HaOOoS3vDNFSKhS

However, the house numbers are positive integers. The value of x will be 35 only.

Therefore, house number 35 is such that the sum of the numbers of houses preceding the house numbered 35 is equal to the sum of the numbers of the houses following it.

*Question 5:

A small terrace at a football ground comprises of 15 steps each of which is 50 m  

long and built of solid concrete. Each step has a rise of m and a tread of

(See figure) calculate the total volume of concrete required to build the terrace.
Vs6sxYS2NxBWm9LpKb468JnNLu iEmIx5gAZptXsiQIze9RRpXU8em GXbqYXEWLbjzPTkBhClrivioTbc2yQWpPMaUEPJl

AnswerMgAYxWiQ45M4djgOi3xBomPzh3PmoDb6P1x0SJCxszMAAIYOgGaPZ58hw9IifTI g7Nki IDNgVM0wvwVnLus oYhmWlzwPIUFV t1L7RPPEF8vOOK5NCIKohZiXwy3GQw0dRGQ

From the figure, it can be observed that

1st step is m wide,

2nd step is 1 m wide,

3rd step is m wide.

Therefore, the width of each step is increasing by m each time whereas their height 1/4 m and length 50 m remains the same.

Therefore, the widths of these steps are

Volume of concrete in 1st step

Volume of concrete in 2nd step

lOhQjWBNpX2wsaFv35vZy 3q5nvcSNYV7AGRJ1rU0OBcz2SSvCEail1Omt

Volume of concrete in 3rd step

It can be observed that the volumes of concrete in these steps are in an A.P.Ndp2v1T2Y4Z8dVs1H4GHMOsOGDSFLxrPkMvNTZCGfLk8nVY vR6x68ZJgZaDIPFXxrPb qchU heGr4X56f1JHnbsPqBB PrrrHCgAYrM4VdgYA1U

Volume of concrete required to build the terrace is 750 m3.EK1q2w kkieuDkpQAah4yVEcgCsZf9874LNo6OphULVx9dKKSegHpLeRQnMT KvOtlG0ZxaqSqq9 nX bIlXP

This chapter comes under unit 3 algebra and this unit has 20 marks allotted in the examination. Students can expect an average of 3 questions from arithmetic progressions. Along with Class 10 examinations, this topic is very important from the point of competitive exams.

Sub-topics of Class 10 Chapter 5 Arithmetic Progression
5.1 Introduction In this chapter, we shall discuss patterns which we come across in our day-to-day life in which succeeding terms are obtained by adding a fixed number to the preceding terms. We shall also see how to find their nth terms and the sum of n consecutive terms, and use this knowledge in solving some daily life problems.

5.2 Arithmetic Progressions The topic describes Arithmetic Progressions, its definition and relatable terms along with fine examples. You will also learn about Finite Arithmetic Progressions and Infinite Arithmetic Progressions. The general form of AP is a, a + d, a + 2d, a + 3d,…

5.3 nth Term of an AP
This topic discusses various methods to determine the nth term of an AP. The concepts are explained with different types of problems solving techniques and finding the nth term of an AP. The examples mentioned in the chapter will help you while solving the exercise problems.

5.4 Sum of First n Terms of an AP
The topics discuss different techniques to find the sum of the first n terms of an AP. It also provides suitable examples which show different techniques to find the sum of the first n terms of AP.

5.5 Summary
It gives an overview of the entire chapter and the important topics explained in the entire chapter. By going through the summary part you can cover the entire chapter in a few points which help in memorizing the essential concepts.

List of Exercise from Class 10 Maths Chapter 5 Arithmetic progression
Exercise 5.1– 4 questions 1 MCQ and 3 descriptive type questions
Exercise 5.2– 20 questions, 1 fill in the blanks, 2 MCQ’s, 7 Short answer questions and 10 Long answer questions
Exercise 5.3– 20 Questions 3 fill in the blanks, 4 daily life examples, and 13 descriptive type questions
Exercise 5.4 5 Questions- 5 Long answer questions

This NCERT solution for class 10 Maths is a perfect study material that will help you solve different kinds of problems. Solving these NCERT Solutions will help you understand the topic completely and help you lay a greater foundation for future studies.

In this chapter, students will discuss patterns in succeeding terms obtained by adding a fixed number to the preceding terms. They also, see how to find nth terms and the sum of n consecutive terms. Students will learn arithmetic progression effectively when they solve daily life problems.

This chapter has Arithmetic Progression Derivation of the nth term and sum of the first n terms of an A.P. and their application in solving daily life problems. This is one of the important chapters from the point of the Class 10 examination. An arithmetic progression is a very basic and important topic to study as almost all the competitive exams will ask questions on arithmetic progression.

Key Features of NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic progressions
Has answers to different types of questions such as MCQs and long answer questions.
Solving this NCERT solution will make you get well versed with important formulas.
Act as a basis to solve arithmetic progression problems asked in competitive examination.
Has answers to all the exercise questions provided in NCERT textbook
Provide you with the necessary practice of solving questions
You can solve different types of questions with varying difficulty.
Different examples taken from day to day life will help you understand the topic thoroughly.
Keep visiting SWC’s to get complete assistance for CBSE class 10 board exams. At SWC’S, students can get several sample papers, question papers, notes, textbooks, videos, animations and effective preparation tips which can help you to score well in the class 10 exams.

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NCERT Exemplar for Class 10 Maths Chapter 5 Arithmetic Progressions
CBSE Notes for Class 10 Maths Chapter 5 Arithmetic Progressions
Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 5
What kind of questions are there in NCERT Solutions for Class 10 Maths Chapter 5?
Chapter 5 of NCERT Solutions for Class 10 Maths has multiple choice questions, descriptive type of questions, long answer type questions, short answer type questions, fill in the blanks and daily life examples. By the end of this chapter students can increase their problem solving skills and time management skills. This helps in procuring high marks in their finals.
Is it necessary to learn all the topics provided in NCERT Solutions for Class 10 Maths Chapter 5?
Yes, it is compulsory to learn all the topics provided in NCERT Solutions for Class 10 Maths Chapter 5 to score high marks in Class 10 board exams.These solutions are designed by subject matter experts who have assembled model questions covering all the exercise questions from the textbook. They also focus on cracking the Solutions of Maths in such a way that it is easy for the students to understand.
Is NCERT Solutions for Class 10 Maths Chapter 5 enough to attend all the questions that come in the board exam?
Yes, it is enough to solve all the questions that come in the board exam of NCERT Solutions for Class 10 Maths Chapter 5. Practising this chapter can make them learn the concepts flawlessly. These questions have been devised, as per the NCERT syllabus and the guidelines. This makes the students to score good marks in the finals.


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NCERT Solutions Class 10 Maths Chapters

  • Chapter 1 Real Numbers
  • Chapter 2 Polynomials
  • Chapter 3 Pair Of Linear Equations In Two Variables
  • Chapter 4 Quadratic Equations
  • Chapter 5 Arithmetic Progressions
  • Chapter 6 Triangles
  • Chapter 7 Coordinate Geometry
  • Chapter 8 Introduction To Trigonometry
  • Chapter 9 Some Applications Of Trigonometry
  • Chapter 10 Circles
  • Chapter 11 Constructions
  • Chapter 12 Areas Related To Circles
  • Chapter 13 Surface Areas And Volumes
  • Chapter 14 Statistics
  • Chapter 15 Probability

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