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# Class 10 NCERT Solutions- Chapter 5 Arithmetic Progressions – Exercise 5.3 | Set 1

### 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

(iv) 1/15, 1/12, 1/10, ……, to 11 terms

Solution:

(i) Given, 2, 7, 12,…, to 10 terms

For this A.P., we have,

first term, a = 2

common difference, d = a2 − a1 = 7−2 = 5

no. of terms, n = 10

Sum of nth term in AP series is,

Sn = n/2 [2a +(n-1)d]

Substituting the values,

S10 = 10/2 [2(2)+(10 -1)×5]

= 5[4+(9)×(5)]

= 5 × 49 = 245

(ii) Given, −37, −33, −29,…, to 12 terms

For this A.P.,we have,

first term, a = −37

common difference, d = a2− a1

= (−33)−(−37)

= − 33 + 37 = 4

no. of terms, n = 12

Sum of nth term in AP series is,

Sn = n/2 [2a+(n-1)d]

Substituting the values,

S12 = 12/2 [2(-37)+(12-1)×4]

= 6[-74+11×4]

= 6[-74+44]

= 6(-30) = -180

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

For this A.P.,

first term, a = 0.6

common difference, d = a2 − a1 = 1.7 − 0.6 = 1.1

no. of terms, n = 100

Sum of nth term in AP series is,

Sn = n/2[2a +(n-1)d]

S12 = 50/2 [1.2+(99)×1.1]

= 50[1.2+108.9]

= 50[110.1]

= 5505

(iv) Given, 1/15, 1/12, 1/10, ……, to 11 terms

For this A.P.,

first term, a = 1/5

common difference, d = a2 –a1 = (1/12)-(1/5) = 1/60

number of terms, n = 11

Sum of nth term in AP series is,

Sn = n/2 [2a + (n – 1) d]

Substituting the values, we have,

= 11/2(2/15 + 10/60)

= 11/2 (9/30)

= 33/20

### Question 2. Find the sums given below:

(i) 7+ 10 1/2 + 14 + ……… + 84

(ii) 34 + 32 + 30 + ……….. + 10

(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)

Solutions:

(i) Given,

First term, a = 7

nth term, an = 84

Common difference, d = 10 1/2 – 7 = 21/2 – 7 = 7/2

Let 84 be the nth term of this A.P.

Then,

an = a(n-1)d

Substituting these values,

84 = 7+(n – 1)×7/2

77 = (n-1)×7/2

22 = n−1

n = 23

We know that, sum of n term is;

Sn = n/2 (a + l), l = 84

Sn = 23/2 (7+84)

= (23×91/2) = 2093/2

= 1046 1/2

(ii) Given,

first term, a = 34

common difference, d = a2−a1 = 32−34 = −2

nth term, an= 10

Let 10 be the nth term of this A.P.,

Now,

an = a +(n−1)d

10 = 34+(n−1)(−2)

−24 = (n −1)(−2)

12 = n −1

n = 13

Sum of n terms is;

Sn = n/2 (a +l), l = 10

= 13/2 (34 + 10)

= (13×44/2) = 13 × 22

= 286

(iii) Given:

First term, a = −5

nth term, an= −230

Common difference, d = a2−a1 = (−8)−(−5)

⇒d = − 8+5 = −3

Let us assume −230 be the nth term of this A.P.

Since,

an= a+(n−1)d

−230 = − 5+(n−1)(−3)

−225 = (n−1)(−3)

(n−1) = 75

n = 76

Sum of n terms, is equivalent to,

Sn = n/2 (a + l)

= 76/2 [(-5) + (-230)]

= 38(-235)

= -8930

### Question 3. In an AP

(i) Given a = 5, d = 3, an = 50, find n and Sn.

(ii) Given a = 7, a13 = 35, find d and S13.

(iii) Given a12 = 37, d = 3, find a and S12.

(iv) Given a3 = 15, S10 = 125, find d and a10.

(v) Given d = 5, S9 = 75, find a and a9.

(vi) Given a = 2, d = 8, Sn = 90, find n and an.

(vii) Given a = 8, an = 62, Sn = 210, find n and d.

(viii) Given an = 4, d = 2, Sn = − 14, find n and a.

(ix) Given a = 3, n = 8, S = 192, find d.

(x) Given l = 28, S = 144 and there are total 9 terms. Find a.

Solutions:

(i) Given values, we have,

a = 5, d = 3, an = 50

The nth term in an AP,

an = a +(n −1)d,

Substituting the given values, we have,

⇒ 50 = 5+(n -1)×3

⇒ 3(n -1) = 45

⇒ n -1 = 15

Obtaining the value of n, we get,

⇒ n = 16

Now, sum of n terms is equivalent to,

Sn = n/2 (a +an)

Sn = 16/2 (5 + 50) = 440

(ii) Given values, we have,

a = 7, a13 = 35

The nth term in an AP,

an = a+(n−1)d,

Substituting the given values, we have,

⇒ 35 = 7+(13-1)d

⇒ 12d = 28

⇒ d = 28/12 = 2.33

Now, Sn = n/2 (a+an)

Obtaining the final value, we get,

S13 = 13/2 (7+35) = 273

(iii) Given values, we have,

a12 = 37, d = 3

The nth term in an AP,

an = a+(n −1)d,

Substituting the given values, we have,

⇒ a12 = a+(12−1)3

⇒ 37 = a+33

Obtaining the value of a, we get,

⇒ a = 4

Now, sum of nth term,

Sn = n/2 (a+an)

= 12/2 (4+37)

Obtaining the final value,

= 246

(iv) Given that,

a3 = 15, S10 = 125

The formula of the nth term in an AP,

an = a +(n−1)d,

Substituting the given values, we have,

a3 = a+(3−1)d

15 = a+2d ………….. (i)

Also,

Sum of the nth term,

Sn = n/2 [2a+(n-1)d]

S10 = 10/2 [2a+(10-1)d]

125 = 5(2a+9d)

25 = 2a+9d …………….. (ii)

Solving eq (i) by (ii),

30 = 2a+4d ………. (iii)

And, by subtracting equation (iii) from (ii), we get,

−5 = 5d

that is,

d = −1

Substituting in equation (i),

15 = a+2(−1)

15 = a−2

a = 17 =

And,

a10 = a+(10−1)d

a10 = 17+(9)(−1)

a10 = 17−9 = 8

(v) Given:

d = 5, S9 = 75

Sum of n terms in AP is,

Sn = n/2 [2a +(n -1)d]

Substituting values, we get,

S9 = 9/2 [2a +(9-1)5]

25 = 3(a+20)

25 = 3a+60

3a = 25−60

a = -35/3

Also,

an = a+(n−1)d

Substituting values, we get,

a9 = a+(9−1)(5)

= -35/3+8(5)

= -35/3+40

= (35+120/3) = 85/3

(vi) Given:

a = 2, d = 8, Sn = 90

Sum of n terms in an AP is,

Sn = n/2 [2a +(n -1)d]

Substituting values, we get,

90 = n/2 [2a +(n -1)d]

⇒ 180 = n(4+8n -8) = n(8n-4) = 8n2-4n

Solving the eq, we get,

⇒ 8n2-4n –180 = 0

⇒ 2n2–n-45 = 0

⇒ 2n2-10n+9n-45 = 0

⇒ 2n(n -5)+9(n -5) = 0

⇒ (n-5)(2n+9) = 0

Since, n can only be a positive integer,

Therefore,

n = 5

Now,

∴ a5 = 8+5×4 = 34

(vii) Given:

a = 8, an = 62, Sn = 210

Since, sum of n terms in an AP is equivalent to,

Sn = n/2 (a + an)

210 = n/2 (8 +62)

Solving,

⇒ 35n = 210

⇒ n = 210/35 = 6

Now, 62 = 8+5d

⇒ 5d = 62-8 = 54

⇒ d = 54/5 = 10.8

(viii) Given :

nth term, an = 4, common difference, d = 2, sum of n terms, Sn = −14.

Formula of the nth term in an AP,

an = a+(n −1)d,

Substituting the values, we get,

4 = a+(n −1)2

4 = a+2n−2

a+2n = 6

a = 6 − 2n …………………. (i)

Sum of n terms is;

Sn = n/2 (a+an)

-14 = n/2 (a+4)

−28 = n (a+4)

From equation (i), we get,

−28 = n (6 −2n +4)

−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

Solving for n,

(n −7)(n +2) = 0

Either n − 7 = 0 or n + 2 = 0

n = 7 or n = −2

Since, we know, n can neither be negative nor fractional.

Therefore, n = 7

From equation (i), we get

a = 6−2n

a = 6−2(7)

= 6−14

= −8

(ix) Given values are,

first term, a = 3,

number of terms, n = 8

sum of n terms, S = 192

We know,

Sn = n/2 [2a+(n -1)d]

Substituting values,

192 = 8/2 [2×3+(8 -1)d]

192 = 4[6 +7d]

48 = 6+7d

42 = 7d

Solving for d, we get,

d = 6

(x) Given values are,

l = 28,S = 144 and there are total of 9 terms.

Sum of n terms,

Sn = n/2 (a + l)

Substituting values, we get,

144 = 9/2(a+28)

(16)×(2) = a+28

32 = a+28

Calculating, we get,

a = 4

### Question 4. How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?

Solution:

Let us assume that there are n terms of the AP. 9, 17, 25 …

For this A.P.,

We know,

First term, a = 9

Common difference, d = a2−a1 = 17−9 = 8

Sum of n terms, is;

Sn = n/2 [2a+(n -1)d]

Substituting the values,

636 = n/2 [2×a+(8-1)×8]

636 = n/2 [18+(n-1)×8]

636 = n [9 +4n −4]

636 = n (4n +5)

4n2 +5n −636 = 0

4n2 +53n −48n −636 = 0

Solving, we get,

n (4n + 53)−12 (4n + 53) = 0

(4n +53)(n −12) = 0

that is,

4n+53 = 0 or n−12 = 0

On solving,

n = (-53/4) or n = 12

We know,

n cannot be negative or fraction, therefore, n = 12 is the only plausible value.

### 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.

Solution:

Given:

first term, a = 5

last term, l = 45

Also,

Sum of the AP, Sn = 400

Sum of AP is equivalent to

Sn = n/2 (a+l)

Substituting the values,

400 = n/2(5+45)

400 = n/2(50)

Number of terms, n =16

Since, the last term of AP series is equivalent to

l = a+(n −1)d

45 = 5 +(16 −1)d

40 = 15d

Solving for d, we get,

Common difference, d = 40/15 = 8/3

### 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?

Solution:

Given:

First term, a = 17

Last term, l = 350

Common difference, d = 9

The last term of the AP can be written as;

l = a+(n −1)d

Substituting the values, we get,

350 = 17+(n −1)9

333 = (n−1)9

Solving for n,

(n−1) = 37

n = 38

Sn = n/2 (a+l)

S38 = 13/2 (17+350)

= 19×367

= 6973

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

Solution:

Given:

Common difference, d = 7

Also,

22nd term, a22 = 149

By the formula of nth term of an AP,

an = a+(n−1)d

Substituting values, we get,

a22 = a+(22−1)d

149 = a+21×7

149 = a+147

a = 2 = First term

Sum of n terms,

Sn = n/2(a+an)

S22 = 22/2 (2+149)

= 11×151

= 1661

### Question 8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18, respectively.

Solution:

Given:

Second term, a2 = 14

Third term, a3 = 18

Also,

Common difference, d = a3−a2 = 18−14 = 4

a2 = a+d

14 = a+4

Therefore,

a = 10 = First term

And,

Sum of n terms;

Sn = n/2 [2a + (n – 1)d]

Substituting values,

S51 = 51/2 [2×10 (51-1) 4]

= 51/2 [2+(20)×4]

= 51 × 220/2

= 51 × 110

= 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.

Solution:

Given:

S7 = 49

S17 = 289

Since, we know

Sn = n/2 [2a + (n – 1)d]

Substituting values, we get,

S7= 7/2 [2a +(n -1)d]

S7 = 7/2 [2a + (7 -1)d]

49 = 7/2 [2a + 6d]

7 = (a+3d)

a + 3d = 7 ………………. (i)

Similarly,

S17 = 17/2 [2a+(17-1)d]

Substituting values, we get,

289 = 17/2 (2a +16d)

17 = (a+8d)

a +8d = 17 ………………. (ii)

Solving (i) and (ii),

5d = 10

Solving for d, we get,

d = 2

Now, obtaining value for a, we get,

a+3(2) = 7

a+ 6 = 7

a = 1

Therefore,

Sn = n/2[2a+(n-1)d]

= n/2[2(1)+(n – 1)×2]

= n/2(2+2n-2)

= n/2(2n)

= n2

(i) an = 3+4n

(ii) an = 9−5n

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

Solutions:

(i) an = 3+4n

Calculating,

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

Now d =

a2 − a1 = 11−7 = 4

a3 − a2 = 15−11 = 4

a4 − a3 = 19−15 = 4

Hence, ak + 1 − ak holds the same value between all pairs of successive terms. Therefore, this is an AP with common difference as 4 and first term as 7.

Sum of nth term is;

Sn = n/2[2a+(n -1)d]

Substituting the value, we get,

S15 = 15/2[2(7)+(15-1)×4]

= 15/2[(14)+56]

= 15/2(70)

= 15×35

= 525

(ii) an = 9−5n

Calculating, we get,

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

Common difference, d

a2 − a1 = −1−4 = −5

a3 − a2 = −6−(−1) = −5

a4 − a3 = −11−(−6) = −5

Hence, ak + 1 − ak holds the same value between all pairs of successive terms. Therefore, this is an AP with common difference as −5 and first term as 4.

Sum of nth term is;

Sn = n/2 [2a +(n-1)d]

S15 = 15/2[2(4) +(15 -1)(-5)]

Substituting values,

= 15/2[8 +14(-5)]

= 15/2(8-70)

= 15/2(-62)

= 15(-31)

= -465

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