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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.

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(i) Given, ∴ n = 16 (ii)Given, 35 = 7 + 12 d 28 = 12d (iii) Given, = 37, d = 3 a12 = a + (12 − 1)3 37 = a + 33 a = 4 (iv) Given, On multiplying equation (1) by 2, we obtain 30 = 2a + 4d (iii) On subtracting equation (iii) from (ii), we obtain d = −1 From...
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(i) Given, 

∴ 

n = 16

 

 

(ii)Given, 

 

35 = 7 + 12 d

28 = 12d

 

(iii) Given,  = 37, d = 3

a12 = a + (12 − 1)3

37 = a + 33

a = 4

 

(iv) Given, 

On multiplying equation (1) by 2, we obtain

30 = 2a + 4d (iii)

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

d = −1

From equation (i),

a = 17

 

(v)Given, 

25 = 3a + 60

 

(vi) Given, 

 

Either  n - 5 = 0 or 4n + 18 = 0

n = 5 or 

However, n can neither be negative nor fractional.

Therefore, n = 5

= 2 + (4) (8)

= 2 + 32 = 34

 

(vii) Given, 

n = 6

62 - 8 = 5d

54 = 5d

 

(viii) Given, 

4 = a + (n - 1)2

4 = a + 2n - 2

a + 2n = 6

= 6 - 2n.............. (i)

 {From equation (i)}

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

= 6 - 14

= -8

 

(ix)Given, 

192 = 4 [6 + 7d]

48 = 6 + 7d

42 = 7d

d = 6

 

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

32 = a + 28

a = 4

 

 

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