Question

In an AP of 50 terms, the sum for first 10 terms is 210 and the sum of last 15 terms is 2565. find the AP.​

Answer 1

Let the first term of the AP be a and common difference be d.

For the sum of first 10 terms: S10 = 210

10/2 [2a + (10-1)d] = 210

5 [2a + 9d] = 210

2a + 9d = 42  -(1)

For the sum of last 15 terms, we can take A as the 36th term, ie, a+35d

15/2 [2A + (15-1)d] = 2565

2(a+35d) + 14d = 2565*2/15

2a + 70d + 14d = 5130/15

2a + 84d = 342 -(2)

subtracting (1) from (2)

2a + 84d = 342

2a + 9d = 42

(-)     (-)      (-)

75d = 300

d = 300/75 = 4

by equation (1)

2a + 9d = 42

2a + 9*4 = 42

2a + 36 = 42

2a = 6

a = 3

Therefore, AP = 3, 7, 11, 15, 19, ......

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Answer 2

Consider a and d as the first term and the common difference of an A.P. respectively.

n th term of an A.P., an = a + ( n – 1)d

Sum of n terms of an A.P., S n = n/ 2 [2a + (n – 1)d]

Given that the sum of the first 10 terms is 210.

⇒ 10 / 2 [2a + 9d ] = 210

⇒ 5[ 2a + 9 d ] = 210

⇒2a + 9d = 42 ----------- (1)

15 th term from the last = ( 50 – 15 + 1 ) th = 36 th term from the beginning

⇒ a36 = a + 35d

Sum of the last 15 terms = 15/2 [2a36 + ( 15 – 1)d ] = 2565

⇒ 15 / 2 [ 2(a + 35d) + 14d ] = 2565

⇒ 15 [ a + 35d + 7d ] = 2565

⇒a + 42d = 171 ----------(2)

From (1) and (2), we have d = 4 and a = 3.

Therefore, the terms in A.P. are 3, 7, 11, 15 . . . and 199.