Q) In an A.P. of 40 terms, the sum of first 9 terms is 153 and the sum of last 6 terms is 687. Determine the first term and common difference of A.P. Also, find the sum of all the terms of the A.P.

Ans: Let’s consider the first terms of AP is a and the common difference is d.

(i) Calculating first term and common difference of AP:

Since the sum of n terms of an AP,

∴ Sum of 9 terms, = [2a + (9 – 1) (d)] = 9 (a + 4 d)

Since it is given that the sum of 9 terms is 153

∴ 153 = 9 (a + 4 d)

∴ 17 = a + 4d

∴ a + 4d = 17 …… (i)

Similarly, Sum of 6 terms, [2a + (6 – 1) (d)] = 3 (2 a + 5 d)

Since it is given that the sum of 9 terms is 687

∴ 687 = 3 (2 a + 5 d)

∴ 229 = 2 a + 5 d

∴ 2 a + 5 d  = 229 ….. (ii)

By multiplying equation (i) by 2 and then subtracting equation (ii), we get:

(2 a + 8 d) – (2 a + 5 d) = 2 x 17 – 229

∴ 3 d = – 195

∴ d =  – 65

By substituting value of d in equation (i), we get:

a + 4d = 17

∴ a + 4 (- 65) = 17

∴ a – 260 = 17

∴ a = 17 + 260

∴  a = 277

Therefore, first term of given AP is 277 and common difference of AP is – 65. 

(ii). Sum of all terms of AP:

Since the sum of n terms of an AP,

It is given that AP has 40 terms, a = 277, d = – 65

∴ Sum of 40 terms, [2 (277) + (40 – 1) (- 65)]

∴ = 20 (554 – (39) (65)] =  20 (554 – 2535) = – 39,620

Therefore sum of its 40 terms is – 39,620

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