In an arithmetic progression (A.P.), the sum of the first ten terms is -150, and the sum of the next ten terms is -550. Find the A.P.

S10 = -150; the sum of next 10 terms = S20 - S10 => -550 = S20 + 150 => S20 = -700; S10 = -150 => 2a1 + 9r = -30; S20 = -700 => 2a1 + 19r = -70; Then 10r = -40 => r = -4 , and, 2a1 -36 = -30 => 2a1 = 6 => a1 = 3;

Answered by crisforp | 2024-06-10

The first term of the arithmetic progression (A.P.) is 3 and the common difference is -4, leading to the sequence 3, -1, -5, and so on. The formula for the nth term of this A.P. is a_n = 3 - 4(n-1). These values satisfy the given conditions about the sums of the terms.
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Answered by crisforp | 2024-12-20