Question

In an A.P., the sum of first ten terms is – 150 and the sum of its next ten terms is-550. Find the A.P.

Answer 1

The AP will be : 3, –1, –5,...

• Let us consider, "a" and "d" as the first term and common difference of A.P.

• Now, S₁₀ = -150 (where S is the sum of first ten terms of (AP)

        (10/2)[2a+(n-1)d] = -150

         2a+9d= -30                                    eq(1)

• Also, a₁₁+a₁₂+.....a₂₀ = -550

       a₁+a₂+...+a₂₀ - (-150) = -550

       S₂₀ = -700

       (20/2)[2a+(20-1)d] = -700

       2a+19d= -70                                     eq(2)

• Upon solving eq(1) and eq(2), we get

        (2a + 19d) – (2a + 9d) = (–70) – (–30)

        10d = – 40

        d = – 4

• When d = –4, we get

        2a + 9 × (–4) = –30                            from eq(1)

        2a = –30 + 36

        2a = 6

        a = 3

• Therefore, the A.P. is 3, –1, –5,...