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

Sum of 10^th term, S₁₀ = -150

Sum of next 10 term = -550

Let, first term = a

Last term = a_n

We know, a_n = a + (n – 1) d

= a + (10 -1) d

= a + 9d

We know, Sum of n terms

S_n = [a + l]

S₁₀ = [a + a + 9d]

-150 = 5 (2a + 9d)

-150 = 5 (2a + 9d)

10a = -150 – 45d

a = (i)

For next 10 term, First term = a₁₁

Last term = a₂₀

11^th term, a₁₁ = a + 10d

20^th term, a₂₀ = a +19d

S’₁₀ = [a + 10d + a + 19d]

-550 = 5 [2a + 29d]

-550 = 10a + 145d

10a = -550 – 145d

a = (ii)

From (i) and (ii), we get

=

-150 – 45d = -550 – 145d

-45d = 145d = -550 + 150

100d = -400

d = -4

Put the value of d in (i), we get

a =

= = = 3

a₁ = 3

a₂ = a + d = 3 + (-4) = -1

a₃ = a + 2d = 3 + 2(-4) = -5

Hence, A.P. is 3, -2, -5,…

And a = 3, d = -4