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

1.	In an A.P. the sum of first ten terms is -150 and the sum of its next 10 term is -550. Find the A.P.
Answer» Let’s take a to be the first term and d to be the common difference. And we know that, sum of first n terms S_n = \(\frac{n}{2}\)(2a + (n − 1)d) Given that sum of the first 10 terms of an A.P. is -150. S₁₀ = -150 And the sum of next 10 terms is -550. So, the sum of first 20 terms = Sum of first 10 terms + sum of next 10 terms S₂₀= -150 + -550 = -700 Now, having S₁₀ = \(\frac{10}{2}\)(2a + (10 − 1)d) ⟹ -150 = 5(2a + 9d) ⟹ -30 = 2a + 9d ⟹ 2a + 9d = -30 . . . . (1) And, S₂₀ = \(\frac{20}{2}\)(2a + (20 − 1)d) ⟹ -700 = 10(2a + 19d) ⟹ -70 = 2a + 19d …. (2) Now, subtracting (1) from (2), we get ⟹ 19d – 9d = -70 – (-30) ⟹ 10d = -40 ⟹ d = -4 Using d in (1), we have 2a + 9(-4) = -30 2a = -30 + 36 a = \(\frac{6}{2}\) = 3 Hence, we have a = 3 and d = -4 So, the A.P is 3, -1, -5, -9, -13,…..

In an A.P. the sum of first ten terms is -150 and the sum of its next 10 term is -550. Find the A.P.

Answer»

Let’s take a to be the first term and d to be the common difference.

And we know that, sum of first n terms

S_n = \(\frac{n}{2}\)(2a + (n − 1)d)

Given that sum of the first 10 terms of an A.P. is -150.

S₁₀ = -150

And the sum of next 10 terms is -550.

So, the sum of first 20 terms = Sum of first 10 terms + sum of next 10 terms

S₂₀= -150 + -550 = -700

Now, having

S₁₀ = \(\frac{10}{2}\)(2a + (10 − 1)d)

⟹ -150 = 5(2a + 9d)

⟹ -30 = 2a + 9d

⟹ 2a + 9d = -30 . . . . (1)

And,

S₂₀ = \(\frac{20}{2}\)(2a + (20 − 1)d)

⟹ -700 = 10(2a + 19d)

⟹ -70 = 2a + 19d …. (2)

Now, subtracting (1) from (2), we get

⟹ 19d – 9d = -70 – (-30)

⟹ 10d = -40

⟹ d = -4

Using d in (1), we have

2a + 9(-4) = -30

2a = -30 + 36

a = \(\frac{6}{2}\) = 3

Hence, we have a = 3 and d = -4

So, the A.P is 3, -1, -5, -9, -13,…..