Question

The sum of the first 20 term of an AP is equal to the sum of first 30 terms show that the sum of the first 50 term of an AP is zero

Answer 1

Explanation:

Sum = (n/2)(2a + (n - 1)d)

(20/2)(2a + 19d)= (30/2)(2a + 29d)

20a + 190d = 30a + 435d

10a = - 245d

a =( - 49/2) d

Sum of 50 first terms = 25(-49d +49d)= 0

Answer 2

Given:

Sum of first 20 terms of an AP = Sum of first 30 terms.

⟹ S₃₀ = S₂₀

Prove:

sum of the first 50 term of an AP is zero

Proof:

We know that,

Sum of first n terms of an AP (Sn) = n/2 * [ 2a + (n - 1)d

So,

★ S₃₀ = 30/2 * [ 2a + (30 - 1)d ]

⟹ S₃₀ = 15 [ 2a + 29d ]

⟹ S₃₀ = 30a + 435d

★ S₂₀ = 20/2 * [ 2a + (20 - 1)d ]

⟹ S₂₀ = 10 (2a + 19d)

⟹ S₂₀ = 20a + 190d

⟹ 30a + 435d = 20a + 190d

⟹ 30a - 20a = 190d - 435d

⟹ 10a = - 245d

⟹ a = - 245d/10 -- equation (1)

Now,

We have to prove:

S₅₀ = 0

★ S₅₀ = 50/2 * [ 2a + (50 - 1)d ]

substitute the value of a from equation (1).

⟹ 0 = 25 [ 2 (- 245d/10) + 49d ]

⟹ 0 = 25 [ - 49d + 49d ]

⟹ 0 = 25(0)

⟹ 0 = 0

Hence, Proved.

I hope it will help you.

Regards.