Question

The 14th term of an AP is twice its 9th term. If its 6th term is -8, then find the sum

of its first 20 terms​

Answer 1

Answer:-

Given:-

In an AP;

a₁₄ = 2 * a₉

We know that,

nth term of an AP (aₙ) a + (n - 1)d

⟹ a + (14 - 1)d = 2 * [ a + (9 - 1)d ]

⟹ a + 13d = 2a + 16d

⟹ 13d - 16d = 2a - a

⟹ - 3d = a -- equation(1).

Also given that;

⟹ a₆ = - 8

⟹ a + 5d = - 8

Substitute the value of a from equation (1).

⟹ - 3d + 5d = - 8

⟹ 2d = - 8

⟹ d = - 8/2

⟹ d = - 4

Substitute the value of d in equation (1).

⟹ a = - 3( - 4)

⟹ a = 12

Now,

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

So,

S₂₀ = 20/2 [ 2(12) + (20 - 1)( - 4) ]

⟹ S₂₀ = 10 [ 24 + 19( - 4) ]

⟹ S₂₀ = 10 (24 - 76)

⟹ S₂₀ = 10(- 52)

⟹ S₂₀ = - 520

∴ The sum of first 20 terms of the given AP is - 520.

Answer 2

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