In an arithmetic series, the sum of the first 12 terms is equal to ten times the sum of the first 3 terms. If the first term is 5, find the common difference and the sum of the first 20 terms

Question

Please explain how you found the answer.

Answer 1

Hi Jessie:

Here's how to tackle sequence questions. We need to find the first term, a, and the common difference, d.

For an arithmetic sequence, the sum of the first n terms, S_n = (n/2)(2a + (n - 10)d)

" " " " 12 " S₁₂ = (12/2) (2a + (12 -1)d)

= 6(2a + 11d)

= 12a + 66d

" " " " 3 " S₃ = (3/2) (2a + (3 -1)d)

= 3a + 3d

We know that S₁₂ = 10 x S₃

∴ 12a + 66d = 10(3a + 3d)

12a + 66d = 30a + 30d

36d = 18a

=> 2d = a . . . . . . . . . . . . . . . . . eqn (1)

We also know that a = 5

Therefore, from eqn (1): 2d = 5

d = 5/2

So, S₂₀ = (20/2) (2 x 5 + (20 - 1)5/2)

= 10 (10 + 19 x 5/2)

= 100 + 19 x 5 x 5) [multiplying through by 10

= 100 + 475

= 575