Question

The fourth term of an arithmetic sequence is 20, and the 13th term is 65. Find the sum of the first 13 terms

Answer 1

S₁₃ = 455

Explanation:

the sum to n terms of an arithmetic is

`S_(n)` =
`(n)/(2)` [ 2a₁ + (n - 1)d ]

where a₁ is the first term and d the common difference

We require to find both a₁ and d

The nth term of an arithmetic sequence is

`a_(n)` = a₁ + (n - 1)d

given a₄ = 20 and a₁₃ = 65 , then

a₁ + 3d = 20 → (1)

a₁ + 12d = 65 → (2)

subtract (1) from (2) term by term to eliminate a₁

9d = 45 ( divide both sides by 9 )

d = 5

substitute d = 5 into (1) and solve for a₁

a₁ + 3(5) = 20

a₁ + 15 = 20 ( subtract 15 from both sides )

a₁ = 5

Then

S₁₃ =
`(13)/(2)` [ (2 × 5 ) + (12 × 5 ) }

= 6.5 (10 + 60)

= 6.5 × 70

= 455