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

User Oesor
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1 Answer

10 votes

Answer:

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

User Ngozi
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