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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

Sagot :

Answer:

S₁₃ = 455

Step-by-step explanation:

the sum to n terms of an arithmetic is

[tex]S_{n}[/tex] = [tex]\frac{n}{2}[/tex] [ 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

[tex]a_{n}[/tex] = 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 = [tex]\frac{13}{2}[/tex] [ (2 × 5 ) + (12 × 5 ) }

     = 6.5 (10 + 60)

     = 6.5 × 70

     = 455

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