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
