Question

Calculate the sum of the first 15 terms of an arithmetic progression is the one where a3= 1 and a7=7. (I need the procedure please)

Answer 1

S₁₅ = 127.5

Explanation:

the nth term of an arithmetic progression is

`a_(n)` = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

given a₃ = 1 and a₇ = 7 , then

a₁ + 2d = 1 → (1)

a₁ + 6d = 7 → (2)

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

0 + 4d = 6

4d = 6 ( divide both sides by 4 )

d = 1.5

substitute d = 1.5 into (1) and solve for a₁

a₁ + 2(1.5) = 1

a₁ + 3 = 1 ( subtract 3 from both sides )

a₁ = - 2

the sum to n terms of an arithmetic progression is

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

with a₁ = - 2 and d = 1.5 , then

S₁₅ =
`(15)/(2)` [ (2 × - 2) + (14 × 1.5) ]

= 7.5(- 4 + 21)

= 7.5 × 17

= 127.5