Question

the 3rd term of an A.P is 6 and the 16th term is 32 find the common difference and the sum of the first 20 term

Answer 1

2 and 420

Explanation:

The n th term of an AP is

`a_(n)` = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Given a₃ = 6 and a₁₆ = 32 , then

a₁ + 2d = 6 → (1)

a₁ + 15d = 32 → (2)

Subtract (1) from (2) term by term

13d = 26 ( divide both sides by 13 )

d = 2

Substitute d = 2 into (1) for corresponding value of a₁

a₁ + 2(2) = 6

a₁ + 4 = 6 ( subtract 4 from both sides )

a₁ = 2

The sum of n terms of an AP is

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

`S_(20)` =
`(20)/(2)` [(2 × 2) + (19 × 2) ]

= 10(4 + 38) = 10 × 42 = 420