Question

5 An arithmetic progression has a second term of -14 and a sum to 21 terms of 84. Find the first term and the 21st term of this progression.​

Answer 1

a₁ = - 16 , a₂₁ = 24

Explanation:

The nth term of an AP is

`a_(n)` = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Given a₂ = - 14 , then

a₁ + d = - 14 → (1)

The sum to n terms of an AP is

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

Given S₂₁ = 84 , then

`(21)/(2)` [ 2a₁ + (n - 1)d ] = 84 ( multiply both sides by 2 )

21(2a₁ + 20d) = 168 ( divide both sides by 21 )

2a₁ + 20d = 8 → (2)

From (1) d = - 14 - a₁ ← substitute into (2)

2a₁ + 20(- 14 - a₁) = 8 , that is

2a₁ - 280 - 20a₁ = 8

- 18a₁ - 280 = 8 ( add 280 to both sides )

- 18a₁ = 288 ( divide both sides by -18 )

a₁ = - 16

Substitute a₁ = - 16 into (1)

- 16 + d = - 14 ( add 16 to both sides )

d = 2

Then

a₂₁ = - 16 + 20(2) = - 16 + 40 = 24