Question

The seventh term of an arithmetic progression is equal to twice the fifth term. The sum

Answer 1

Answer:

a₁ = - 24

Step-by-step 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 a₇ = 2a₅ , then

a₁ + 6d = 2(a₁ + 4d) = 2a₁ + 8d ( subtract 2a₁ + 8d from both sides )

- a₁ - 2d = 0 → (1)

The sum to n terms of an AP is

$S_{n}$ = $\frac{n}{2}$ [ 2a₁ + (n - 1)d ]

Given $S_{7}$ = 84 , then

$\frac{7}{2}$ (2a₁ + 6d) = 84

3.5(2a₁ + 6d) = 84 ( divide both sides by 3.5 )

2a₁ + 6d = 24 → (2)

Thus we have 2 equations

- a₁ - 2d = 0 → (1)

2a₁ + 6d = 24 → (2)

Multiplying (1) by 3 and adding to (2) will eliminate d

- 3a₁ - 6d = 0 → (3)

Add (2) and (3) term by term to eliminate d

- a₁ = 24 ( multiply both sides by - 1 )

a₁ = - 24