Question

A) If the 5th term of an A.P. is double the 7th term, show that the sum of the 17 terms zero.​

Answer 1

Answer:

see explanation

Step-by-step 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 a₅ is double a₇ , then

a₁ + 4d = 2(a₁ + 6d) , that is

a₁ + 4d = 2a₁ + 12d ( subtract a₁ from both sides )

4d = a₁ + 12d ( subtract 12d from both sides )

- 8d = a₁

The sum of n terms of an AP is

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

$S_{17}$ = $\frac{17}{2}$ ( 2(- 8d) + 16d)

     = 8.5(- 16d + 16d)

     = 8.5 × 0

     = 0