Question

The 9th term of an arithmetic series is 5/2

Answer 1

Answer:

- 8050

Step-by-step explanation:

The n th term of an arithmetic sequence is

$a_{n}$ = a + (n - 1)d

where a is the first term and d the common difference.

We require to find both a and d

Given the 9 th term is 2.5 , then

a + 8d = 2.5 → (1)

Given the sum of the second and fifth term is 27, then

a + d + a + 4d = 27, that is

2a + 5d = 27 → (2)

Multiply (1) by - 2 and add to (2) to eliminate a

- 2a - 16d = - 5 → (3)

Add (2) and (3) term by term

- 11d = 22 ( divide both sides by - 11 )

d = - 2

Substitute d = - 2 into (1) and solve for a

a - 16 = 2.5 ( add 16 to both sides )

a = 18.5

The sum to n terms of an arithmetic sequence is

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

$S_{100}$ = 50 [ (2 × 18.5) + (99 × - 2) ] = 50(37 - 198) = 50(- 161) = - 8050