Question

Can someone help me answer this?

The first and third terms of an arithmetic series are 21 and 27 respectively.

a) Find the common difference of the series.

b) Find the sum of the first 40 terms of the series.​

Answer 1

a. Common differences (d) = 3

b. S40 = 3,180

Explanation:

Sn = a + (n - 1) d

S1 = 21 + (1 - 1) d

S1 = 21 + 0

S1 = 21......(1)

S3 = 21 + ( 3- 1) d

S3 = 21 + (2) d

S3 = 21 + 2d..,. (2)

.: d = 3

So that 21 + 2(3) = 21 + 6 = 27

For the first 40 terms

Sn = n/2 ( 2a + [n - 1] d])

S40 = 40/2 ( 2 × 21 + ( 40- 1) 3

S40 = 20 ( 42 + (39) 3]

S40 = 20 ( 42+ 117)

S40 =20 × 159

S40 = 3,180