Question

The sums of first n terms of three

a.Ps' are s1, s2 and s3. The first term of each is 5 and their common differences are 2,4 and 6 respectively. Prove that s1 s3

Answer 1

Step-by-step explanation:

1st series

5, 7, 9 .........

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

$= \frac{n}{2} (2 \times 5 + (n - 1)2) \\ = \frac{n}{2} (10 + 2n - 2) \\ = \frac{n}{2} (2n + 8) \\ = \frac{n}{2} \times 2(n + 4) \\ = {n}^{2} + 4n$

2nd series

5, 9, 13.........

sum of n terms

$= \frac{n}{2} (2 \times 5 + (n - 1)4) \\ = \frac{n}{2}( 10 + 4n - 4) \\ = \frac{n}{2} (4n + 6) \\ = \frac{n}{2} \times 2(2n + 3) \\ = {2n}^{2} + 3n$

3rd series

5, 11, 17.......

$= \frac{n}{2} (2 \times 5 + (n - 1)6) \\ = \frac{n}{2} (10 + 6n - 6) \\ = \frac{n}{2} (6n + 4) \\ = \frac{n}{2} \times 2(3n + 2) \\ = 3 {n}^{2} + 2n$

prove that

(S1 + S3)/2 = S2

$\frac{ {n}^{2} + 4n + 3 {n}^{2} + 2n }{2} \\ = \frac{ {4n}^{2} + 6n }{2} \\ = \frac{2(2 {n}^{2} + 3n) }{2} \\ = 2 {n}^{2} + 3n \: \: sum \: of \: n \: terms \: of \: 2nd \: series$

hence proved

hope you get your answer