Question

If the sum of first m terms of an ap is am2 + bm find its common difference

Answer 1

given:.Sm=am²+bm

let m=1
S1=a(1)²+b(1)
S1=a+b {S1=a1, a1=a+b}

let m=2
S2=a(2)²+b(2)
S2=4a+2b

S2=a1+a2
4a+2b=(a+b)+a2
a2=4a+2b-a-b
a2=3a+b

d=a2-a1
d=3a+b-a-b
d=2a

Answer 2

Answer:

2a+2b

Step-by-step explanation:

Given : $S_m =am^2+bm$

To find:Common difference

Solution:

$S_m =am^2+bm$

Substitute m = 1

$S_1 =a(1)^2+b(1)$

$S_1 =a+b$

Thus the first term is a+b

Now substitute m =2

$S_2 =a(2)^2+b(2)$

$S_2 =4a+2b$

Thus the sum of first two terms is 4a+2b

Second term = Sum of first two terms - first term

Second term = $4a+2b-a-b$

                      = $3a+b$

Common difference = second term - first term

                                  = 3a+b -(a+b)

                                  = 2a+2b

Hence the common difference is 2a+2b