If the sum of n terms of an AP is given by Sn = (2n2 + 3n), then find

1.	If the sum of n terms of an AP is given by Sn = (2n2 + 3n), then find its common difference.
Answer» Given: S_n = (2n² + 3n) To find: find common difference Put n = 1 we get S₁ = 5 OR we can write a = 5 …equation 1 Similarly put n = 2 we get S₂ = 14 OR we can write 2a + d = 14 Using equation 1 we get d = 4 We are given S_n=(2n²+3n) ------(1) To find the common difference 'd', d=a₂-a₁ ------(2) First, to find the second term of the A.P., a₂=S₂- S₁ --------(3) Now, To find S₁, put n=1 in equation 1, we get, S₁=(2.1²+3.1)=2+3=5 -----(4) Here S₁=a₁ Similarly for S₂,put n=1 in equation 1, we get, S₂=(2.2²+3.2)=8+6=14 ------(5) Here S₂=a₂+a₁ Using equation (3),(4) and (5) we get, a₂=14-5=9 Now we know a₂=9 and a₁=5 ∵ a₁=S₁ From equation (2), d=9-5=4 Hence, the common difference for the A.P. is 4.

If the sum of n terms of an AP is given by Sn = (2n2 + 3n), then find its common difference.

Answer»

Given: S_n = (2n² + 3n)

To find: find common difference

Put n = 1 we get

S₁ = 5 OR we can write

a = 5 …equation 1

Similarly put n = 2 we get

S₂ = 14 OR we can write

2a + d = 14

Using equation 1 we get

d = 4

We are given S_n=(2n²+3n) ------(1)

To find the common difference 'd', d=a₂-a₁ ------(2)

First, to find the second term of the A.P., a₂=S₂- S₁ --------(3)

Now, To find S₁, put n=1 in equation 1, we get,

S₁=(2.1²+3.1)=2+3=5 -----(4)

Similarly for S₂,put n=1 in equation 1, we get,

S₂=(2.2²+3.2)=8+6=14 ------(5)

Using equation (3),(4) and (5) we get,

a₂=14-5=9

Now we know a₂=9 and a₁=5 ∵ a₁=S₁

From equation (2),

d=9-5=4

Hence, the common difference for the A.P. is 4.