If $S_{n} = (3 n^{2} + 2 n),$ find the first term and the common difference of the AP.

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Solution

Given:
S_n = (3n² + 2n) ...(i)
Replacing n by (n − 1) in (i), we get:
S_n-1 = 3(n − 1)² + 2(n − 1)
= 3(n² − 2n + 1) + 2n − 2
= 3n² − 4n + 1
∴ T_n = ( S_n − S_{n -1})
= (3n² + 2n) − (3n² − 4n + 1) = 6n − 1
∴ n^th term, T_n = (6n − 1) ...(ii)

Putting n = 1 in (ii), we get:
T₁ = (6 ⨯ 1) − 1 = 5
Putting n = 2 in (ii), we get:
T₂ = (6 ⨯ 2) − 1 = 11
∴ Common difference, d = T₂ − T₁ = 11 − 5 = 6
Thus, a = 5 and d = 6

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