Question

If Sn=2n²+3n, then d = ......,select a proper option (a), (b), (c) or (d) from given options so that the statement becomes correct.(All the problems refer to A.P.)
(a) 13
(b) 4
(c) 9
(d) -2

Answer 1

Given, $\bf{S_n=2n^2+3n}$
but we know according to formula of sum of n terms in AP is given as $\bf{S_n=\frac{n}{2}[2a+(n-1)d]}$
where a is the first term and d is the common difference of given AP.
Let's resolve Sn into standard format.

Sn = 2n² + 3n
= n[ 2n + 3 ]
= n/2[ 4n + 6 ]
= n/2 [ 4n - 4 + 4 + 6]
= n/2[ (6 + 4) + 4n - 4]
= n/2[10 + 4(n - 1)]
= n/2[2 × 5 + 4(n - 1)] .......(i)
now compare eq. (i) to formula,
we get, a = 5 and d = 4

therefore, option (b) is correct.

Answer 2

$S _{n} \: = \: 2n {}^{2} + \: 3n \\ \\ S _{1} = \: 2( {1}^{2} ) + 3(1) = 5 = a_{1} \\ S _{2} = \: 2( {2}^{2} ) + 3(2) = 14 = a_{1} + a_{2} \\ \\ a_{2 } = 9 \\ \\ \large \boxed { d = a_{2} - a_{1} = 4}$