Find the $n^{t h}$ term if sum of first n terms of an AP is $(3 x^{2} + n)$

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Solution

$S_{n} = \frac{n}{2} [2 a + (n - 1) d] = (3 n^{2} + n)$

$\Rightarrow \frac{n}{2} [2 a + (n - 1) d] = n (3 n + 1)$

$\Rightarrow a + (\frac{n - 1}{2}) d = 3 n + 1$

$\Rightarrow (\frac{n - 1}{2}) d = 3 n + 1 - a$

$\Rightarrow (n - 1) d = (3 n - a + 1) 2$

$\Rightarrow (n - 1) = \frac{6 n - 2 a + 2}{d}$ ……..(i)

$T_{n} = a + (n - 1) d$

Put equation (i)

$T_{n} = a + (\frac{6 n - 2 a + 2}{d}) d$

$= a + 6 n - 2 a + 2$

$T_{n} = 6 n - a + 2$ .

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Find the nth term if sum of first n terms of an AP is (3x2+n)

Sn=n2[2a+(n−1)d]=(3n2+n)⇒n2[2a+(n−1)d]=n(3n+1)⇒a+(n−12)d=3n+1⇒(n−12)d=3n+1−a⇒(n−1)d=(3n−a+1)2⇒(n−1)=6n−2a+2d ……..(i)Tn=a+(n−1)dPut equation (i)Tn=a+(6n−2a+2d)d=a+6n−2a+2Tn=6n−a+2.

Find the $n^{t h}$ term if sum of first n terms of an AP is $(3 x^{2} + n)$