The sum of n terms of a series is $3 n^{2} + 4 n$ . Show that the series is an A.P. and find the first term and common difference. What will be its nth term?

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Solution

$T_{n} = S_{n} - S_{n - 1}$ . $(1)$
Now $S_{n} = 3 n^{2} + 4 n$
$∴ S_{n - 1} = 3 (n - 1)^{2} + 4 (n - 1) = 3 n^{2} - 2 n - 1$
$∴ T_{n} = (3 n^{2} - 4 n) - (3 n^{2} - 2 n - 1)$ , by $(1)$
$= 6 n + 1$
$∴ T_{1} = 7, T_{2} = 13, T_{3} = 19$ ,______
$∴ a = 7, d = 6$ .

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The sum of n terms of a series is 3n2+4n. Show that the series is an A.P. and find the first term and common difference. What will be its nth term?

Tn=Sn−Sn−1 .(1)Now Sn=3n2+4n∴Sn−1=3(n−1)2+4(n−1)=3n2−2n−1∴Tn=(3n2−4n)−(3n2−2n−1), by (1)=6n+1∴T1=7,T2=13,T3=19,______∴a=7,d=6.

The sum of n terms of a series is $3 n^{2} + 4 n$ . Show that the series is an A.P. and find the first term and common difference. What will be its nth term?

$T_{n} = S_{n} - S_{n - 1}$ . $(1)$
Now $S_{n} = 3 n^{2} + 4 n$
$∴ S_{n - 1} = 3 (n - 1)^{2} + 4 (n - 1) = 3 n^{2} - 2 n - 1$
$∴ T_{n} = (3 n^{2} - 4 n) - (3 n^{2} - 2 n - 1)$ , by $(1)$
$= 6 n + 1$
$∴ T_{1} = 7, T_{2} = 13, T_{3} = 19$ ,______
$∴ a = 7, d = 6$ .