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

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Solution

Sum of first $n$ terms, $S_{n} = 3 n^{2} + 2 n$

$r$ th term is $a_{r} = S_{r} - S_{r - 1}$

$= 3 r^{2} + 2 r - (3 (r - 1)^{2} + 2 (r - 1))$

$= 3 r^{2} + 2 r - 3 (r - 1)^{2} - 2 (r - 1)$

$= 3 r^{2} + 2 r - 3 r^{2} - 3 + 6 r - 2 r + 2$

$= 6 r - 1$

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Find the rth term of an AP, the sum of whose first n terms is (3n2+2n)

Sum of first n terms, Sn=3n2+2nrth term is ar=Sr−Sr−1=3r2+2r−(3(r−1)2+2(r−1))=3r2+2r−3(r−1)2−2(r−1)=3r2+2r−3r2−3+6r−2r+2=6r−1

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

Sum of first $n$ terms, $S_{n} = 3 n^{2} + 2 n$

$r$ th term is $a_{r} = S_{r} - S_{r - 1}$

$= 3 r^{2} + 2 r - (3 (r - 1)^{2} + 2 (r - 1))$

$= 3 r^{2} + 2 r - 3 (r - 1)^{2} - 2 (r - 1)$

$= 3 r^{2} + 2 r - 3 r^{2} - 3 + 6 r - 2 r + 2$

$= 6 r - 1$