the sum of the first n terms of an AP is given by s_{n}=3n^{2}-n , determine the AP and its 25^{th}term.

Dear Student!

Given, sum of first *n* terms of an AP, *S _{n} *= 3

*n*

^{2}–

*n*

*S _{n} *= 3

*n*

^{2}

^{ }–

*n*

Replacing *n* by *n* – 1, we get

Let the *n*^{th} term of AP be *a _{n}*.

$\mathrm{Now},\mathrm{put}\mathrm{n}=1\mathrm{in}{\mathrm{a}}_{\mathrm{n}},\mathrm{we}\mathrm{get}\mathrm{a}=6\times 1-4=2\phantom{\rule{0ex}{0ex}}\mathrm{put}\mathrm{n}=2\mathrm{in}{\mathrm{a}}_{\mathrm{n}},\mathrm{we}\mathrm{get}{\mathrm{a}}_{2}=6\times 2-4=8\phantom{\rule{0ex}{0ex}}\mathrm{put}\mathrm{n}=3\mathrm{in}{\mathrm{a}}_{\mathrm{n}},\mathrm{we}\mathrm{get}{\mathrm{a}}_{3}=6\times 3-4=14\phantom{\rule{0ex}{0ex}}\mathrm{and}\mathrm{so}\mathrm{on}.\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{required}\mathrm{AP}\mathrm{is},2,8,14,.....$

Putting *n* = 25, we get

*a*_{25} = 6 × 25 – 4 = 150 – 4 = 146

Thus, the 25^{th} term of the given A.P. is 146.

Cheers!

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