Question

if the sum of first n term of a n AP is 4n²-n, find the 12th term.
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Answer 1

Answer:

I think the question will be the n term of an AP is 4n²-n, find sum of 12 th term.

please check it once

Answer 2

${\bold{\huge{\underline{\blue{\sf{Given:}}}}}}$

${\bold{\sf{The\:sum\:of\:first\:'n'\:terms=4n²-n}}}$

${\bold{\huge{\underline{\pink{\sf{To \: Find:}}}}}}$

${\bold{\sf{The\:{12}^{th}\:term\:of\:AP}}}$

${\bold{\huge{\underline{\red{\sf{Formula:}}}}}}$

${\boxed{\sf{Sum\:of\:'n'\:terms= \frac{n}{2} (2a+(n-1)d)}}}$

${\boxed{\sf{{n}^{th} \: term\:(t_n) = a+(n-1)d}}}$

${\bold{\sf{Here,\: d\: is \: common\: difference}}}$

${\bold{\huge{\underline{\green{\sf{Solution:}}}}}}$

${\bold{\sf{\frac{n}{2} (2a+(n-1)d)=4n²-n}}}$

${\bold{\sf{\frac{1}{2} (2a+(n-1)d)=4n-1}}}$

${\bold{\sf{2a+(n-1)d=8n-2}}}$

${\bold{\sf{2a=8n-2-nd+d}}}$

________________________________

${\bold{\sf{t_n = a+(n-1)d}}}$

${\bold{\sf{t_{12} = \frac{8n-2-nd+d}{2} +(n-1)d}}}$

${\bold{\sf{t_{12} = \frac{8n-2-nd+d+2nd-2d}{2}}}}$

${\bold{\sf{t_{12} = \frac{8n-2+nd-d}{2}}}}$

${\bold{\sf{t_{12} = \frac{8(12)-2+(12)d-d}{2}}}}$

${\bold{\sf{t_{12} = \frac{94+11d}{2}}}}$