Question

the sum of first 3 terms of an ap is 33 if the product of the first and the third term exceeds the second term by 29.find the ap

Answer 1

$\huge\purple{\underline{\underline{\pink{Ans}\red{wer:-}}}}$

$\sf{The \ required \ A.P. \ is \ 2,11,20,…}$

$\sf\orange{Given:}$

$\sf{\implies{S3=33}}$

$\sf{\implies{t1\times \ t3=t2+29}}$

$\sf\pink{To \ find:}$

$\sf{The \ A.P.}$

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

$\sf{Let \ three \ terms \ of \ A.P. \ be}$

$\sf{(a-d), \ a \ and \ (a+d)}$

$\sf{According \ to \ first \ condition}$

$\sf{(a-d)+a+(a+d)=33}$

$\sf{\therefore{3a=33}}$

$\sf{\therefore{a=\frac{33}{3}}}$

$\sf{\implies{a=11...(1)}}$

$\sf{According \ to \ second \ condition}$

$\sf{(a-d)(a+d)=a+29}$

$\sf{But, \ a=11 \ from \ eq(1)}$

$\sf{a^{2}-d^{2}=11+29}$

$\sf{11^{2}-d^{2}=40}$

$\sf{121-d^{2}=40}$

$\sf{d^{2}=121-40}$

$\sf{d^{2}=81}$

$\sf{On \ taking \ square \ root \ of \ both \ sides}$

$\sf{\implies{d=9}}$

$\sf{\therefore{t1=(a-d)=11-9=2}}$

$\sf{t2=a=11}$

$\sf{t3=(a+d)=11+9=20}$

$\sf\purple{\tt{\therefore{The \ required \ A.P. \ is \ 2,11,20,…}}}$