Question

In an AP the sum of first 9 term is 14 more then 5 time the 8th term 8th and 2rd term are in the ratio 11:2 find the sums of first 20 terms of the A.P​

Answer 1

Answer:

Answer. Answer: The sum of the first 20 terms of the AP is 1180.

Step-by-step explanation:

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Answer 2

We have given the information about an AP.

We have to find the sum of first 20 terms of the AP.

Now,

$\displaystyle{\sf\:t_8\::t_2\:=\:11\::2\:\:\quad\:\:\:-\:[\:Given\:]}$

$\displaystyle{\red\implies\sf\orange{\:\dfrac{t_8}{t_2}\:=\:\dfrac{11}{2}}}$

$\displaystyle{\blue\implies\sf\green{\:t_8\:=\:\dfrac{11}{2}\:\times\:t_2}}$

$\displaystyle{\pink\implies\sf\purple{\:t_8\:=\:\dfrac{11\:t_2}{2}}}$

$\displaystyle{\orange\implies\sf\red{\:t_8\:=\:\dfrac{11\:\times\:[\:a\:+\:(\:2\:-\:1\:)\:d\:]}{2}}}$

$\displaystyle{\green\implies\sf\blue{\:t_8\:=\:\dfrac{11\:\times\:(\:a\:+\:d\:)}{2}}}$

$\displaystyle{\purple\implies\sf\pink{\:t_8\:=\:\dfrac{11a\:+\:11d}{2}\:\quad\:\:\:-\:-\:-\:(\:1\:)}}$

Now,

$\displaystyle{{\blue\sf\:S_n\:=\:\dfrac{n}{2}\:[\:2a\:+\:(\:n\:-\:1\:)\:d\:]}\sf\:\quad\:-\:\:[\:Formula\:]}$

Now, from the given condition,

$\displaystyle{\sf\:S_9\:=\:5\:\times\:t_8\:+\:14}$

$\displaystyle{\blue\implies\sf\green{\:\dfrac{9}{2}\:[\:2a\:+\:(\:9\:-\:1\:)\:d\:]\:=\:5\:\times\:\left[\:\dfrac{11a\:+\:11d}{2}\:\right]\:+\:14\:\quad\:\:\:[\:From\:(\:1\:)\:]}}$

$\displaystyle{\red\implies\sf\orange{\:\dfrac{9}{2}\:(\:2a\:+\:8d\:)\:=\:\dfrac{55a\:+\:55d}{2}\:+\:14}}$

$\displaystyle{\pink\implies\sf\purple{\:\dfrac{18a\:+\:72d}{\cancel{2}}\:=\:\dfrac{55a\:+\:55d\:+\:28}{\cancel{2}}}}$

$\displaystyle{\green\implies\sf\blue{\:18a\:+\:72d\:=\:55a\:+\:55d\:+\:28}}$

$\displaystyle{\orange\implies\sf\red{\:18a\:+\:72d\:-\:55a\:-\:55d\:=\:28}}$

$\displaystyle{\purple\implies\sf\pink{\:18a\:-\:55a\:+\:72d\:-\:55d\:=\:28}}$

$\displaystyle{\blue\implies\sf\green{\:-\:37a\:+\:17d\:=\:28\:\quad\:\:\:-\:-\:-\:(\:2\:)}}$

Now, we know that,

$\displaystyle{\red{\sf\:t_n\:=\:a\:+\:(\:n\:-\:1\:)\:\times\:d}\sf\:\quad\:\:-\:[\:Formula\:]}$

Now,

$\displaystyle{\sf\:t_8\::t_2\:=\:11\::2\:\:\quad\:\:\:-\:-\:-\:[\:Given\:]}$

$\displaystyle{\red\implies\sf\orange{\:\dfrac{t_8}{t_2}\:=\:\dfrac{11}{2}}}$

$\displaystyle{\pink\implies\sf\purple{\:\dfrac{a\:+\:(\:8\:-\:1\:)\:d}{a\:+\:(\:2\:-\:1\:)\:d}\:=\:\dfrac{11}{2}}}$

$\displaystyle{\green\implies\sf\blue{\:\dfrac{a\:+\:7d}{a\:+\:d}\:=\:\dfrac{11}{2}}}$

$\displaystyle{\orange\implies\sf\red{\:2\:(\:a\:+\:7d\:)\:=\:11\:(\:a\:+\:d\:)}}$

$\displaystyle{\purple\implies\sf\pink{\:2a\:+\:14d\:=\:11a\:+\:11d}}$

$\displaystyle{\blue\implies\sf\green{\:14d\:-\:11d\:=\:11a\:-\:2a}}$

$\displaystyle{\red\implies\sf\orange{\:3d\:=\:9a}}$

$\displaystyle{\pink\implies\sf\purple{\:d\:=\:\dfrac{\cancel{9}\:a}{\cancel{3}}}}$

$\displaystyle{\green\implies\boxed{\sf\:d\:=\:3a}\sf\:\quad\:\:-\:(\:3\:)}$

Now, by substituting d = 3a in equation ( 2 ), we get,

$\displaystyle{\sf\:-\:37a\:+\:17d\:=\:28\:\quad\:\:\:-(\:2\:)}$

$\displaystyle{\green\implies\sf\blue{\:-\:37a\:+\:17\:\times\:3a\:=\:28}}$

$\displaystyle{\orange\implies\sf\red{\:-\:37a\:+\:51a\:=\:28}}$

$\displaystyle{\purple\implies\sf\pink{\:14a\:=\:28}}$

$\displaystyle{\blue\implies\sf\green{\:a\:=\:\cancel{\dfrac{28}{14}}}}$

$\displaystyle{\red\implies\boxed{\green{\sf\:a\:=\:2}}}$

Now, we know that,

$\displaystyle{\blue{\sf\:S_n\:=\:\dfrac{n}{2}\:[\:2a\:+\:(\:n\:-\:1\:)\:d\:]}\sf\:\quad\:\:-\:[\:Formula\:]}$

$\displaystyle{\red\implies\sf\orange\:S_{20}\:=\:\dfrac{20}{2}\:[\:2a\:+\:(\:20\:-\:1\:)\:d\:]}$

$\displaystyle{\pink\implies\sf\purple{\:S_{20}\:=\:10\:(\:2a\:+\:19d\:)}}$

$\displaystyle{\green\implies\sf\blue{\:S_{20}\:=\:10\:(\:2a\:+\:19\:\times\:3a\:)\:\quad\:-[\:From\:(\:3\:)\:]}}$

$\displaystyle{\orange\implies\sf\red{\:S_{20}\:=\:10\:\times\:(\:2a\:+\:57a\:)}}$

$\displaystyle{\blue\implies\sf\green{\:S_{20}\:=\:10\:\times\:59a}}$

$\displaystyle{\red\implies\sf\orange{\:S_{20}\:=\:590\:\times\:2}}$

$\displaystyle{\pink\implies\underline{\boxed{\purple{\sf\:S_{20}\:=\:1180}}}}$

$\orange\therefore\sf\red{The\: sum \:of\: the \: first\:20\:terms}$

$\sf \red{of \: the \:AP\: is\: 1180.}$