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

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Answer 2

Answer:

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

Step-by-step-explanation:

We have given the information about an AP.

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

Now,

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

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

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

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

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

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

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

Now, we know that,

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

Now, from the given condition,

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

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

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

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

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

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

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

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

Now, we know that,

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

Now,

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Now, we know that,

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

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

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

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

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

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

$\displaystyle{\implies\sf\:S_{20}\:=\:10\:\times\:59\:\times\:2}$

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

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

∴ The sum of the first 20 terms of the AP is 1180.