Question

if the eleventh term of an arithmetic progression is 24 find s21​

Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:S_{21}\:=\:504}}}$

Step-by-step-explanation:

We have given that, $\displaystyle{\sf\:t_{11}\:=\:24}$ for a arithmetic progression.

We have to find $\displaystyle{\sf\:S_{21}}$

Now, we know that,

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

$\displaystyle{\implies\sf\:t_{11}\:=\:a\:+\:(\:11\:-\:1\:)\:\times\:d}$

$\displaystyle{\implies\sf\:24\:=\:a\:+\:10\:d}$

$\displaystyle{\implies\sf\:a\:+\:10\:d\:=\:24\:\:\:-\:-\:-\:(\:1\:)}$

Now, we know that,

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

$\displaystyle{\implies\sf\:S_{21}\:=\:\dfrac{21}{2}\:[\:2a\:+\:(\:21\:-\:1\:)\:\times\:d\:]}$

$\displaystyle{\implies\sf\:S_{21}\:=\:\dfrac{21}{2}\:[\:2a\:+\:20\:d\:]}$

$\displaystyle{\implies\sf\:S_{21}\:=\:\dfrac{21}{2}\:[\:2\:\times\:(\:a\:+\:10\:d\:)\:]}$

$\displaystyle{\implies\sf\:S_{21}\:=\:\dfrac{21}{2}\:(\:2\:\times\:24\:)\:\:\:-\:-\:-\:[\:From\:(\:1\:)\:]}$

$\displaystyle{\implies\sf\:S_{21}\:=\:\dfrac{21}{\cancel{2}}\:\times\:\cancel{2}\:\times\:24}$

$\displaystyle{\implies\sf\:S_{21}\:=\:21\:\times\:24}$

$\displaystyle{\implies\underline{\boxed{\red{\sf\:S_{21}\:=\:504}}}}$

∴ The sum of the first 21 terms of the arithmetic progression is 504.

Answer 2

Answer:

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