In an AP, if $S_{5} +\ S_{7} \ =\ 167$ and $S_{10} \ =\ 235$, then find the A.P., where $S_{n}$ denotes the sum of its first $n$ terms

Academic Mathematics NCERT Class 10

Given: $S_{5}+ S_{7} = 167$ and $S_{10} = 235$, in an A.P. Where $S_{n}$ denotes the sum of its first n terms.

To do: To find the A.P.

Solution:

As known if a is the first term and d is the common difference of an A.P.

Then sum of n terms, $S_{n}=\frac{n}{2}\left[ 2a+\left( n-1\right) d\right]$

$\therefore S_{5}=\frac{5}{2}\left[ 2a+\left( 5-1\right) d\right]$

$S_{5}=\frac{5}{2}\left( 2a+4d\right) .............( 1)$

And $S_{7}=\frac{7}{2}\left[ 2a+\left( 7-1\right) d\right]$

$S_{7}=\frac{7}{2}\left( 2a+6d\right) \ ..............( 2)$

As given ,

$S_{5}+ S_{7}=167$

$\Rightarrow \frac{5}{2}\left( 2a+4d\right) +\frac{7}{2}\left( 2a+6d\right)=167$

$\Rightarrow \frac{5\left( 2a+4d\right) +7\left( 2a+6d\right)}{2}=167$

$\Rightarrow \frac{10a+20d+14a+42d}{2}=167$

$\Rightarrow \frac{24a+62d}{2}=167$

$\Rightarrow 24a+62d=167\times2$

$\Rightarrow 24a+62d=334 \ \ .................( 3)$

And $S_{10}=\frac{10}{2}\left[ 2a+\left( 10-1\right) d\right]$

$\Rightarrow 235=\frac{10}{2}\left[ 2a+\left( 10-1\right) d\right]$

$\Rightarrow 2a+9d=47\ \ \ \ \ \ \ \ \ \ \ \ \ .................\left( 4\right)$

Multiplying equation $( 4)$ by 12, we get

$24a+108d=\ \ \ 564\ \ \ \ \ \ \ .................\left( 5\right)$

Subtracting $( 3)$ from $( 5)$, we get,

$24a+108d-24a-62d=564-33$

$\Rightarrow 46d=230$

$\Rightarrow d=\frac{230}{46} =5$

Substituting value of $d$ in $( 4)$, we have,

$2a+9\times 5=47$

$\Rightarrow 2a+45=47$

$\Rightarrow 2a=47-45=2$

$\Rightarrow 2a=2$

$\Rightarrow a=1$

Thus, The given A.P. is 1, 6, 11, 16,..........

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Updated on: 10-Oct-2022

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