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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
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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