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# The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. If the sum of the first ten terms of this AP is 235 , find the sum of its first twenty terms.

Given:

The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167.

The sum of the first ten terms of this AP is 235.

To do:

We have to find the sum of its first $20$ terms.

Solution:

Let the first term be $a$ and the common differnce be $d$.

We know that,

Sum of $n$ terms$ S_{n} =\frac{n}{2}(2a+(n-1)d)$

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

$=\frac{5}{2}(2a+4d)$

$=5a+10d$

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

$=\frac{7}{2}(2a+6d)$

$=7a+21d$

According to the question,

$S_5+S_7=(5a+10d)+(7a+21d)$

$167=12a+31d$........(i)

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

$235=5(2a+9d)$

$47=2a+9d$.......(ii)

Multiplying (ii) by 6 and subtracting it from (i), we get,

$(12a+31d)-6(2a+9d)=167-6(47)$

$12a-12a+31d-54d=167-262$

$-23d=-115$

$d=\frac{-115}{-23}$

$d=5$

This implies,

$2a=47-9(5)$ [From (ii)]

$2a=47-45$

$a=\frac{2}{2}$

$a=1$

The sum of $20$ terms $S_{20}=\frac{20}{2}[2(1)+(20-1)5]$

$=10[2+19(5)]$

$=10(2+95)$

$=10(97)$

$=970$

Hence, the sum of $20$ terms is $970$.

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