The sum of the first five terms of an AP

Solution:

Let the first term, common difference and the number of terms of an AP are a, d and n, respectively.

$\because$ Sum of first $n$ terms of an AP, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $\ldots$ (i)

$\therefore$ Sum of first five terms of an AP, $S_{5}=\frac{5}{2}[2 a+(5-1) d]$ [from Eq. (i)]

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

$\Rightarrow \quad S_{5}=5 a+10 d$ ....(ii)

and sum of first seven terms of an AP, $S_{7}=\frac{7}{2}[2 a+(7-1) d]$

$=\frac{7}{2}[2 a+6 d]=7(a+3 d)$

$\Rightarrow$ $S_{7}=7 a+21 d$ ....(iii)

Now, by given condition,

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

$\Rightarrow \quad 5 a+10 d+7 a+21 d=167$

$\Rightarrow \quad 12 a+31 d=167$ ....(iv)

Given that, sum of first ten terms of this AP is 235 .

$\therefore \quad S_{10}=235$

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

$\Rightarrow \quad 5(2 a+9 d)=235$

$\Rightarrow \quad 2 a+9 d=47$ .....(v)

On multiplying Eq. (v) by 6 and then subtracting it into Eq. (vi), we get

$\Rightarrow \quad d=5$

Now, put the value of $d$ in Eq. (v), we get

$2 a+9(5)=47 \Rightarrow 2 a+45=47$

$\Rightarrow$ $2 a=47-45=2 \Rightarrow a=1$

Sum of first twenty terms of this AP, $S_{20}=\frac{20}{2}[2 a+(20-1) d]$

$=10[2 \times(1)+19 \times(5)]=10(2+95)$

 

$=10 \times 97=970$

Hence, the required sum of its first twenty terms is 970 .