Question

Find the sum of the first $5$ terms of an arithmetic progression in which the $6^{th}$ term is $14$ and the ${11}^{th}$ term is $22$.

• A. $2.2$
• B. $6.0$
• C. $12.4$
• D. $32.6$
• E. $46.0$

Answer 1

The correct option is E $46.0$

Let '$a$' is the first term and '$d$' is the common difference between the given AP

According to the question $6^{th}$ term is $14$

$\Rightarrow a+(6-1)d=14$

$\Rightarrow a+5d=14$......eq (1)

and ${11}^{th}$ term is $22$.

$\Rightarrow a+(11-1)d=22$

$\Rightarrow a+10d=22$.....eq (2)

Subtract eq (1) from eq (2), we get

$\Rightarrow -5d=-8$

$\Rightarrow d=\frac{8}{5}$

Substitute the value of $d$ in eq (1)

$\Rightarrow a+5\times \frac{8}{5}=14$

$\Rightarrow a=14-8$

$\Rightarrow a=6$

Sum of first $n$ terms of AP $=$ $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$\therefore S_5=\frac{5}{2}[2\times 6+(5-1\left)\frac{8}{5}\right]$

$\Rightarrow S_5=\frac{5}{2}[2\times 6+4\times \frac{8}{5}]$

$\Rightarrow S_5=\frac{5}{2}[12+\frac{32}{5}]$

$\Rightarrow S_5=\frac{5}{2}\left[\frac{60+32}{5}\right]$

$\Rightarrow S_5=\frac{5}{2}\times \frac{92}{5}=46$