Question

The 4th term of an AP is 9 and the 9th term of the AP is 34. The sum of the first 10 terms of the AP is .

• A. 156
• B. 165
• C. 561
• D. 167

Answer 1

The correct option is B 165

Given, the 4th term of an AP is 9 and the 9th term of the AP is 34.
The $n^{th}$ term of an AP with the first term a and common difference d is given by
$t_n=a+(n-1)d$
$\Rightarrow t_4=a+(4-1)d=9\Rightarrow a+3d=9$.......(i)
$\Rightarrow t_9=a+(9-1)d=34\Rightarrow a+8d=34$......(ii)
Solving the above equations:
Subtraction equation(i) from equation (ii), we get
$a+8d-a-3d=34-9$
$\Rightarrow 5d=25$
$\therefore d=5$
Substitute $d=5$ in equation(i), we get $a+3\left(5\right)=9$
$\Rightarrow a+15=9$
$\therefore a=9-15=-6$

The sum of n terms of an AP with first term $a$ and common difference $d$ is given by $S_n=\frac{n}{2}(2a+(n-1\left)d\right)$

Hence, the sum of 10 terms $=S_{10}=\frac{10}{2}\left(2\right(-6)+(10-1\left)5\right)$
$\Rightarrow S_{10}=5(-12+45)=5\times 33=165.$