Question

If the sum of ${12}^{th}$ and ${22}^{th}$ terms of an AP is $100$, then the sum of the first $33$ terms of an AP is

• A. $1700$
• B. $1650$
• C. $3300$
• D. $3400$

Answer 1

The correct option is B

$1650$

Explanation for the correct option:

Step 1: Solve for the relation between first term $\left(a\right)$ and common difference $\left(d\right)$ of the given A.P.

We know that $n^{th}$ term of the A.P. is given by, $a_n=a+(n-1)d$

$$\begin{gathered} a_{12}=a+11d...\left(1\right) \\ {a}_{22}=a+21d...\left(2\right) \end{gathered}$$

Given: The sum of ${12}^{th}$ and ${22}^{th}$ terms of an AP is $100$.

$$\begin{gathered} a+11d+a+21d=100 \\ \Rightarrow 2a+32d=100 \end{gathered}$$

Divide the equation by $2,$

$\Rightarrow a+16d=50$

Step 2: Solve for the required sum.

We know that sum of first $n$ term is given by, ${\mathbf{S}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{2}}\left[2a+\left(n-1\right)d\right]$

$$\begin{gathered} S_{33}=\frac{33}{2}\left[2a+32d\right] \\ \Rightarrow S_{33}=\frac{33}{2}\times 2\left[a+16d\right] \\ \Rightarrow S_{33}=33\left[a+16d\right]\left[\because a+16d=50\right] \\ \Rightarrow S_{33}=33\left(50\right) \\ \Rightarrow S_{33}=1650 \end{gathered}$$

Hence option(B) is the correct option.