Question

If the $2^{nd}$ term in AP is $13$ and the $5^{th}$ term is $25$, what is the $7^{th}$ term?

• A. $30$
• B. $33$
• C. $37$
• D. $38$

Answer 1

The correct option is B

$33$

Step 1: Find the equation

Let the first term $=a_1$

Common difference $=d$

Number of terms $=n$

$\therefore n^{th}$ term $\left(a_n\right)$ $=a_1+(n-1)d$

The $2^{nd}$ term $=a_1+(2-1)d$

$=a_1+d$

The $5^{th}$ term $=a_1+(5-1)d$

$=a_1+4d$

The given condition is, the $2^{nd}$ term of is $13$

$\therefore a_1+d=13.....\left(1\right)$

Another given condition, is the $5^{th}$ term is $25$.

$\therefore a_1+4d=25.....\left(2\right)$

Step 2. Finding the first term and common difference.

Subtracting equation $\left(2\right)$ from equation $\left(1\right)$ we get

$$\begin{gathered} a_1+d=13 \\ \underline{-a_1-4d=-25} \\ -3d=-12 \\ \Rightarrow d=4 \end{gathered}$$

Therefore the common difference is $4$

Putting the value of $d$ in equation $\left(1\right)$ we get

$$\begin{gathered} a_1+d=13 \\ \Rightarrow a_1+4=13 \\ \Rightarrow a_1=13-4 \\ \Rightarrow a_1=9 \end{gathered}$$

Therefore the first term is $9$

Step 3. Finding the value of the $7^{th}$ term

The $7^{th}$ term is

$a_n=a_1+\left(n-1\right)d$

Now substituting $a_1=9,d=4$ and $n=7$ , we get

$a_n=9+\left(7-1\right)4$

$\begin{array}{rcl}{a}_7& =& 9+6\times 4\\ & =& 33\end{array}$

Hence, the correct option is $\mathrm{Option}\left(\mathrm{B}\right)$