Question

If $m^{th}$ term of the series $63+65+67+69+....$and $3+10+17+24+...$ be equal, then $m=$

• A. $11$
• B. $12$
• C. $13$
• D. $15$

Answer 1

The correct option is C

$13$

Explanation for correct option

Given series is $63+65+67+69+....$ and $3+10+17+24+...$

For the series $63+65+67+69+....$ the first term is $a_1=63$ and second term is $a_2=65$

Therefore the common difference is $d=a_2-a_1=65-63=2$

The $m^{th}$ term of the series is $T_m=a_1+d\left(m-1\right)=63+2\left(m-1\right)$

For the series $3+10+17+24+...$ the first term is $b_1=3$ and second term is $b_2=10$

Therefore the common difference is $d\text{'}=b_2-b_1=10-3=7$

The $m^{th}$ term of the series is $T{\text{'}}_m=b_1+d\text{'}\left(m-1\right)=3+7\left(m-1\right)$

Given that the $m^{th}$ term of both the series is equal .

$$\begin{gathered} \therefore T_m=T{\text{'}}_m \\ \Rightarrow 63+2\left(m-1\right)=3+7\left(m-1\right) \\ \Rightarrow 63+2m-2=3+7m-7 \\ \Rightarrow 63-2-3+7=7m-2m \\ \Rightarrow 65=5m \\ \Rightarrow m=13 \end{gathered}$$

Hence, option (C) i.e. $13$ is correct