Question

The $10^{th}$ common term between the series $3+7+11+...$ and $1+6+11+...$is

• A. $191$
• B. $193$
• C. $211$
• D. $186$

Answer 1

Answer: (A)

The first common term is $11$.
Now, the next common term is obtained by adding $L.C.M$. of the common difference $4$ and $5$, i.e., $20$.
Therefore, $10^{th}$ common term $=T_{10}$ of the $A.P$.
whose $a=11$ and $d=20$
$\therefore T_{10}=a+9d$
$=11+9\left(20\right)$
$=191$