Question

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

• A. 191
• B. 193
• C. 211
• D. none of these.

Answer 1

Answer: (A)

$T_n$ of $3+7+11+.....=3+\left(n-1\right)4=4n-1$
$T_{m}of1+6+11+.....=1+\left(m-1\right)5$
$=5m-4$
For a common term, $4n-1=5m-4$
i.e., $4n=5m-3$
For $n=3,m=3$ gives the first common term i.e. $11$.
$n=8,m=7$ gives the second common term i.e., $31$.
$n=13,m=11$ gives the third common term i.e., $51$.
$n=18,m=15$ gives the fourth common term i.e., $71$.
Now, $10$th term of $3,8,13,18,....$
$=3+\left(10-1\right)5=48$
For $10$ th common term, $n=48$
$\therefore$ reqd. common term $=4\left(48\right)-1$
$=192-1=191$