Question

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

• A. $191$
• B. $190$
• C. $192$
• D. $194$

Answer 1

Answer: (A)

$191$

Series are $3+7+11+15+19+23+27+\mathrm{31...}$ and $1+6+11+16+21+26+31+...$

$1^{st}$ common term is $11$ and next will be $31$ so $d=20,a=11$

$T_{10}=a+(n-1)d$

$\Rightarrow T_{10}=11+9\times 20=191$