Question

Mark the correct alternative in the following question:

The 10th common term between the A.P.s 3, 7, 11, 15, ... and 1, 6, 11, 16, ... is

(a) 191 (b) 193 (c) 211 (d) none of these

Answer 1

$$\begin{gathered} \mathrm{As},\mathrm{the}\mathrm{common}\mathrm{difference}\mathrm{of}\mathrm{the}\mathrm{A}.\mathrm{P}.3,7,11,15,...=7-3=4\mathrm{and} \\ \mathrm{the}\mathrm{common}\mathrm{difference}\mathrm{of}\mathrm{the}\mathrm{A}.\mathrm{P}.1,6,11,16,...=6-1=5 \\ \mathrm{And},\mathrm{the}\mathrm{common}\mathrm{terms}\mathrm{of}\mathrm{both}\mathrm{the}\mathrm{A}.\mathrm{P}.\mathrm{s}\mathrm{will}\mathrm{be}\mathrm{in}\mathrm{A}.\mathrm{P}. \\ \mathrm{So},\mathrm{the}\mathrm{common}\mathrm{difference}\mathrm{of}\mathrm{the}\mathrm{A}.\mathrm{P}.\mathrm{of}\mathrm{the}\mathrm{common}\mathrm{terms},d=\mathrm{LCM}\left(4,5\right)=4\times 5=20\mathrm{and} \\ \mathrm{its}\mathrm{first}\mathrm{common}\mathrm{term},a=11 \\ \mathrm{Now},\mathrm{the}\mathrm{tenth}\mathrm{common}\mathrm{term},a_{10}=a+\left(10-1\right)d \\ =11+9\times 20 \\ =11+180 \\ =191 \end{gathered}$$

Hence, the correct alternative is option (a).