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)

191

Answer 2

$T_{n}$ of $3+7+11+.....= 3+\left(n-1\right)4 = 4n-1$ $T_{m} of 1+6+11+.....=1+\left(m-1\right)5$ $= 5m-4$ 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 $= 4\left(48\right)-1$ $= 192-1 = 191$