Question

The largest term common to the sequences $1, 11, 21, 31,...$to $100$ terms and $31, 36, 41, 46,.....$ to $100 $ terms is

• A. 381
• B. 471
• C. 281
• D. none of these.

Answer 1

The Correct Option is D

Let mth term of the first sequence be equal to the n term of the second sequence. Then $1 + (m - 1)10 = 31 + (n - 1)5$ $\Rightarrow 10m - 9 = 5n + 26$ $\Rightarrow 10m - 35 = 5n$ $\Rightarrow 2m -7 = n \le 100 $ $\Rightarrow 2m \le 107$ $\Rightarrow m \le 53 \frac{1}{2}$ $\therefore$ largest value of $m = 53$ largest term $= 1 + 52 \times 10 = 521$