Question

Let $a_n=46+8n$ and $b_n=98+4n$ be two sequences for natural numbers $n ≤ 100$ . Then, the sum of all terms common to both the sequences is

• A. 14900
• B. 15000
• C. 14798
• D. 14602

Answer 1

Answer: (A)

14900

Answer 2

The first series goes as follows: $46, 54, 62, 70, 78, 86, 94, 102,....$
The second series follows this format: Numbers $98, 102, 106, 110,..$
The difference between them will be hcf(4,8) = 8, which is the first common term (term of the common terms).
The necessary order is $102,110,118$, (the final term must be less than 468, which is the second series' hundredth term).
$[102 + (n-1)(8) ≤ 498]$
If n is equal to or less than 50.5, then n = 50.

Applying the A.P. formula summation, we get: Required sum = $\frac{n}{2}(2a + (n - 1)d) = \frac{50}{2}(2 \times 102 + 49 \times 8) = 14900$
The correct option is (A): 14900.