Question

Let 3, 7, 11, 15, ...., 403 and 2, 5, 8, 11, . . ., 404 be two arithmetic progressions. Then the sum, of the common terms in them, is equal to _________.

Answer 1

The first arithmetic progression (AP) is:

3, 7, 11, 15, ..., 403

The second arithmetic progression (AP) is:

2, 5, 8, 11, ..., 404

To find the common terms, we first find the least common multiple (LCM) of the common differences of both progressions:

$\text{LCM}(4, 3) = 12$

The sequence of common terms is:

11, 23, 35, ..., 403

This is an AP with first term $a = 11$ and common difference $d = 12$. We need to find the number of terms ($n$) in this AP such that the last term is 403:

$403 = 11 + (n - 1) \times 12$

$392 = (n - 1) \times 12 \implies n - 1 = \frac{392}{12} = 32 \implies n = 33$

The sum of the common terms is given by:

$S_n = \frac{n}{2} [2a + (n - 1) \times d]$

Substituting the values:

$S_{33} = \frac{33}{2} [2 \times 11 + (33 - 1) \times 12]$

$= \frac{33}{2} [22 + 32 \times 12]$

$= \frac{33}{2} \times 406 = 6699$