# The sum of all the terms common to the arithmetic progressions 1, 3, 5, ....., 1991 and 1, 6, 11, ......., 1991, is? (1) 199100 (2) 199200 (3) 199300 (4) 200196

##### 1 Answer
Dec 18, 2016

(2) $199200$

#### Explanation:

Given:

$1 , 3 , 5 , \ldots , 1991$

$1 , 6 , 11 , \ldots , 1991$

Note that the common difference of the first sequence is $2$ and that of the second is $5$.

Since these have no common factor greater than $1$, their least common multiple is $10$, which is the common difference of the intersection of the two sequences:

$1 , 11 , 21 , 31 , \ldots , 1991$

This sequence has $200$ terms, with average value:

$\frac{1}{2} \cdot \left(1 + 1991\right) = \frac{1992}{2}$

So the sum is:

$200 \cdot \frac{1992}{2} = 199200$