Question

the sum of all the terms common to the arithmetic progression
1,3,5,....1991 and 1,6,11,....1991 is :
(a) 199100 , (b) 199300
(c) 199200 , (d) 200196​

Answer 1

Answer:

its is so easy answer is 199200 dul i am not sure

Answer 2

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

,

...

,

1991

This sequence has

200

terms, with average value:

1

2

⋅

(

1

+

1991

)

=

1992

2

So the sum is:

200

⋅

1992

2

=

199200