|
|
Progressions - Quant/Math - CAT 2009
Question 4 the day:
July 14, 2003
The question for the day is from the topic of Progressions.
- Find the sum of all the integers which are multiples of 7 and lie between 200 and 400.
| (1) |
8729 |
| |
(2) |
8700 |
| (3) |
8428 |
| |
(4) |
None of these |
Correct Answer - (1)
Solution:
Since 4 is left as remainder when we divide 200 by 7 the least number greater than 200 divisible by 7 is 203.
Again if we divide 400 by 7, 1 is left as remainder.
This implies that the greatest number less than 400, which is divisible by 7 is 399.
Here t1 = a = 203, d = 7, l = 399
Let n be the total number of terms in this series.
Then 399 = 203 + (n – 1) (7)
=>7n = 399 – 203 + 7 = 406 – 203 = 203 => n = 29.
Hence the required sum 
= (29) (301) = 8729
|
|
|
|
