Find the sum of all the integers between 120 and 400 (both inclusive) which are divisible by 7.

9600

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11000

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11500

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8750

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10500

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Solution

The correct option is E 10500
Solution: Option (e)

In between 120 and 400

The first number divisible by 7 = 126, Last number divisible by 7 = 399

The other numbers divisible by 7 are 126, 133, 140, 147... 399; All the numbers are in an AP with common difference of 7. Number of terms in the AP = $\frac{(399 - 126)}{7 + 1} + 1 = 40$

Sum of all the terms = $(\frac{40}{2}) (126 + 399) = 10500.$

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