there will be 28 multiples of 7 between 200 and 400,,,among them 1st one is 203 and last is 399..so the average is 301.......there by sum of all those is 28*301 that is 8428.....

What is the sum of all positive integers that are multiples of 7 from 200 to 400? 8729 8700 8428 8278

Correct Choice is (1) and Correct Answer is 8729

Explanatory Answer

Because 4 is the remainder when we divide 200 by 7 the least number greater than 200 divisible by 7 is 203. When we divide 400 by 7, we get a remainder of 1.

This implies that the greatest number less than 400, which is divisible by 7 is 399.

Therefore, this is an Arithmetic Progression in which the first term, t1 = a = 203, the common difference 'd' = 7, and the last term 'l' = 399

Let 'n' be the total number of terms in this series.

The nth term of an Arithmetic Sequence is an = a + (n - 1)d

8700 hi hoga pta nhi is chutiye ne itna explain karke bhi galat bataya.... 200/7 quotient will be 28 ,,,,,400/7 quotient will be 57......then 57-28 = 29 hence the value of n is 29 then apply the formula n(n+1) ....29(29+1) = 8700 okkkkkkkkkkkkk bosss

## 12 comments:

any one give solution and formula for this

what is the sum of all positive integers lying between 200 and 400 that are multiples of 7 ?

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there will be 28 multiples of 7 between 200 and 400,,,among them 1st one is 203 and last is 399..so the average is 301.......there by sum of all those is 28*301

that is 8428.....

What is the sum of all positive integers that are multiples of 7 from 200 to 400?

8729

8700

8428

8278

Correct Choice is (1) and Correct Answer is 8729

Explanatory Answer

Because 4 is the remainder when we divide 200 by 7 the least number greater than 200 divisible by 7 is 203.

When we divide 400 by 7, we get a remainder of 1.

This implies that the greatest number less than 400, which is divisible by 7 is 399.

Therefore, this is an Arithmetic Progression in which the first term, t1 = a = 203, the common difference 'd' = 7, and the last term 'l' = 399

Let 'n' be the total number of terms in this series.

The nth term of an Arithmetic Sequence is an = a + (n - 1)d

Then 399 = 203 + (n – 1) (7)

Simplifying we get, 7n = 399 – 203 + 7

Or 406 – 203 = 203 or n = 29.

Hence, the required sum is n/2[a + l]

Or 29/2[203 + 399] = (29) * (301) = 8729

thanq

thanq

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Its an ap series....a=203,d=7,l=399.....putting the formula n/2(a+l) u'll get ans=8729.

Rahul, 301 is d avg of 203, 204, 205...399 and nt of 203,210...399. I hd also made d same mistake in hurry.

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thank you so much for ur help

8700

8700 hi hoga pta nhi is chutiye ne itna explain karke bhi galat bataya....

200/7 quotient will be 28 ,,,,,400/7 quotient will be 57......then 57-28 = 29

hence the value of n is 29 then apply the formula n(n+1) ....29(29+1) = 8700

okkkkkkkkkkkkk bosss

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