Find the sum of all multiples of 9 lying between 400 and 800 ?
Answers
Answered by
105
The first multiple of 9 between 400 and 800 is 405 and the last multiple is 792.
We know that An = a+(n-1)*d
792 = 405(n-1) * 9
387 = (n-1) * 9
n = 44.
Sum = n/2 * (2a+(n-1)*d)
= 44/2(2 * 405 + (44-1) * 9)
= 22(810 + 387)
= 22 * 1197
= 26334.
We know that An = a+(n-1)*d
792 = 405(n-1) * 9
387 = (n-1) * 9
n = 44.
Sum = n/2 * (2a+(n-1)*d)
= 44/2(2 * 405 + (44-1) * 9)
= 22(810 + 387)
= 22 * 1197
= 26334.
Answered by
30
your answer is in attachment....
Hope its help you
Hope its help you
Attachments:
Similar questions