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Find the sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7.?
SUCH NUMBERS ARE 106,119,135,------------------------,791. THEN, Tn=a+(n-1)d gives n=44 so,Sn=n/2*(a+Tn) gives Sn=19668 so,answer is 19668
SUCH NUMBERS ARE 106,119,135,------------------------,791.
THEN, Tn=a+(n-1)d gives n=44
so,Sn=n/2*(a+Tn) gives Sn=19668
so,answer is 19668
Any number that leaves a remainder of 7 on division with 16 can be represented as 16k+7. In this case the number lies between 100 and 800. So 6≤k≤49. Now the sequence 16k+7 is an AP with common difference 16 starting from 103 and ending with 791, containing (49-6)+1=44 terms. So the required sum is (n/2)(a+l)=(44/2)(103+791)= 19668 Please like the answer if you understood the solution. SOHINI PAUL Dept of Electrical engg IIT Kharagpur
Any number that leaves a remainder of 7 on division with 16 can be represented as 16k+7. In this case the number lies between 100 and 800. So 6≤k≤49. Now the sequence 16k+7 is an AP with common difference 16 starting from 103 and ending with 791, containing (49-6)+1=44 terms.
So the required sum is (n/2)(a+l)=(44/2)(103+791)= 19668
Please like the answer if you understood the solution.
SOHINI PAUL
Dept of Electrical engg
IIT Kharagpur
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