Join now for JEE/NEET and also prepare for Boards Join now for JEE/NEET and also prepare for Boards. Register Now
Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
Find the sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7.? 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
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -