arithmetic progression?

Question

Find the sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7.

Answer 1

100/16 is 6 with a remainder of 4. So 103/16 leaves a quotient of 6 and a remainder of 7.
800/16 is 50 with a remander of 0. So 791/16 is 49 with a remainder of 7.

Therefore the required sum is:

Sum(i=6,49) (16i+7)
= Sum(i=6,49)16i + Sum(i=6,49)7
= 16 * Sum(i=6,49)i + 7*Sum(i=6,49)1
= 16 * [Sum(i=1,49)i - Sum(i=1,5)i] + 7*(49-6+1)
= 16*[49*50/2-5*6/2]+7*44
= 16*(1225-15)+308
= 16*1210 + 308
= 19,360+308
= 19,668.

I used the formula that Sum(i=1,n)i = n(n+1)/2, the formula for a triangular number.

Answer 2

The first number is 103. To produce the others, just keep adding 16. So you get the sequence 103,119,135,151,...791. Now use the formula for the sum of an arithmetic series. a(1) =103, a(n) = 791 so you need to find n. Use the formula for an arithmetic sequence to find n: a(n)=a(1) + (n-1)d. We know a(1) =103, a(n) = 791 and d =16. Solve for n and use the formula (n/2) (a(1) + a(n)) to find the sum.

Answer 3

find the first and last number which is divisible by 16 and leaves the remainder 7
here we getting 103 is the first term and 791 is the last term and difference is 16
find the number of term between 103 and 791
we get 44 using tn= a+ (n-1)d
791=103+(n-1)16
we get, 688 =16(n-1)
44 = n
sum of the term = n/2(a+l)
= 44/2(103+791)
= 22* 894
= 19668.
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