Question

Find the sum of positive integers below 100,whose remainder is 1 when divided by 5

Answer 1

here the first integer will be 6. and the last term below 100 will be 96.
by using the formula,
t = a+ (n-1)d
here t= 96 ,a= 6, D= 5
96 = 6 + (n-1) 5
96-6=(n-1) 5
90 ÷ 5 = n-1
n-1 = 18
n= 19
therefore 96 is the 19th term of the a.p
therefore , sum of 19 terms will be,
n÷2(a+l)
= 19÷2 (6 + 96)
= 9.5 × 102
= 969
therefore the sum of the terms will be 969.