Question

Find the sum of all natural numbers divisible by8 from 100 to 500

Answer 1

the sequence will be 104,112,120....496
the sequence will be in AP with a=104 d=8
we know that
TN=a+(n-1)d
496=104+(n-1)8
496-104=(n-1)8
392=(n-1)8
392/8=n-1
49=n-1
n=50
now,
SN=n/2(2a+(n-1)d
s50=50/2(2×104+(50-1)8]
s50=25(208+49×8)
=25(208+392)
=25(600)
=15000
therefore sum of all no is 15000