Question

Find the middle term of the sequence formed by all 3 digit number which leaves remainder 5 when divided by 7

Answer 1

Answer:

hehehehehehehehehehehehehe

Answer 2

according to the question

103,110,117.......999 is the Ap

an=a+(n-1)d

( an formula)

999=103+(n-1)7

999-103=(n-1)7

896/7=n-1

128+1=n

129=n this is odd then (129+1)/2

=130/2=65th term

a65=103+64(7)

=103+448

=551

S64=64/2(2(103)+63(7) )

=32(206+441)

=32(647)

=20704

S129=129/2(206+128(7))

=129/2(206+896)

=129/2(1102)

=71079

S129-(S64+551)

71079-(20704+551)

71079-21255

49824

hope it helps ...