Find the middle term of the sequence by 3digit number which leave of Manderson 5 when leaved by 7 then find sum of all numbers of the middle term
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Answer:
103,110,117.......999
an=a+(n-1)d
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
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