Question

find the middle term of the sequence formed by all three-digit no. which leave a remainder when divide by 7 .Also find the sum of all no. on both sides of the middle term separate ly​

Answer 1

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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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Answer 2

Answer:

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