Question

Find the middle term of the sequence formed by all three-digit numbers which leave a remainder 5 when divided by 7. Also find the sum of all numbers on both sides of the middle term separately

Answer 1

A.P: 103, 110, 117, ...., 999

a = 103  d = 7

n be the number of terms

$a_{n}$ = 999

103 + (n - 1)7 = 999

103 + 7n - 7 = 999

7n + 96 = 999

7n = 903

n = 129

Middle term = $\frac{n+1}{2} ^{th}$ term = $65^{th}$ = a + 64d = 103 + 64 x 7 = 551

$S_{n}$ = $\frac{n}{2}$ [2a + (n - 1)d]

$S_{64}$ = $\frac{64}{2}$ [2 x 103 + 63 x 7] = 20704

$S_{129}$ = $\frac{129}{2}$ [2 x 103 + 128 x 7] = 71079

$S_{129}$ - ($S_{64}$ + 551) = 71079 - (20704 + 551) = 49824

Hope I helped you! :)

Answer 2

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