0.28 Find the middle term of the sequence formed by all three-digit nu

1.	0.28 Find the middle term of the sequence formed by all three-digit numberswhich leave a remainder 5 when divided by 7. Also find the sum of allnumber on both sides of the middle term separately
Answer» A.P: 103, 110, 117, ...., 999 a = 103 d = 7 n be the number of terms an= 999 103 + (n - 1)7 = 999 103 + 7n - 7 = 999 7n + 96 = 999 7n = 903 n = 129 Middle term term = (n+1)/2=65th= a + 64d = 103 + 64 x 7 = 551 Sn =n/2 [2a + (n - 1)d] =64/2 [2 x 103 + 63 x 7] = 20704 = 129/2[2 x 103 + 128 x 7] = 71079 S129 - ( S64+ 551) = 71079 - (20704 + 551) = 49824 Like if you find it useful

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

Answer»

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

a = 103 d = 7

n be the number of terms

an= 999

103 + (n - 1)7 = 999

103 + 7n - 7 = 999

7n + 96 = 999

7n = 903

n = 129

Middle term term = (n+1)/2=65th= a + 64d = 103 + 64 x 7 = 551

Sn =n/2 [2a + (n - 1)d]

=64/2 [2 x 103 + 63 x 7] = 20704

= 129/2[2 x 103 + 128 x 7] = 71079

S129 - ( S64+ 551) = 71079 - (20704 + 551) = 49824

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