Do this all math lovers it's a challenge to all Q 10)
Answers
Answer:
20704, 49824
Step-by-step explanation:
The three digit numbers which leaves 5 as remainder when divided by 7 are
103, 110, ...999
The given Series forms an A.P.
First term a = 103.
Common difference, d = 7.
Last term = an = 999.
We know that nth term of an AP an = a + (n - 1) * d
⇒ 999 = 103 + (n - 1) * 7
⇒ 999 - 103 = 7n - 7
⇒ 903 = 7n
⇒ n = 129.
(i) Middle-term:
= (n + 1/2)th term
= (129 + 1/2) th term
= 65th term.
So,
a₆₅ = 103 + (65 - 1) * 7
= 103 + 64 * 7
= 551.
(ii) Sum of all numbers before the middle term(a₁ to a₆₄):
a₁ = 103, d = 7, a₆₄ = 551 - 7 = 544
S₆₄ = (n/2)[a₁ + a₆₄]
= (64/2)[103 + 544]
= 20704.
(iii) Sum of all numbers after the middle term(a₆₆ to a₁₂₉):
a₆₆ = 551 + 7 = 558, d = 7, a₁₂₉ = 999
S₁₂₉ - S₆₅ = (n/2)[a₆₆ + a₁₂₉]
= (64/2)[558 + 999]
= 49824.
Hope it helps!
sum of all numbers before middle term is 20704
and
sum of all numbers after middle term is 49854
