Asked by Vijay
For any series 5+7+9+.....the sum of how many terms will be 480 ?
Answers
Answered by
Reiny
use your formula
a = 5, d = 2 , n = ? , sum(n) = 480
(n/2)(10 + 2(n-1)) = 480
n(8 + 2n) = 960
8n + 2n^2 = 960
n^2 + 4n - 480 = 0
(n+24)(n-20) = 0
n = -24 or n = 20 , but n must be a whole number
the sum of 20 terms would be 480
check:
sum(20) = (20/2)(10 + 2(19)) = 480
a = 5, d = 2 , n = ? , sum(n) = 480
(n/2)(10 + 2(n-1)) = 480
n(8 + 2n) = 960
8n + 2n^2 = 960
n^2 + 4n - 480 = 0
(n+24)(n-20) = 0
n = -24 or n = 20 , but n must be a whole number
the sum of 20 terms would be 480
check:
sum(20) = (20/2)(10 + 2(19)) = 480
Answered by
Steve
or, since the sum of the first n odd numbers is n^2,
n^2 - 2^2 = 480
n^2 = 484
n = 22
The sum of the first 22 odd numbers is 484. Subtract the first 2, leaving 20, as above.
n^2 - 2^2 = 480
n^2 = 484
n = 22
The sum of the first 22 odd numbers is 484. Subtract the first 2, leaving 20, as above.
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