The number 3136 is the sum of the first how many odd whole numbers?
3 answers
56
how did you solve
so we have
1+3+5+7+... = 3136
let the number of terms of this AS be n
(n/2)(2a +(n-1)d) = Sum(n)
So we have a=1, d=2 , and n = ?
(n/2)(2 + (n-1)2) = 3136
times 2 ....
n(2 + 2n - 2) = 6272
2n^2 = 6272
n^2 = 3136
n = ± 56 , but n has to be a positive whole number,
there are 56 terms
1+3+5+7+... = 3136
let the number of terms of this AS be n
(n/2)(2a +(n-1)d) = Sum(n)
So we have a=1, d=2 , and n = ?
(n/2)(2 + (n-1)2) = 3136
times 2 ....
n(2 + 2n - 2) = 6272
2n^2 = 6272
n^2 = 3136
n = ± 56 , but n has to be a positive whole number,
there are 56 terms
1 answer