Question

How many terms of the arithmetic sequence `{1,3,5,7,...}` will give a sum of `961`?

Answer 1

`31` terms

Explanation:

For an arithmetic sequence with initial value `a` and a difference between terms of `d`
the sum of the first `n` terms is given by the formula
`color(white)("XXX")Sigma = n/2*(2color(cyan)(a)+(n-1)color(green)(d))`

For the given sequence
`color(white)("XXX")color(cyan)(a=1)`
and
`color(white)("XXX")color(green)(d=2)`

We are told that the required sum is `961`

So
`color(white)("XXX")961=n/2*(2(color(cyan)(1))+(n-1)(color(green)(2)))`

`color(white)("XXX")961=n/color(red)(cancel(color(black)(2)))*color(red)(cancel(color(black)(2))) +n/color(blue)(cancel(color(black)(2)))(n-1)(color(blue)(cancel(color(black)(2))))`

`color(white)("XXX")961 = n+n^2 - n`

`color(white)("XXX")n^2 =961`

`color(white)("XXX")n=31`