how many terms of the arithmetic sequence {1,3,5,7,...} will give a sum of 961?
n/2 (2+(n-1)*2) = 961
or, just recall that the sum of the first n odd numbers is n^2 ...
To find out how many terms of the arithmetic sequence {1, 3, 5, 7, ...} will give a sum of 961, we can use the formula for the sum of an arithmetic sequence.
The general formula for the sum of an arithmetic sequence is:
Sn = (n/2) * (a1 + an),
where Sn is the sum of the first n terms, n is the number of terms, a1 is the first term, and an is the nth term.
In this case, the first term (a1) is 1, and the common difference between terms is 2.
Using the formula, we can rewrite the problem as:
961 = (n/2) * (1 + an),
where we need to find the value of n.
First, let's find the value of an (the nth term). We know that the common difference is 2, so the nth term will be a1 + (n-1)d, where d is 2:
an = 1 + (n-1) * 2 = 1 + 2n - 2 = 2n - 1.
Substituting an = 2n - 1 into the equation:
961 = (n/2) * (1 + 2n - 1).
Now, we can simplify the equation:
961 = (n/2) * 2n,
or
961 = n^2.
To solve for n, we take the square root of both sides of the equation:
n = √961.
Taking the square root of 961, we get:
n = 31.
Therefore, 31 terms of the arithmetic sequence {1, 3, 5, 7, ...} will give a sum of 961.