How many terms of the series 5 + 7 + 9 must be taken so that sum may be 480?

Answer 1

Let's first find the common difference of the arithmetic sequence.

d = 7 - 5 = 2

Now we can find the nth term of the sequence:

a_n = a_1 + (n-1)d

We want the sum of the first n terms to be 480, so we can use the formula for the sum of an arithmetic sequence:

S_n = (n/2)(a_1 + a_n)

Substituting in the appropriate values:

480 = (n/2)(5 + a_1 + 2(n-1))

Simplifying:

960 = n(2a_1 + 2n - 1)

We know a_1 = 5 and d = 2, so we can substitute that in to get:

960 = n(10 + 2n - 1)

960 = n(2n + 9)

2n^2 + 9n - 960 = 0

Using the quadratic formula:

n = (-9 ± sqrt(9^2 - 4(2)(-960))) / (2(2))

n = (-9 ± sqrt(9665)) / 4

We can disregard the negative root since n must be a positive integer. So:

n = (-9 + sqrt(9665)) / 4 ≈ 23.57

Since we can't take a partial term, we know we need to take at least 24 terms to get a sum of 480. We can check this:

5 + 7 + 9 + ... + 115 + 117 = 480

So the answer is 24 terms.