Question

How many terms of the AP : 9, 17, 25, ..... must be taken to give a sum of 636?

Answer 1

Let there be \(n\) terms in this Arithmetic Progression (A.P.).
For this A.P., the first term is \(a = 9\) and the common difference is \(d = a_2 - a_1 = 17 - 9 = 8\).
The sum of the first \(n\) terms is given by \(S_n = \frac{n}{2}[2a + (n-1)d]\).

Given \(S_n = 636\), we have:
\(636 = \frac{n}{2}[2 \times 9 + (n-1)8]\)

\(636 = \frac{n}{2}[18 + 8n - 8]\)
\(636 = \frac{n}{2}[10 + 8n]\)
\(636 = n[5 + 4n]\)
\(636 = 5n + 4n^2\)
\(4n^2 + 5n - 636 = 0\)
To solve for \(n\), we can factor the quadratic equation. We look for two numbers that multiply to \(4 \times -636 = -2544\) and add up to 5. These numbers are 53 and -48.
\(4n^2 + 53n - 48n - 636 = 0\)
Factor by grouping:
\(n(4n + 53) - 12(4n + 53) = 0\)
\((4n + 53)(n - 12) = 0\)
This gives two possible solutions:
Either \(4n + 53 = 0\) or \(n - 12 = 0\).
If \(4n + 53 = 0\), then \(n = -\frac{53}{4}\).
If \(n - 12 = 0\), then \(n = 12\).
Since the number of terms (\(n\)) cannot be negative or a fraction, we discard \(n = -\frac{53}{4}\).

Therefore, the number of terms is \(n = 12\).