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June 2009 p1 q7
868
The first two terms in an arithmetic progression are 5 and 9. The last term in the progression is the only term which is greater than 200. Find the sum of all the terms in the progression.
Solution
The first term of the arithmetic progression is given as \(a = 5\), and the second term is \(9\), so the common difference \(d = 9 - 5 = 4\).
The general formula for the \(n\)-th term of an arithmetic progression is \(a_n = a + (n-1)d\).
We need to find the smallest \(n\) such that \(a_n > 200\).
Set \(a_n = 200\):
\(200 = 5 + (n-1) imes 4\)
\(200 = 5 + 4n - 4\)
\(200 = 4n + 1\)
\(199 = 4n\)
\(n = \frac{199}{4} = 49.75\)
Since \(n\) must be an integer, we take \(n = 50\) as the smallest integer greater than 49.75.
Now, calculate the sum of the first 50 terms using the sum formula for an arithmetic progression:
