An arithmetic progression has a first term of 32, a 5th term of 22 and a last term of -28. Find the sum of all the terms in the progression.
Solution
Let the first term be \(a = 32\) and the common difference be \(d\).
The 5th term is given by \(a + 4d = 22\).
Substituting \(a = 32\), we have:
\(32 + 4d = 22\)
\(4d = 22 - 32\)
\(4d = -10\)
\(d = -2.5\)
The last term is \(-28\), so:
\(a + (n-1)d = -28\)
\(32 + (n-1)(-2.5) = -28\)
\(32 - 2.5(n-1) = -28\)
\(32 - 2.5n + 2.5 = -28\)
\(34.5 - 2.5n = -28\)
\(-2.5n = -28 - 34.5\)
\(-2.5n = -62.5\)
\(n = 25\)
The sum of an arithmetic progression is given by:
\(S_n = \frac{n}{2} (a + l)\)
where \(l\) is the last term.
\(S_{25} = \frac{25}{2} (32 + (-28))\)
\(S_{25} = \frac{25}{2} (4)\)
\(S_{25} = 25 \times 2\)
\(S_{25} = 50\)
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