The first, second and last terms in an arithmetic progression are 56, 53 and -22 respectively. Find the sum of all the terms in the progression.
Solution
Let the first term be \(a = 56\) and the common difference be \(d = 53 - 56 = -3\).
The last term is given as \(-22\). Using the formula for the \(n\)-th term of an arithmetic progression, \(a_n = a + (n-1)d\), we have:
\(-22 = 56 + (n-1)(-3)\)
\(-22 = 56 - 3(n-1)\)
\(-22 = 56 - 3n + 3\)
\(-22 = 59 - 3n\)
\(3n = 59 + 22\)
\(3n = 81\)
\(n = 27\)
The sum of the first \(n\) terms of an arithmetic progression is given by:
\(S_n = \frac{n}{2} (a + l)\)
where \(l\) is the last term.
\(S_{27} = \frac{27}{2} (56 + (-22))\)
\(S_{27} = \frac{27}{2} (34)\)
\(S_{27} = 27 \times 17\)
\(S_{27} = 459\)
Log in to record attempts.
