9709 P12 - Nov 2011 - Q10
858
An arithmetic progression contains 25 terms and the first term is -15. The sum of all the terms in the progression is 525. Calculate
(i) the common difference of the progression,
(ii) the last term in the progression,
(iii) the sum of all the positive terms in the progression.
Solution
(i) The sum of an arithmetic progression is given by the formula:
\(S_n = \frac{n}{2} (2a + (n-1)d)\)
Given \(S_n = 525\), \(n = 25\), and \(a = -15\), we have:
\(525 = \frac{25}{2} (2(-15) + 24d)\)
\(525 = \frac{25}{2} (-30 + 24d)\)
\(525 = 25(-15 + 12d)\)
\(21 = -15 + 12d\)
\(36 = 12d\)
\(d = 3\)
(ii) The last term \(l\) in the progression is given by:
\(l = a + (n-1)d\)
\(l = -15 + 24 \times 3\)
\(l = -15 + 72\)
\(l = 57\)
(iii) The positive terms start from 3 and go up to 57. The first positive term is 3, which is the second term in the sequence:
\(a + d = 3\)
\(-15 + 3 = 3\)
The number of positive terms is 19 (from 3 to 57). The sum of these terms is:
\(S_{19} = \frac{19}{2} (3 + 57)\)
\(S_{19} = \frac{19}{2} \times 60\)
\(S_{19} = 570\)
