0606 P11 - Nov 2022 - Q5 - 6 marks
7851
An arithmetic progression is such that the fourth term is \(25\) and the ninth term is \(50\).
(a) Find the first term and the common difference.
(b) Find the least number of terms for which the sum of the progression is greater than \(25000\).
Solution
Answer: (a) first term \(10\), common difference \(5\); (b) \(99\).
Answer: (a) first term \(10\), common difference \(5\); (b) \(99\).
(a) Let the first term be \(a\) and the common difference be \(d\).
The fourth term is
\(a+3d=25.\)
The ninth term is
\(a+8d=50.\)
Subtract the first equation from the second:
\(5d=25,\)
so
\(d=5.\)
Then
\(a+3(5)=25,\)
so
\(a=10.\)
(b) The sum of the first \(n\) terms is
\(S_n=\frac n2\left(2a+(n-1)d\right).\)
Using \(a=10\) and \(d=5\),
\(S_n=\frac n2(20+5n-5)=\frac n2(5n+15).\)
We need
\(\frac n2(5n+15)\gt 25000.\)
This is
\(5n^2+15n-50000\gt 0.\)
The positive root of \(5n^2+15n-50000=0\) is approximately \(98.5\). Therefore the least integer \(n\) giving a sum greater than \(25000\) is
\(n=99.\)
