Arithmetic & Geometric Progressions: GP: 99% of \(S_\infty\); modified AP — RI 2025 H2 Math Prelim Paper 1
What this question tests
Question
- A geometric series has first term 2 and common ratio 0.8. Find algebraically the least value of \(m\) for the sum of the first \(4m\) terms of the series to be greater than 99% of the sum to infinity.
- A finite arithmetic progression \(A\) has \(n\) terms, first term \(a\) and common difference \(d\).
In another progression \(B\), the \(k\)th term is obtained by adding the \(k\)th positive odd integer
to the corresponding term of \(A\). That is, the first term of \(B\) is obtained by adding 1 to
the first term of \(A\), the second term of \(B\) is obtained by adding 3 to the second term of \(A\),
and so on.
It is given that the twelfth term of \(A\) is 25, the sum of all the terms of \(A\) is 676, and the sum of all the terms of \(B\) is twice the sum of all the terms of \(A\).
- Find the values of \(n\), \(a\) and \(d\).
- Obtain the sum of the first ten terms of \(B\).
Show full worked solution▾
(a) Sum of first \(4m\) terms: \(S_{4m} = \dfrac{2(1-0.8^{4m})}{0.2} = 10(1-0.8^{4m})\).
Sum to infinity: \(S = \dfrac{2}{1-0.8} = 10\).
Require \(S_{4m} > 0.99\,S\): \[\begin{aligned} 10(1-0.8^{4m}) &> 9.9\\ 0.8^{4m} &< 0.01\\ 4m\ln 0.8 &< \ln 0.01\\ 4m &> \frac{\ln 0.01}{\ln 0.8} = 20.637\ldots \end{aligned}\] Since \(4m\) must be a positive integer, \(4m \ge 21\).
(b)(i) Sum of all terms of \(A\): \(\dfrac{n}{2}[2a+(n-1)d] = 676\) \(\cdots(1)\)
Sum of all terms of \(B = \) sum of \(A\) \(+\) sum of first \(n\) positive odd integers \(= 676 + (1+3+\cdots+(2n-1)) = 676 + n^2\).
Require this to equal \(2(676) = 1352\), so: \[\begin{aligned} 676 + n^2 &= 1352\\ n^2 &= 676\\ n &= 26 \end{aligned}\]
Sub. \(n = 26\) into (1): \(\dfrac{26}{2}(2a+25d) = 676 \Rightarrow 2a + 25d = 52\) \(\cdots(2)\)
12th term of \(A\): \(a + 11d = 25\) \(\cdots(3)\)
From GC (or solving (2) and (3)): \(n = 26\), \(a = \dfrac{53}{3}\), \(d = \dfrac{2}{3}\).
(b)(ii) \(B\) is an AP with first term \((a+1) = \dfrac{56}{3}\) and common difference \((d+2) = \dfrac{8}{3}\).
Sum of first 10 terms of \(B\): \[\begin{aligned} &= \frac{10}{2}\!\left[2\!\left(\frac{56}{3}\right)+9\!\left(\frac{8}{3}\right)\right] = 5\cdot\frac{112+72}{3} = \frac{920}{3} \end{aligned}\]