SPM 2021 Add Maths Trial Paper (Selangor) – Paper 1 (Question 5 & 6)

Question 5:
(a) Given a geometric progression with terms a, ar, ar², ar³, … , ar^n-2, ar^n-1 and the sum of first n terms is S_n. 
Derive the formula for the sum of the first n terms, S_n for the geometric progression when |r| < 1. [3 marks]

(b) Hence, find the value of n for which the sum of the first 2n terms is 127/128 times of the sum of first n terms of a geometric progression which has a common ratio of -½. [4 marks]


Solution:
(a)

$\begin{array}{l}{S}_n=a+ar+a{r}^2+a{r}^3+\mathrm{\dots }+a{r}^{n-2}+a{r}^{n-1}\mathrm{\dots }\mathrm{\dots }\mathrm{\dots }\mathrm{\dots }(1)\\ \\ \left(1\right)\times r:r{S}_n=ar+a{r}^2+a{r}^3+a{r}^4+\mathrm{\dots }+a{r}^{n-1}+a{r}^n\mathrm{\dots }\mathrm{\dots }\mathrm{\dots }\mathrm{\dots }(2)\\ \\ \left(1\right)-\left(2\right),\\ S_n-r{S}_n=a-a{r}^n\\ S_n\left(1-r\right)=a\left(1-r^n\right)\\ S_n=\frac{a\left(1-r^n\right)}{\left(1-r\right)},\left|r\right|<1\end{array}$


(b)

$\begin{array}{l}{S}_{2n}=\frac{127}{128}{S}_n\\ \frac{a\left(1-{\left(-\frac{1}{2}\right)}^{2n}\right)}{1-\left(-\frac{1}{2}\right)}=\frac{127}{128}\left(\frac{a\left(1-{\left(-\frac{1}{2}\right)}^n\right)}{1-\left(-\frac{1}{2}\right)}\right)\\ \\ \frac{\cancel{a}\left(1-{\left(-\frac{1}{2}\right)}^{2n}\right)}{\cancel{\frac{3}{2}}}=\frac{127}{128}\left(\frac{\cancel{a}\left(1-{\left(-\frac{1}{2}\right)}^n\right)}{\cancel{\frac{3}{2}}}\right)\\ \\ 128\left(1-{\left(-\frac{1}{2}\right)}^{2n}\right)=127\left(1-{\left(-\frac{1}{2}\right)}^n\right)\\ \\ 128-128{\left(-\frac{1}{2}\right)}^{2n}=127-127{\left(-\frac{1}{2}\right)}^n\\ \\ \text{Let }{\left(-\frac{1}{2}\right)}^n=x\\ 128-128x^2=127-127x\\ 128x^2-127x-1=0\\ \left(128x+1\right)\left(x-1\right)=0\\ x=-\frac{1}{128}\text{ or }x=1\\ \\ {\left(-\frac{1}{2}\right)}^n=-\frac{1}{128}\\ {\left(-\frac{1}{2}\right)}^n={\left(-\frac{1}{2}\right)}^7\\ n=7\\ \\ \text{or }{\left(-\frac{1}{2}\right)}^n=1\\ {\left(-\frac{1}{2}\right)}^n={\left(-\frac{1}{2}\right)}^0\\ n=0\text{ (not accepted)}\end{array}$