2020 H2 Mathematics Paper 1 Question 8

Arithmetic and Geometric Progressions (APs, GPs)

Answers

{u_{30}=\frac{95}{2}}

{\frac{3345}{2}=1672.5}

{S_{\infty}= 20}

Smallest possible value of

{n=18}

(ai)

\begin{gather*} u_5 = 10 \\ a+4d = 10 \\ 4+4d = 10 \\ d = \frac{3}{2} \end{gather*}

\begin{align*} u_{30} &= a + \left( n - 1 \right) d \\ &= 4 + \left( 30 - 1 \right) \frac{3}{2} \\ &= \frac{95}{2} \; \blacksquare \end{align*}

(aii)

Sum of the 21st term to the 50th term inclusive

\begin{align*} &= S_{50} - S_{20} \\ &= \frac{50}{2}\Big(2(4)+49(\frac{3}{2})\Big) - \frac{20}{2}\Big(2(4)+19(\frac{3}{2})\Big) \\ &= \frac{3345}{2} \; \blacksquare \end{align*}

(bi)

\begin{gather*} u_5 = \frac{1024}{625} \\ ar^4 = \frac{1024}{625} \\ 4r^4 = \frac{1024}{625} \\ r^4 = \frac{256}{625} \\ \end{gather*}

Since the common ratio is positive,

r=\frac{4}{5}

\begin{align*} S_{\infty} &= \frac{a}{1-r} \\ &= \frac{4}{1-\frac{4}{5}} \\ &= 20 \; \blacksquare \end{align*}

(bii)

\begin{gather*} S_n > 19.6 \\ \frac{a\left(1-r^{n}\right)}{1-r} > 19.6 \\ \frac{4\Big( 1 - 0.8^n \Big)}{1-0.8} > 19.6 \\ 0.8^n - 1 > 0.98 \\ 0.8^n < 0.02 \; \blacksquare \end{gather*}

\begin{gather*} n \ln 0.8 < \ln 0.02 \\ n > \frac{\ln 0.02}{\ln 0.8} \\ n > 17.531 \end{gather*}

Smallest possible value of

{n=18 \; \blacksquare}