Given that the sum of the first $n$ terms of a series is ${{S}_{n}}$, where \({{S}_{n}}=3{{

Question

Given that the sum of the first $n$ terms of a series is ${{S}_{n}}$, where ${{S}_{n}}=3{{n}^{2}}-n$, state the value of its first term and write down an expression for ${{S}_{n-1}}$ for $n\ge 2$. Hence show that the terms in the series form an arithmetic progression.[4] The first term of an infinite geometric progression is $a$ and the common ratio is $r$, where $\left| r \right|<1$. Given that the sum of all the terms after the $n$th term is equal to the $n$th term of the progression, show that $r=\frac{1}{2}$ and hence find the sum to infinity of the progression in terms of $a$.[4] A bank has an account for investors. Compound interest is added to the account at the end of each year at a fixed rate of $3\%$ of the amount in the account at the beginning of that year. Mr Ang decides to invest $\$2\mathrm{} \, 000$ on 1 January 2023 and a further $\$2\mathrm{} \, 000$ on 1 January of each subsequent year. How much compound interest has his original $\$2\mathrm{} \, 000$ earned at the end of 31 December 2030?[2] By considering the total amount in the account at the end of the $n$ th year, find the year in which the total in the account will first exceed $\$40\mathrm{} \, 000$. Explain whether this occurs at the beginning or the end of the year.[6]

Accepted Answer

(a) First term $2$; $S_{n-1}=3n^2-7n+4$; AP with common difference $6$. (b) $r=\frac12$, $S_\infty=2a$. (c)(i) $\$533.54$. (ii) Smallest value of $n$ is $16$. The total amount first exceeds $\$40,000$ in $2038$. When $n=15$ on $31^\mathrm{st}$ December $2037$, Amount$2000\left[ \frac{1.03({{1.03}^{15}}-1)}{1.03-1} \right]=38313.76$. On $1$ January $2038$, $38313.76+2000=40313.76>40000$.

2023 YIJC Promo Q12

2023 YIJC Promo Q12

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