Question

a cumulative deposit of monthily instalment of Rs.1200at 9% pe annum simple interest earns an interest of Rs.5994. find the number of instalments paid?.

Answer 1

$$\begin{gathered} Simpleinterest=\frac{P\times R\times T}{100}=P\times \frac{r}{100}\times \frac{n(n+1)}{2\times 12} \\ NowprincipalisgivenasRs1200andr=9\%andinterest=Rs5994 \\ Letthenumberofinstallmentsben \\ Henceputtingthesevaluesweget \\ 5994=1200\times \frac{9}{100}\times \frac{n(n+1)}{2\times 12} \\ \Rightarrow n(n+1)=\frac{5994\times 2\times 12\times 100}{1200\times 9} \\ \Rightarrow n(n+1)=1332 \\ \Rightarrow n^2+n-1332=0 \\ \Rightarrow n^2+37n-36n-1332=0 \\ \Rightarrow n(n+37)-36(n+37)=0 \\ \Rightarrow (n-36)(n+37)=0 \\ \Rightarrow n=-37or36 \\ Asnumberofinstallmentscannotbenegative \\ Hencenumberofinstallmentspaid=36 \end{gathered}$$

Derivation of the expression for total time

$$\begin{gathered} LetthemonthlyinstallmentbePandnumberofinstallmentsbenandrateofinterestber\%perannum \\ Nowinterestiscalculatedonfirstinstallmentfornmonths,onsecondinstallmentfor(n-1)months..........,onlastinstallmentfor1month \\ HencethisformsanA.Pwhosesum=n+(n-1)+(n-2)+(n-3)......+2+1 \\ Hencetotaltime=sumoftheA.P=\frac{n(n+1)}{2}months=\frac{n(n+1)}{2\times 12}years \\ \therefore Interest=\frac{p\times r\times t}{100}=\frac{p\times r\times n(n+1)}{100\times 2\times 12} \end{gathered}$$