A fund of 3000\$ is to be accumulated by n annual payments of 65\$ at the end of each year, followed by n annual payments of 135\$ at the end of each year, plus a smaller final payment made one year after the last regular payment. If the effective rate of interest is 6% per year, find n and the amount of the final irregular payment
Can someone help me with this problem, I started something but I need help to solve the rest .. What I did from now is that 3000 = (65x S[n,6%] ) + (135x S[n,6%]) + X Where the x is the final irregular payment thank you
In my view we have $n$ payments of $p=65\$$ and then other $n$ payments of $q=135\$$ and a last payment $x$. Let be $i=6\%$, $S=3,000\$$ and $n=\lfloor N\rfloor$ the number of year (integer).
The last payment $x$ will be made $1$ year after $n$ payments $p$ and after other $n$ payments $q$, i.d. at year $2n+1$.The first payment $p$ will produce the value $p s_{\overline{n}|i}$ at year $n$ and this amount will produce the sum $p s_{\overline{n}|i}(1+i)^{n+1}$ after other $n+1$ years, i.e. at year $2n+1$. Analogously for the stream $q$ we have $q s_{\overline{n}|i}$ after $n$ years and this amount will produce the sum $q s_{\overline{n}|i}(1+i)$ after another $1$ year.
The equation for $S$ is $$ S=p s_{\overline{n}|i}(1+i)^{n+1}+q(1+i)s_{\overline{n}|i}+x\tag 1 $$
Without the last payment $x$ and for a not integer number of years $N$ we would have after $2N+1$ years the sum $$ S=p s_{\overline{N}|i}(1+i)^{N+1}+q(1+i)s_{\overline{N}|i}=p\,\frac{y-1}{i}y\,(1+i)+q\,y\,\frac{y-1}{i}\tag 2 $$ where $y=(1+i)^N>0$. The equation (2) becomes $$ y(y-1)=\frac{iS}{p(1+i)+q}\tag 3 $$ So we have that the positive root of eq. (3) is $ y^+\approx 1.564$ and then $$ N=\frac{\log(y^+)}{\log(1+i)}\approx7.8 $$ So we have $$n=\lfloor N\rfloor=7$$ and from eq. (1) we find $$ x\approx 426.53\$ $$