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A Sum of Rs.13,360 was borrowed at 8[3/4] % per annum compound interest and paid back in two years in two equal annual installments . what was the amount of each installment?

I have tried:

Let each installment be x

(present worth of X due 1 yr hence) + (Present worth of rs x due 2 year hence) = 13360

x/(1+ 35/400) + x/(1+35/400)^2 =13360

I have got the Answer x = 7369

This is correct Answer

It is taking Large amount of steps in calculationg the answer, is there any alternate way to find out the answer for these type of sums, or shortcut manner , please anyone guide me answer

Generally how to approach these type of Installment sums,anyone Guide me the Answer

2 Answers 2

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The problem is of the form of

$$ \sum_{i=1}^n \frac{x}{(1+r)^n} = N$$

$$ x\sum_{i=1}^n \frac{1}{(1+r)^n} = N$$

$$ x\sum_{i=1}^n \left(\frac{1}{1+r} \right)^n= N$$

This is a geometric series with common ratio $R=\frac{1}{1+r}.$

$$x\frac{R(1-R^n)}{1-R}=N$$

$$x=\frac{N(1-R)}{R(1-R^n)}=\frac{N(\frac1R-1)}{1-R^n}=\frac{Nr}{1-\frac{1}{(1+r)^n}}$$

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    Is there any shortcut method is there? for these type of sums – 2017-02-11
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    I just did it once using geometric series. Now you just have to apply the formula at the very last line next time isn't it? – 2017-02-11
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Let's start with the equation that you provided

$$\frac{x}{1+\frac{35}{400}}+\frac{x}{(1+\frac{35}{400})^2} = 13360$$ $$x\cdot \frac{1}{1+\frac{35}{400}}+x\cdot\frac{1}{(1+\frac{35}{400})^2} = 13360$$ $$x\cdot \frac{1}{1+\frac{35}{400}}+x\cdot\bigg(\frac{1}{1+\frac{35}{400}}\bigg)^2 = 13360$$

To simplify things, lets represent $1/(1+35/400)$ symbolically as $t$ $$x t+x t^2 = 13360$$ $$x(t+t^2) = 13360$$ Further suppose that instead of being paid back over two years, it was paid back over ten years. Then the equation would be $$x(t+t^2+t^3+t^4+t^5+t^6+t^7+t^8+t^9+t^{10}) = 13360$$

Can you find a way of simplifying this expression?

If the interest rate $i$ equals $35/400$, how would you express $t$ in terms of $i$?