A computer is sold either for Rs.19200 cash or for Rs.4800 cash down payment together with five equal monthly installments. If the rate of interest charged is 12% per annum, then the amount of each installment (nearest to a rupee) is:

Amount on which interest will be charged = 19200 - 4800 = 14400

The total amount paid will be equal to the sum of all monthly instalments. Therefore, we have

$$14400*k^{5a} = I (k^{4a} +k^{3a}+k^{2a}+k^{a}+1 )$$   .....(1)

where, k = $$1+\frac{12}{100}$$ & a = $$\frac{1}{12}$$

We know that, $$k^{5a}-1 = (k-1)(k^{4a} +k^{3a}+k^{2a}+k^{a}+1)$$

=>$$k^{4a} +k^{3a}+k^{2a}+k^{a}+1$$ = $$\frac{k^{5a}-1}{k-1}$$ ....(2)

Substituting in equation (1) we get

I = $$14400*k^{5a}*\left[\frac{k-1}{k^{5a}-1}\right]$$ ....(3)

On substituting the values of k and a in equation (3) we get
I $$\approx$$ 2965

Hence, option B.