Value of the last installment.

Question

The price of a TV set is ₹20,000 to be paid in 20 installments of ₹1,000 each. Rate of interest=6% p.a

The first installment is to be paid at the time of purchase, then what will be the value of the last installment covering the interest as well?

There are two explanations I found:

1. Money paid in cash = ₹ 1000 Balance payment = (20000 - 1000) = ₹ 19000 (the above is from R.S. Aggarwal book and doesn't seem complete)
2. These links explain and conclude the answer to be 1950 but I didn't exactly understand: link 1 and link 2

I thought something like below (assuming monthly installments, hence taking 6/12 as rate of interest):

• Value after $20$ installments $= 20000 + (20000\cdot 20\cdot 6)/(12\cdot 100) = 22000$
• So, value after 19 installments $= 1000+(1000\cdot 20\cdot 6)/(12\cdot 100) +....+ 1000+(1000\cdot 2\cdot 6)/(12\cdot 100) = 19000 + (1000\cdot 6/(12\cdot 100))\cdot(19\cdot 20/2) = 19950$
• Thus, value of last installment $= 22000-19950=2050$

Can someone explain what am I doing wrong?

Answer 1

Let be $P=1,000$ the installment, $n=20$ number of monthly installments, $S=20,000$ the value of the TV.

Observing that $P\times n=1,000\times 20=20,000=S$, we have that installments are for the principal amount $S$ only and interest is yet to be paid.

The interest $I$ is to be paid with the amount of last installment $L$ and hence last installment will be $$ L=P+I $$ Let be $i=6\%$ the annual interest rate. the duration of each month is $1/12$ of a year. With the simple interest we have $$ I=\text{principal}\times \text{interest rate}\times \text{time} $$ At month $1$, the principal is $S_1=S-P=19,000$ and the interest to be paid at month $20$ is $$I_1=S_1\times i\times 1=19,000\times0.06\times 1/12$$

At month $2$, the principal is $S_2=S_1-P=18,000$ and the interest to be paid at month $20$ is $$I_2=S_2\times i\times 1=18,000\times0.06\times 1/12$$

At month $19$, the principal is $S_{19}=S_{18}-P=1,000$ and the interest $$I_{19}=S_{19}\times i\times 1=1,000\times0.06\times 1/12$$

At month $20$, the principal is $S_{20}=S_{19}-P=0$ and the interest is zero.

Summing up all the interests we have $$ I=\sum_{k=1}^{20}I_k=\sum_{k=1}^{19}I_k=\sum_{k=1}^{19} S_{k}\times i\times 1/12 $$ Observing $S_k=1000\times k$ we have $$ I=\frac{1}{12}\sum_{k=1}^{19} S_{k}\times i=\frac{1}{12}\sum_{k=1}^{19} 1000k\times i=\frac{1000\times 0.06}{12}\sum_{k=1}^{19} k=5\sum_{k=1}^{19} k=5\frac{19\times 20}{2}=950 $$ beacause $\sum_{k=1}^n k=\frac{n(n+1)}{2}$.

so we have that the last installmenet is $$ L=P+I=1,000+950=1,950 $$