How the formula for EMI is derived I was looking for a formula to calculate EMI (Equated Monthly Installments). I have some fixed known parameters like, Principal Amount, Rate of Interest and No. Of Installments. By googling, I came across the formula,
$$Installment Amount = \frac {P*i*(1 + i)^n}{(1 + i)^n - 1}$$
      where i  =  interest rate per installment payment period,  
            n =  number of Installments,  
            P  = principal amount of the loan

This formula does my job, but I actually want to understand the formula in detail, that how it derived. I have done googling to decode it but no luck.
Can anybody help me to understand the formula? Like, what each operation in the formula stands for?
 A: Hi while googling I found the below detailed answer posted by someone, how EMI is computed.
"Thanks to the person who posted it".
Below is the link to reach that post
http://rmathew.com/2006/calculating-emis.html
A: This question is old, but I'd like to provide an alternative answer that provides more intuition. The geometric series is good for rigor.
In an installment account, the goal is to have equal payments $A$ (installment amount) such that the loan (initial amount $P$) is paid off after $n$ periods. If there were no interest, this would be simple and we could define $A = \frac{P}{n}$. However, if we include interest and it compounds, things become more complicated. Payments made earlier on prevent interest from compounding later and we need to take this into account.
If this loan were a pure compound interest loan, the amount due at the end would be $P(1+i)^n$. Note the $(1+i)^n$ term appears in the top and bottom of the installment formula.
The Original Formula is
$$EMI = \frac{(1+i)^n}{[(1+i)^n -1]} \times (P \times i)$$
I'm going to rewrite the formula a bit so we can break it up. We have
$$EMI = \frac{P \times (1+i)^n}{P \times [(1+i)^n -1]} \times (P \times i)$$
Notice I have multiplied by $\frac{P}{P}$. Now the top of the fraction $P \times (1+i)^n$ is the same as in the compound interest formula. It is the total amount due with interest. The bottom part of the fraction is similar, except it takes away $P$. We have
$$
\begin{align*}
P \times [(1+i)^n -1] &= P(1+i)^n - P
\end{align*}
$$
So the denominator is the total amount of interest due (because we subtracted the loan amount). So the fraction gives us how many total dollars are needed per interest dollar.
Lastly, we have $(P \times i)$ which is the amount of interest that the original loan amount gains every month. When we multiple that by the fraction we get the amount of total dollars that need to be paid every month. In words
$$
\text{Monthly Payment} = \frac{\text{Total Amount with Interest}}{\text{Total Amount of Interest}} \times \text{Principle Interest Per Month}
$$
A: Let $I$ be the installment payment and $B(j)$ the balance remaining after $j$ payments.  We want to choose $I$ so that $B(n)=0$.  We are given $B(0)=P$.  Each month, the interest is applied and the payment deducted to get the new balance, so $B(j)=(1+i)\cdot B(j-1)-I$  If we write this out we get:
$B(0)=P \\ B(1)=(1+i)P-I \\ B(2)=(1+i)((1+i)P-I)-I=(1+i)^2P-(1+i)I-I \\ B(j)=(1+i)^jP-(1+i)^{(j-1)}I-(1+i)^{j-2}I-\ldots I=(1+i)^jP-\frac {(1+i)^j-1}{i}I \\ 0=(1+i)^nP-\frac{(1+i)^n-1}{i}I$
where the second equals sign in $B(j)$ comes from summing the geometric series
A: I prefer to think of the EMI formula in the following way.
First, I have the remuneration of the capital:
$\text{Future Value} = VF = P(1+i)^n$
where P = principal amount.
Then, I have a series of installments to be paid:
$\text{Installments} = I = I_1 + I_2 + ... + I_n$
Each installment is corrected retroactively by the number of months until the final payment is due, using the same interest rate:
$I_1 = f(1+i)^{n-1}$
$I_2 = f(1+i)^{n-2}$
...
$I_n = f$
where $f = \text{fixed payment}$.
So:
$I =  f(1+i)^{n-1} +f(1+i)^{n-2} ... + f = f\frac{(1+i)^n - 1}{i}$
(If you have trouble understanding the above part, just look at the sum of geometric series with finite terms and substitute $a = f$, $r = (1+i)$)
Basically, the formula is the point where those two curves intersect:
$P(1+i)^n = f\frac{(1+i)^n - 1}{i}$
After separating the terms we have:
$f = P(1+i)^n\frac{i}{(1+i)^n-1}$
A: First thing you should realise is that gp is at heart of CI
$\text I*(1+r)^n+I*(1+r)^{n-1}+......+I*(r)=P*(1+r)^n$
Here each Instalment's future value is being caculated i.e. if the interest is charged on intalment what would be the resultant. In the end the sum of resultant would to equal if it is applied to Principal itself p.s. like conservation of momentum
Just apply sum of gp to lhs and reaarange and voila
A: Let's understand this with an example.
We have taken a loan:

*

*Principal: P

*monthly interest rate: r (annual interest rate / 12)

*for the time period: n months

*Monthly EMI instalment: I

Now after paying n instalments, i.e. at month n+1, we'll be free from loan. So we'll calculate the future value of each instalment i.e. after n+1 months
Future value of the First-month instalment ( I1 ) : I1 * (1 + r)n
Future value of the Second-month instalment ( I2 ) : I2 * (1 + r)n - 1
Similarly, Future value of the last-month instalment ( In ) : In * (1 + r)1
$\Sigma$ ( future value of monthly instalment ) = Final amount = P * (1 + r) ^ n
$I∗(1 + r)^n + ∗(1 + r)^(n - 1) +...... + ∗(1 + r) = ∗(1 + r)^ n$
Now using the sum of the GP formula, we can get the final result.
Answer inspired by: https://math.stackexchange.com/a/4512409/1086435
