I'm not sure if this is the right place to ask this kind of a question, but here it goes.

I have the following information:

  1. A = Total amount needed
  2. B = Number of periods (monthly, 12 payments of equal value then increased by the percentage increase repeated for B number of periods or annually (once per year) increasing by percentage increase every time for B number of periods)
  3. C = Percentage Increase (whether increasing every payment or every 12th payment)

I'm tasked with finding:

  1. The starting amount payment value (X) to equal the total amount needed (A) if paid out for (B) periods, and increasing by (C) percent

I know I can find the starting value in today's dollars or the end value according to

A * ((1 + C/100) ^ B) = X or A = X / ((1 + C/100) ^ B)

I also know you can quickly calculate something like compound interest with this equation:

A * ((1 + C/100) ^ (B / 12)) = X

for 12 variable, compounding payments in a period split to equal (C) rate of return.

What is the proper equation to find the starting payment amount (then increasing by (C) percent every period) for my inquiry? I'm looking for two conditions, the starting payment value whilst receiving 12 payments of equal value and then increasing by (C) percent for the next 12 payments (yearly) or receiving one payment per year, each increasing by (C) percent over the previous payment.

Excuse my ignorance if this is a duplicate question, but I'm having a hard time finding an applicable answer on this site given my lack of terminology.


I'm looking around and I came across this site: http://www.hughchou.org/calc/formula_deriv.php

The page above is about paying off a mortgage, which isn't quite what I'm looking for. The value I'm trying to find is static, not an increasing loan amount. It's just receiving increasing payments to meet that defined specific value. I think that would be on the right track, but omitting the increase in principal.

  • Does this mortgage with escalated payments have an interest rate?
    – DJohnM
    Dec 18, 2013 at 20:05
  • What you seek is easily done with a spreadsheet. The formula for each cell isn't tough, and if the increase is constant it can be triggered from a cell. So the sheet builds from the balance, rate, original amortization, and then the payment cell can increase based on the formula you define. Dec 19, 2013 at 0:41
  • I need an equation, not to work this out on a spreadsheet. Trial and error is not what I'm looking for. I need an equation or a for-loop summation to be used in an application I'm building.
    – slickplaid
    Dec 19, 2013 at 14:10
  • Also, calling it a mortgage is a misnomer. It's more like an annuity and the ending total payout has no real interest rate. I have the ending total payout balance and I need to figure out the starting payment dollar amount given X periods and Y rate that the payments will increase to equal exactly Z in payout.
    – slickplaid
    Dec 19, 2013 at 14:19

2 Answers 2


The following formula will calculate the constant regular payment amount to accumulate a total amount q. The formula to use an inflation-linked payment amount is added further below.

x = (c*q)/((1 + c)*((1 + c)^n - 1))


x is periodic payment

c is the periodic interest rate as a decimal

n is the number of periods

q is the total amount needed


c = 0.01
n = 4
q = 1000

x = (c*q)/((1 + c)*((1 + c)^n - 1)) = 243.843

Reversing the calculation to check: 4 payments accumulating varying interest according to the time on deposit :-

x*(1 + c) + x*(1 + c)^2 + x*(1 + c)^3 + x*(1 + c)^4 = 1000

Here the payment is made at the beginning of each period, so by the end of four periods a total of 1000 is accumulated. The first payment accumulates four cycles of interest, i.e. x*(1 + c)^4 and the last payment only accumulates one cycle of interest: x*(1 + c).

For monthly calculations the figures will work out if the monthly interest is calculated from the annual percentage rate (apr) like so :-

apr = 5.0

c = (1 + apr/100)^(1/12) - 1 = 0.00407412


q = 1000
n = 12

x = (c*q)/((1 + c)*((1 + c)^n - 1)) = 81.1519

Checking by accumulating payments and interest :-

x*(1 + c) + x*(1 + c)^2 + x*(1 + c)^3 +
x*(1 + c)^4 + x*(1 + c)^5 + x*(1 + c)^6 +
x*(1 + c)^7 + x*(1 + c)^8 + x*(1 + c)^9 +
x*(1 + c)^10 + x*(1 + c)^11 + x*(1 + c)^12 = 1000


Adding in an inflation term, inflating the payment at each period by z.

First an example calculation showing what is being accumulated at each period :-

Initial payment x = 10

Periodic interest rate c = 0.01

Inflation rate z = 0.005

p1 = ( 0 + x*(1 + z)^0) * (1 + c) = 10.1
p2 = (p1 + x*(1 + z)^1) * (1 + c) = 20.3515
p3 = (p2 + x*(1 + z)^2) * (1 + c) = 30.7563
p4 = (p3 + x*(1 + z)^3) * (1 + c) = 41.3161
p5 = (p4 + x*(1 + z)^4) * (1 + c) = 52.0328

So the total amount required in five periods is 52.0328.

The formula to work out the initial payment (x) is :-

x = (q*(c - z))/((1 + c)*((1 + c)^n - (1 + z)^n))

where q is the total amount.


c = 0.01
z = 0.005
n = 5
q = 52.0328

x = (q*(c - z))/((1 + c)*((1 + c)^n - (1 + z)^n)) = 10

Standard annuity formulas can determine variables when a flow is changing over time.

In this case, the continually changing annuity can be applied to determine the initial payment or the future value, setting r=0, FV_GA=A, n=B, g=C, and P=X.

  • I'm not sure why someone downvoted you. This is actually very applicable to what I'm looking for. I think all I need to do is rework the equation to solve for the starting payment of the growing annuity. Thank you! If this works out, I'll accept your answer. This page might be more relevant: financeformulas.net/… I'll look at it here in a little while.
    – slickplaid
    Dec 19, 2013 at 14:14
  • @slickplaid - When I put my demo figures into the Growing Annuity Payment formula in your link I get x = q*((c - z)/((1 + c)^n - (1 + z)^n)) = 10.1 instead of 10.0 as I would expect. Jan 19, 2014 at 18:52

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