How do I make an algebraic formula for a number that's increasing by 74%?

Starting number:

93.13, I need a formula to go up by 74% 100 times.

Step-by-step explanation:

If the number is increasing at a constant rate of 74%, this rate of increment is in geometric progression. The formula for the nth term of a geometric sequence can be expressed as

Tn = ar^ (n - 1)

Where

n represents the number of terms in there sequence.

a represents the first term of the sequence.

r represents the common ratio

From the information given,

The number increases by 74%, it means that

r = 1 + 74/100 = 1 + 0.74 = 1.74

Starting number is 93.13 so

a = 93.13

It would increase 100 times so

n = 100

Therefore, the formula will be

Tn = 93.13 * 1.74^ (n - 1)

When n = 100,

T100 = 93.13 * 1.74^ (100 - 1)

T100 = 93.13 * 1.74^99

T100 = 6521.92 * 10^20