6.0k views
4 votes
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.

1 Answer

3 votes

Answer:

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

User JelleP
by
8.1k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories