What is the formula to find a random term's number in a geometric progression using the random term?

Question

I think it is the inverse of this formula: RANDOM TERM = (SUM * (1-COMMON RATIO) / (1-COMMON RATIO^x).

Answer 1

nth term of GP

an = a1(r^(n-1))

take a specific example

you know an=32 and next term = 64

the GP is 1, 2, 4, 8 16, 32, 64, ...

64/32 = 2 = r = common ratio

an = 32 = a1(2^(n-1))

32/a1 = 2^(n-1)

ln(32/a1) = (n-1)ln2

n-1 = ln(32/a1)/ln2

n = 1 + ln(32/a1)/ln2

n = 1 + ln32/ln2

n = 1 + log base 2 of 32

n = 1 +5

n = 6 = the 6th term

Answer 2

Using common G.P. notation:

a_n = a₁r^n-1

a_n / a₁ = r^n-1

ln(a_n / a₁) = ln r^n-1

ln(a_n / a₁) = (n - 1) ln r

[ln(a_n / a₁)] / ln r = n - 1

{[ln a_n / a₁] / ln r} + 1 = n

Answer 3

The formula to find the position or number of a term in a geometric progression (GP) when you know the value of the term (random term) is given by:

n = ( [log(r/a)] / [log(r)] ) + 1

Where:

- \( n \) is the position or number of the term,

- \( T \) is the value of the random term,

- \( a \) is the first term of the geometric progression, and

- \( r \) is the common ratio of the geometric progression.

In this formula, \(\log\) denotes the logarithm. You can use either the natural logarithm (base \(e\)) or the logarithm with any other base, as long as you consistently use the same base for both logarithms.

Make sure that the value inside the logarithms is positive.

This formula is derived from the formula for the n-th term of a geometric progression, which is given by:

T_n = a * r^(n-1)

By rearranging this formula, you can solve for n, resulting in the formula mentioned earlier.