What is the least number of terms of a GP 2,3,9/2 to give a total great than 30

The formula for the sum of a geometric series is given by:

S = a * (r^n - 1) / (r - 1),

where:
S = sum of the series,
a = first term of the series,
r = common ratio of the series,
n = number of terms in the series.

In this case, the first term (a) is 2 and the common ratio (r) is 3/2. We want to find the least number of terms (n) that give a sum greater than 30.

Let's start by solving the inequality:

S > 30.

Substituting the values of a and r into the sum formula, we get:

2 * (3/2)^n - 1 / (3/2 - 1) > 30.

Simplifying further:

2 * (3/2)^n - 1 / (1/2) > 30.

Multiplying through by 2:

(3/2)^n - 1 > 60.

Now, we can begin solving the inequality:

(3/2)^n > 61.

To find the least number of terms that satisfy this inequality, we can start with n = 1 and keep increasing it until the inequality is satisfied.

Let's calculate the left side of the inequality for n = 1:

(3/2)^1 = 3/2 = 1.5.

Since 1.5 is less than 61, we need to try a higher value of n.

For n = 2:

(3/2)^2 = 9/4 = 2.25.

Again, 2.25 is less than 61. We continue this process until we find the least value of n that satisfies the inequality.

For n = 3:

(3/2)^3 = 27/8 = 3.375.

3.375 is also less than 61.

For n = 4:

(3/2)^4 = 81/16 = 5.0625.

5.0625 is still less than 61.

Finally, for n = 5:

(3/2)^5 = 243/32 = 7.59375.

7.59375 is greater than 61, which means the least number of terms needed to give a sum greater than 30 is 5 terms.

Therefore, to get a total greater than 30 in the geometric progression 2, 3, 9/2, you need a minimum of 5 terms.