How many terms in the geometric progression 1, 1:1, 1:21, 1:331,.... Will be needed so the sum of the first n Terms is greater than 20?

To find the number of terms needed so that the sum of the first n terms is greater than 20, we need to first find the common ratio of the geometric progression.

The common ratio can be found by dividing any term by the previous term. In this case, we can see that the common ratio is 1:1:10.

The sum of the first n terms of a geometric progression can be calculated using the formula:

S_n = a(1 - r^n) / (1 - r)

where:
S_n is the sum of the first n terms
a is the first term
r is the common ratio
n is the number of terms

We want to find the smallest value of n such that S_n is greater than 20. Let's substitute the values into the formula:

20 < 1(1 - 10^n) / (1 - 10)

Solving for n:

20 < (1 - 10^n) / -9
-180 < 1 - 10^n
-181 < -10^n
10^n < 181
n > log(181) / log(10)
n > 2.257

Since n must be a whole number, we need at least 3 terms for the sum of the first n terms to be greater than 20.

The sum of the first n terms of a geometric progression is given by:

$$S_n = \frac{a(1 - r^n)}{1 - r}$$

where a is the first term and r is the common ratio.

In this case, a = 1 and r = 1/11. Substituting these values into the formula, we get:

$$S_n = \frac{1(1 - (1/11)^n)}{1 - (1/11)} = \frac{1 - (1/11)^n}{10/11}$$

We want to find the smallest value of n such that:

$$\frac{1 - (1/11)^n}{10/11} > 20$$

$$\Rightarrow 1 - (1/11)^n > 200/11$$

$$\Rightarrow (1/11)^n < 1/11$$

$$\Rightarrow n > \log_{1/11}(1/11) = 1$$

Therefore, we need at least 2 terms so that the sum of the first n terms is greater than 20.