How many terms in the GP, 4, 3.6, 3.24,...... are needed so the sum exceeds 35?

To find the number of terms needed for the sum of the geometric progression to exceed 35, we can use the formula for the sum of a geometric progression:

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

where:
S_n = sum of the first n terms
a = first term of the geometric progression
r = common ratio
n = number of terms

In this case, the first term (a) is 4 and the common ratio (r) is 3.6 / 4 = 0.9. We want to find the number of terms (n) needed for the sum to exceed 35.

Plugging in the values, we get:

35 < 4 * (1 - 0.9^n) / (1 - 0.9)

Simplifying:

35 < 4 * (1 - 0.9^n) / 0.1

8.75 < 1 - 0.9^n

0.9^n < 0.75

Taking the natural logarithm of both sides:

ln(0.9^n) < ln(0.75)

n * ln(0.9) < ln(0.75)

n > ln(0.75) / ln(0.9)

n > 6.79

Therefore, we need at least 7 terms in the geometric progression for the sum to exceed 35.

Let there be n terms in the GP. Then,

$$S_n=4\left(\frac{1-\left(\frac34\right)^n}{1-\frac34}\right)=4-4\left(\frac34\right)^n$$

$$S_n>35\\\ \Rightarrow 4-4\left(\frac34\right)^n>35\\\ \Rightarrow 4\left(\frac34\right)^n<4-35\\\ \Rightarrow \left(\frac34\right)^n<\frac{-31}{4}=-7.75\\\ \Rightarrow n<\log_{\frac34}(-7.75)\approx5.43$$

So, n=5 is the required number of terms.