Question

Найдите первый член геометрической прогрессии, если S4=40; q=3 пожалуйсто помогите.

Answer 1

$S_{4} = \frac{ b_{1}*(3^4-1) }{3-1} =40 \\ \\ b_{1}= \frac{40*2}{81-1} =1$

Answer 2

Finding the First Term of a Geometric Progression

To find the first term (b1) of a geometric progression when S4=40 and q=3, we can use the formula for the sum of the first n terms of a geometric progression:

S_n = b_1 * (1 - q^n) / (1 - q)

Where: - S_n is the sum of the first n terms - b_1 is the first term - q is the common ratio - n is the number of terms

We are given S4=40 and q=3. We can use these values to find b1.

Calculation

Using the given values, we can calculate the first term (b1) as follows:

S4 = b1 * (1 - q^4) / (1 - q)

Substitute the given values: 40 = b1 * (1 - 3^4) / (1 - 3)

Solve for b1: 40 = b1 * (1 - 81) / -2 40 = b1 * (-80) / -2 40 = 40 * b1

So, the first term (b1) of the geometric progression is 1.

Conclusion

The first term of the geometric progression is 1. This is calculated using the given sum of the first 4 terms (S4=40) and the common ratio (q=3).