Let's denote the first term of the geometric progression as "a" and the common ratio as "r".

Given that the sum of the first three terms is 39, we can write the equation:

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Also, the sum of the first six terms is 156, so we have:

$true$

To find the first term and common ratio, we can solve these two equations simultaneously.

Subtracting equation (1) from equation (2), we get:

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$true$ .....(3)

Dividing equation (2) by equation (1), we get:

$true$

Substituting equation (3) into equation (4), we have:

$true$

Now, we can solve this quadratic equation to find the values of "r".

Using the quadratic formula, we have:

r=(-1±√((1^2-4(1)(-2)) ))/(2(1))

r=(-1±√((1+8) ))/2

r=(-1±√9)/2

$r=\frac{(-1\pm \sqrt(1^2-4(1)(-2)))}{2\times1}=(-1\pm \sqrt9)/2$

Simplifying further, we have:

r = (-1 ± 3) / 2

Therefore, we have two possible values for "r":

r1 = (-1 + 3) / 2 = 1

r2 = (-1 - 3) / 2 = -2

Now, we substitute each value of "r" back into equation (1) to find the corresponding value of "a".

For r = 1:

$a+a\times1 + a\times1^2=39$

a + a + a = 39

3a = 39

a = 39 / 3

a = 13

For r = -2:

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a - 2a + 4a = 39

3a = 39

a = 39 / 3

a = 13

Therefore, we have two possible solutions for the first term and the common ratio:

Solution 1: a = 13, r = 1

Solution 2: a = 13, r = -2

Hence, the first term of the geometric progression is 13, and the common ratio can be either 1 or -2.

Absolute Value