Three numbers form a GP. If the first and last term are 5 and 245 respectively,find two possible values for the middle number

r^2 = 245/5 = 81

so r = ±9
now finish it off

To find the two possible values for the middle number of a geometric progression (GP) with first term 5 and last term 245, we can use the formula for the nth term of a GP:

an = a1 * r^(n-1),

where a1 is the first term, an is the nth term, and r is the common ratio.

First, let's find the common ratio (r). We can use the formula for finding the common ratio of a GP:

r = (an / a1)^(1/(n-1)).

In this case, a1 = 5 and an = 245, so the common ratio (r) can be calculated as:

r = (245 / 5)^(1/(n-1)).

Since we are looking for two possible values for the middle number, we can consider two cases: one with n = 3, and one with n = 4.

Case 1: n = 3
For n = 3, we have the three terms: a1, a2, and a3.
a1 = 5
a2 = a1 * r = 5 * r
and
a3 = a2 * r = 5 * r * r = 5 * r^2.

Since a3 should be 245, we can set up the following equation:

5 * r^2 = 245.

Simplifying the equation:

r^2 = 245 / 5,
r^2 = 49,
r = √49,
r = ±7.

Thus, for n = 3, the two possible values for the middle number are 5 * 7 = 35 and 5 * (-7) = -35.

Case 2: n = 4
For n = 4, we have the four terms: a1, a2, a3, and a4.
a1 = 5
a2 = a1 * r = 5 * r
a3 = a2 * r = 5 * r * r = 5 * r^2
and
a4 = a3 * r = 5 * r^2 * r = 5 * r^3.

Since a4 should be 245, we can set up the following equation:

5 * r^3 = 245.

Divide both sides by 5:

r^3 = 245 / 5,
r^3 = 49,
r = ∛49,
r = ±3.

Thus, for n = 4, the two possible values for the middle number are 5 * 3 = 15 and 5 * (-3) = -15.

Therefore, the two possible values for the middle number of the geometric progression are 35, -35, 15, and -15.