Find two geometric progressions having: 54 as third term and 27/32 as ninth term.
T9 is the 6th term after T3, so
27/32 = 54*r^6
1/64 = r^6
r = ± 1/2
I think you can do the rest.
Expand (1-x)^4.Hence find S if S = (1-x^3)^4 - (1-x^3)^3 + 6(1-x^3)^2 - 4(1-x^3) + 1.
To find two geometric progressions with specific terms, we need to know at least two terms of each progression and the common ratio (r) between the terms.
Let's assume the first geometric progression has a common ratio of r1 and its third term is 54.
The general formula for the terms of a geometric progression is given by:
an = a1 * r^(n-1)
For the first progression, the third term (a3) is given as 54, and we need to find the second term (a2), first term (a1), and the common ratio (r1).
a3 = a1 * r1^(3 - 1)
54 = a1 * r1^2
Next, let's assume the second geometric progression has a common ratio of r2 and its ninth term is 27/32.
For the second progression:
a9 = a1 * r2^(9 - 1)
27/32 = a1 * r2^8
So, we have two equations with two unknowns:
Equation 1: 54 = a1 * r1^2
Equation 2: 27/32 = a1 * r2^8
You can solve these equations simultaneously to find the values of a1, r1, a1, and r2.