4,x,y,32 are in GP, find x,y and the 11th term of the progression.
r = x/4
r = y/x
r = 32/y
x/4 = y/x
x^2 = 4y ----> y = x^2/4
32/y = y/x
y^2 = 32x
then : x^4/16 = 32x
x^3 = 512
x = 8
y = 64/4 = 16
the sequence is 4,8, 16, 32, ...
term11 = ar^10 = 4(2)^10 = 4096
OR, even simpler ....
a = 4
ar^4 = 32
4r^3 = 32
r^3 = 8
r = 2
so the sequence is 4, 8, 16, 32, ...
x = 8, y = 16
etc
To find x, y, and the 11th term of the geometric progression (GP) 4, x, y, 32, we can use the formula for the nth term of a GP:
\[a_n = a_1 \cdot r^{(n-1)} \]
where \(a_n\) is the nth term, \(a_1\) is the first term, r is the common ratio, and n is the position of the term in the progression.
In this case, we are given the first term, a_1 = 4, and the fourth term, a_4 = 32. We need to find the common ratio (r) and the 11th term (a_11).
Since a_4 = 4 \cdot r^{(4-1)} = 32, we can solve for the common ratio:
\[r^3 = \frac{32}{4} = 8 \]
Taking the cube root of both sides, we find that r = 2.
Now that we have the common ratio, we can find the second term (x) using the formula:
\[a_2 = a_1 \cdot r^{(2-1)} = 4 \cdot 2^1 = 8 \]
Therefore, x = 8.
Similarly, we can find the third term (y) using the formula:
\[a_3 = a_1 \cdot r^{(3-1)} = 4 \cdot 2^2 = 16 \]
Therefore, y = 16.
Finally, we can find the 11th term (a_11):
\[a_{11} = a_1 \cdot r^{(11-1)} = 4 \cdot 2^{10} = 4 \cdot 1024 = 4096 \]
So, the 11th term of the progression is 4096.