Problem Solving

If the first term of a geometric progression is 1/32 and 11th term is 32, what is the 6th term?

Don't have an OA or the answer choices. Detailed explanations would be appreciated. Thanks

knight247 wrote:If the first term of a geometric progression is 1/32 and 11th term is 32, what is the 6th term?

Don't have an OA or the answer choices. Detailed explanations would be appreciated. Thanks

Here, using the geometric progression formula may come in handy.

Gn = G1[r^(n-1)], Where, Gn= the nth. term
r= common ratio, G1= the first term
n= the total number of terms under consideration.

Accordingly,

G11 = G1[r^10]
32 = [1/32][r^10]
32 * 32 = r^10
2^5 *2^5 = r^10
2^10 = r^10
r=2

Hence, G6= G1[r^5]
= [1/32][2^5]
= 1.
Therefore, the answer is 1.

let 'a' be the first term and 'r' be the common ratio
nth term = ar^(n-1)
11th term = 32
1/32*r^(11-1) = 32
r = 2.

putting in formula above,
6th term = 1/32*2^(6-1)= 1

Hi,
Just to add to the above posts, r can be -2 or +2. So, 6th term will be either -1 or +1.
a = 1/32
ar^10 = 32
So, a^2*r^10 = 1 =>(ar^5)^2 = 1
So, Sixth term, ar^5 = -1 or +1

Frankenstein wrote:Hi,
Just to add to the above posts, r can be -2 or +2. So, 6th term will be either -1 or +1.
a = 1/32
ar^10 = 32
So, a^2*r^10 = 1 =>(ar^5)^2 = 1
So, Sixth term, ar^5 = -1 or +1

Appreciated , I did not notice that r can also be -2 .

32 = (1/32)* r^10

gives r= 2

6th term = 1/32 * r^5 = 1