A sequence Q consists of 15 numbers arranged in ascending order.

Given:
1. A sequence Q consists of 15 numbers arranged in ascending order.
2. The first term in this sequence is 25.
3. In this sequence, for the first 14 terms, the ratio of the term to the next term is a fixed constant.
4. The last term of the sequence is four times the first term.

Asked: What is the 8th term in the sequence?

1. A sequence Q consists of 15 numbers arranged in ascending order.
n = 15
\(a_1<a_2<a_3<.....<a_{15}\)

2. The first term in this sequence is 25.
\(a_1 = 25\)

3. In this sequence, for the first 14 terms, the ratio of the term to the next term is a fixed constant.
Let the ratio be r
\(a_2 = a_1*r\)
\(a_n = a_1*r^{n-1} \)

4. The last term of the sequence is four times the first term.
\(a_{15} = 4 * a_1 = r^{14} * a_1\)
\(r^7 = 2\); \(r^7 = -2\) is not feasible since \(a_1<a_2<a_3<.....<a_{15}\)

The 8th term in the sequence = \(a_ 1 * r^7 = 25 * 2 = 50 \)

IMO A

Thanking everyone for the clear understanding provided.

Solution:

We see that the sequence is a geometric sequence since the ratio between every two consecutive terms is a constant, also known as the common ratio. Therefore, if a(1) is the first term and r is the common ratio, then a(2) = a(1) * r and a(3) = a(1) * r^2, and so on. Notice that the last term, a(15), is a(1) * r^14. Since we are given that a(1) = 25 and a(15) = 4 * a(1) = 100. We have:

a(15) = a(1) * r^14

100 = 25r^14

4 = r^14

2 = r^7

Notice that the eighth term, a(8), is a(1) * r^7. So a(8) = 25 * 2 = 50.

Answer: A

chetan2u

In this question would it be correct to say that since 8th term is exactly halfway in the series, it will be 2 times the first term?
because 15th term is 4 times the first term and the ratio throughout the series in constant.

Yes, that would be correct.

The terms are \(a,ar,ar^2,....ar^{14}\)
15th term is \(ar^{14}=4a....r^{14}=(r^7)^2=4......r^7=2\)........(i)
Eighth term is \(ar^7\)
Now, using (i), the 8th term is \(ar^7=a*2\), so 2 times the first term = 2*25 = 50.

Did not understand how we solved r^7 = 2 ? Can someone pls explain

The question is testing the concept of geometric progression (GP).

In a GP the expression for \(n^{th}\) term is \(a.r^{n-1}\)

a = first term
r = common ratio
n = number of terms

\(15^{th}\) term = \(25.r^{14}\)

From the question stem we also know that 15th term in 4 times the first term

\( 25.4 = 25.r^{14}\)

Dividing the right hand side and left hand side by 25

\(r^{14} = 4\)

\((r^7)^2 = 2^2\)

\(r^7 = 2\)

\(8^{th}\) term = \(25.r^7\)
= 50


Hope that helps.

You will never require to solve for \(r^7=2\) in GMAT. You may have to replace r^7 with 2 somewhere.
For example value of 25r^7=25*2=50.

But if you want to know what is r then
\(r^7=2........r=2^{\frac{1}{7}}=\sqrt[7]{2}\)

This is a seriously flawed question. It restricts the geometic relationship to the first 14 terms. So if the 15th term happens to follow the same condition, which is what most people believed in their responses above, then the answer could be 50.

However, it is also possible, as an counterexample, that r =1.000000001 and the 15th term does not follow the relationship of the first 14 terms. In this case, each of the first 14 terms would be roughly 25 and the 15th term would be 100. Following this logic, there are infinitely many values of r>1 such that all conditions are still met, and each gives a different 8th term ranging from 25 to 50.