Suppose the ratio of the term to the next term is r

\(a_1 = 25\)

\(a_{15} = a_1* r^{14} = 100\)

\(r^{14}= \frac{100}{25} = 4\)

\(r^7 = 2\)

\(a_8 = a_1 * r^7 = 25*2 = 50\)

the ratio of the term to the next term is a fixed constant - highlights that the sequence is a geometric progression.

\(T_n = a*(r)^{n-1}\) (a - first term, r - common ratio, n - number of terms)
\(T_{15} = 25*(r)^{14}=4*25\)
\(T_{15} = (r)^{7}=2\)

\(T_8 = a*(r)^{8-1}= 25*r^{7}=50\)

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
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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
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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.
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Yes, that would be correct.

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