Question

In a certain sequence of all positive terms, {a1, a2, a3, …} each term equals the previous term times a constant factor. If (a1)(a5) = 900, what is the value of a3?

Answer 1

Let $k$ be the constant factor between terms, so that

$a_2=ka_1$

$a_3=ka_2=k^2a_1$

and so on, with

$a_n=ka_{n-1}=\cdots=k^{n-1}a_1$

(notice how the exponent and the subscript add to $n$)

Then

$a_5=k^4a_1$

so if

$a_1a_5=900$

then

$k^4{a_1}^2=900\implies k^4=\dfrac{900}{{a_1}^2}=\left(\dfrac{30}{a_1}\right)^2\implies k^2=\dfrac{30}{a_1}$

Now,

$a_3=k^2a_1=\dfrac{30a_1}{a_1}=30$

so the third term in the sequence is 30.

Answer 2

Using a geometric sequence, it is found that the value is $a_3 = 30$

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Geometric sequence:

In a geometric sequence, the quotient between consecutive terms is always the same, that is, each term equals the previous term times a constant factor, and this factor is called common ratio.

The nth term of a geometric sequence is given by:

$a_n = a_1q^{n-1}$

In which $a_1$ is the first term and q is the common ratio.

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Considering $a_1a_5 = 900$:

The 5th term is: $a_5 = a_1q^{4}$

Thus:

$a_1a_1q^{4} = 900$

$a_1^2q^4 = 900$

$a_1^2 = \frac{900}{q^4}$

$a_1 = \sqrt{\frac{900}{q^4}}$

$a_1 = \frac{30}{q^2}$

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The third term is:

$a_3 = a_1q^2$

Since $a_1 = \frac{30}{q^2}$

$a_3 = \frac{30}{q^2}\times q^2$

$a_3 = 30$

The value is $a_3 = 30$

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