If the third term is 3 of a geometric is 36 and the sixth term is 9/2, what is the explicit formula for the sequence ?

Question

Ariana Alvarez · Accepted Answer

The formula will be #144(1/2)^(n - 1)# where #n ≥ 1#.

If #36# is the 3rd term, than the 4th term is #36r#, the 5th term is #36(r)(r) = 36r^2# and the 6th term is #36r(r)(r) = 36r^3#. Thus: #36r^3 = 9/2# #r^3 = 9/72# #r^3 = 1/8# #r = 1/2# The first term will be #144#, because the third term is #36#, so by the common ratio the second term is #36/(1/2) = 72# and the first is #72/(1/2) = 144#. So the formula for the sequence is #144(1/2)^(n - 1)#, because the count starts at #n = 1# for term #1#. Hopefully this helps!

Elias Ackerman · Answer

#a_n = 144*(1/2)^(n-1)#

The general (#n^(th)#) term of a geometric sequence is given by:

#a_n = a_1*r^(n-1)#

Where #a_1# is the first term and (#r#) the common ratio

In this example we are given two terms, as follows:

#a_3 = 36 and a_6 = 9/2#

Applying the formula for the general term above:

#36 = a_1r^2# [A]

#9/2 = a_1r^5# [B]

[A] #-> r^2 = 36/a_1# [C]

[C] in [B] #-> 9/2 = a_1 * (36/a_1)^(5/2)#

Square both sides:

#81/4 = a_1^2 * (36/a_1)^(5)#

#81/4 = 36^5/a_1^3#

Expressing #a_1^3# in prime factors

#a_1^3 = (2*2*(3*3*2*2)^5)/(3*3)^2#

#= (2*2)^6*(3*3)^3#

#:. a_1 = (2*2)^2*(3*3)#

#= 16xx9 = 144#

#a_1 = 144# in [C] #-> r^2 = 36/144 = 1/4#

Hence, #r = +-sqrt(1/4) = =+-1/2#

Since #a_3 and a_6>0 -> r>0#

#:. r= 1/2#

Thus our general term for the sequence is:

#a_n = 144*(1/2)^(n-1)#

Liam Brennan · Answer

undefined To find the explicit formula for the geometric sequence, we can use the formula:

$$ a_n = a_1 \times r^{(n-1)} $$

where:
- $ a_n $ is the nth term of the sequence
- $ a_1 $ is the first term of the sequence
- $ r $ is the common ratio of the sequence

Given that the third term is 36 and the sixth term is $ \frac{9}{2} $, we can form two equations using the formula:

For the third term:
$$ a_3 = a_1 \times r^{(3-1)} = 36 $$

For the sixth term:
$$ a_6 = a_1 \times r^{(6-1)} = \frac{9}{2} $$

We can then solve these equations simultaneously to find the values of $ a_1 $ and $ r $. Once we find these values, we can write the explicit formula for the sequence.

Let's solve the equations:

From the third term equation:
$$ a_1 \times r^2 = 36 $$

From the sixth term equation:
$$ a_1 \times r^5 = \frac{9}{2} $$

Divide the equation for the sixth term by the equation for the third term to eliminate $ a_1 $:
$$ \frac{r^5}{r^2} = \frac{\frac{9}{2}}{36} $$

$$ r^3 = \frac{1}{8} $$

$$ r = \sqrt[3]{\frac{1}{8}} $$

$$ r = \frac{1}{2} $$

Substitute $ r = \frac{1}{2} $ into the third term equation to find $ a_1 $:
$$ a_1 \times (\frac{1}{2})^2 = 36 $$

$$ a_1 \times \frac{1}{4} = 36 $$

$$ a_1 = 36 \times 4 $$

$$ a_1 = 144 $$

Therefore, the explicit formula for the sequence is:

$$ a_n = 144 \times (\frac{1}{2})^{(n-1)} $$