Question

In a geometric sequence, a4=54 and a7=1458. What is the 12th term?
a. 5,616
b. 354,294
c. 14,350,365
d. 234,812,358

Answer 1

Answer-

$\boxed{\boxed{a_{12}=354294}}$

Solution-

If there are 2 terms of a Geometric Sequence $a_n$ and $a_k$ where (n > k), then

$a_n=a_k\cdot r^{n-k}$

where, r = common ration of the G.P.

Here,

$a_4 = 54\\\\a_7=1458$

Putting in the equation,

$\Rightarrow a_7=a_4\cdot r^{7-3}$

$\Rightarrow a_7=a_4\cdot r^3$

$\Rightarrow 1458=54\cdot r^3$

$\Rightarrow r^3=\dfrac{1458}{54}$

$\Rightarrow r^3=27$

$\Rightarrow r=3$

Again putting the values in the equation,

$\Rightarrow a_{12}=a_7\cdot r^{12-7}$

$\Rightarrow a_{12}=a_7\cdot r^5$

$\Rightarrow a_{12}=1458\cdot 3^5$

$\Rightarrow a_{12}=354294$

Answer 2

The 12th term of the geometric sequence is 354,294

The nth term of a geometric sequence

The nth term of a geometric progression is given by the formula:

$a_n = ar^{n-1}$

For n = 4

$a_4=ar^{4-1}\\\\a_4=ar^3\\\\ar^3=54..................(1)$

For n = 7

$a_7=ar^{7-1}\\\\a_7=ar^6\\\\ar^6=1458................(2)$

Divide equation (2) by equation (1)

$\frac{ar^6}{ar^3} =\frac{1458}{54} \\\\r^3=27\\\\r^3=3^3\\\\r = 3$

Substitute r = 3 into equation (1)

$a(3)^3=54\\\\27a=54\\\\a=\frac{54}{27}\\\\a=2$

The 12th term of the sequence is therefore calculated below

$a_{12}=ar^{11}\\\\a_{12}=2(3^{11})\\\\a_{12}=354294$

The 12th term of the geometric sequence is 354,294

Learn more on geometric sequences here: https://brainly.com/question/1662572