Question

The geometric sequence 20, 60, 180, 540,………, 393660 has how many terms?

Answer 1

It is given then the given sequence is a geometric sequence, so first of all we are going to write the general expression for the given geometric sequence which is equal to $a{{r}^{n-1}}$. In this formula, “a” is the first term of the sequence and “r” is the common ratio. Now, the first term of the sequence is self explanatory what it is and the common ratio is calculated by dividing the number by its previous number. Then to find the number of total terms given in the sequence, we will take the last term of the sequence and equate it to the formula $a{{r}^{n-1}}$ and substitute the value of “a'' and “r” and then find the value of “n”. Here, “n” is the total number of terms.

Complete step-by-step solution:
The geometric sequence given in the above problem is as follows:
20, 60, 180, 540,………, 393660
We know that the general term for any geometric sequence is as follows:
$a{{r}^{n-1}}$
In the above formula, “a” is the first term of the sequence and “r” is the common ratio of the geometric sequence. Now, the first term in the given sequence is equal to 20.
And the common ratio is calculated by dividing any term by its previous term in the following way:
Let us take a term say 180 then we are going to divide this term by its previous term (which is equal to 60) so dividing 180 by 60 we get the common ratio of the geometric sequence:
$\begin{aligned} & \Rightarrow \dfrac{180}{60} \\\ & =3 \\\ \end{aligned}$
From the above, we got the common ratio of the geometric sequence as 3. Now, substituting the values of “a” and “r” in the general term formula we get,
$\begin{aligned} & \Rightarrow a{{r}^{n-1}} \\\ & =20{{\left( 3 \right)}^{n-1}} \\\ \end{aligned}$
Now, to find the total number of terms we are going to equate the above formula to 393660 and we get,
$\Rightarrow 20{{\left( 3 \right)}^{n-1}}=393660$
Dividing 20 on both the sides of the above equation we get,
$\begin{aligned} & \Rightarrow {{\left( 3 \right)}^{n-1}}=\dfrac{393660}{20} \\\ & \Rightarrow {{\left( 3 \right)}^{n-1}}=19683 \\\ \end{aligned}$
Now, we should see which power of 3 will give us the R.H.S of the above equation and we get,
$\Rightarrow {{\left( 3 \right)}^{n-1}}={{3}^{9}}$
As base of the powers written in the L.H.S and R.H.S is same so we can equate the power of the two bases and we get,
$\begin{aligned} & \Rightarrow n-1=9 \\\ & \Rightarrow n=10 \\\ \end{aligned}$
Hence, the number of terms in the above sequence is 10.

Note: If unfortunately you forgot the formula for the general term of the given sequence then also you can find the number of terms. As you know that in geometric sequence, ratio is common between two consecutive terms and by dividing the two consecutive terms, you will get the common ratio then multiply 10 to 540 then multiply 10 to the result of the multiplication of 540 to 10 and then see how many ten multiplications starting from the first term will give you the last term.