Question

The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.

Answer 1

Here, we will first use the formula to calculate the sum of first $n$ terms of a geometric progression is $S = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$ to find the number of terms. Then we will use the formula of the last term of the G.P. using the formula ${a_n} = a{r^{n - 1}}$, where $r$ is the common ratio and $n$ is the number of terms to find the required values.

Complete step-by-step answer:
We are given that the sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively.
We know that a geometric progression is a sequence of numbers in which each is multiplied by the same factor to obtain the next number in the sequence.
We know that the formula to calculate the sum of first $n$ terms of a geometric progression is $S = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$.
First, we will find the common ratio $r$ and the first term $a$ in the above series.
$\Rightarrow r = 2$
$\Rightarrow a = 5$
Using the above values of sum of given terms of G.P., $r$ and $a$ in the above formula of sum of geometric progression, we get

$$\Rightarrow 315 = \dfrac{{5\left( {{2^n} - 1} \right)}}{{2 - 1}} \\\ \Rightarrow 315 = 5\left( {{2^n} - 1} \right) \\\$$

Dividing the above equation by 5 on both sides, we get

$$\Rightarrow \dfrac{{315}}{5} = \dfrac{{5\left( {{2^n} - 1} \right)}}{5} \\\ \Rightarrow 63 = {2^n} - 1 \\\$$

Adding the above equation by 1 on both sides, we get

$$\Rightarrow 63 + 1 = {2^n} - 1 + 1 \\\ \Rightarrow 64 = {2^n} \\\ \Rightarrow {2^6} = {2^n} \\\$$

We know that when the base are same, then the powers are also same, we get
$\Rightarrow n = 6$
Thus, there are 6 terms of G.P.

We will now find the last term of the G.P. using the formula ${a_n} = a{r^{n - 1}}$, where $r$ is the common ratio and $n$ is the number of terms.
Using the values of the first terms and number of terms $n$ in the above formula, we get

$$\Rightarrow {a_6} = 5 \times {2^{6 - 1}} \\\ \Rightarrow {a_6} = 5 \times {2^5} \\\ \Rightarrow {a_6} = 5 \times 32 \\\ \Rightarrow {a_6} = 160 \\\$$

Hence, the last term of the G.P. is 160.

Note: While solving this question, be careful when we find the sum of the geometric series using the formula of first $n$ terms of a geometric progression, $S = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$. This is the key point of the question, some students try to solve this equation directly and end up with a long solution, which is time consuming and mostly leads to wrong answers.