Question

What is $a_4$ when $a_1=2$ , $r=-3?$
A. $27$
B. $-27$
C. $-54$
D. $54$

Answer 1

Hint: As we know that above question is related to GP series or Geometric progression series. It is a sequence of non zero numbers where each term after the first is found by multiplying the previous one by a fixed number called the common ratio. We know the GP formula for $n^{th}$ term i.e. $a_n=a{r}^{n-1}$ .

Complete step-by-step answer:
In the given question we have been given $a_1=2$ , $r=-3$ . We have to find $a_4$ .
We know that the general form of the Geometric series is
 $a_1+a_{2}r+a_3{r}^2+...a{r}^n$ , where $a_1$ is the first term, $a_2$ is the second term and so on… and $r$ is the common ratio.
So in the given series we have $a_1=2$ and common ratio $r=-3$ . And our $nth$ term i.e. $n=4$ .
Now by applying the formula we can write $a_4=2\times (-3{)}^{4-1}$ .
On solving we have $2\times -3^3=2\times (-27)$ . It gives us the value $-54$ .
Hence the correct option is (c) $-54$ .
So, the correct answer is “Option C”.

Note: We should note that if the geometric series is finite then we take the formulas for finding the sum as $S_n={\displaystyle \frac{a(r^n-1)}{r-1}};r>1$ and if $r<1$ , then the formula is $S_n={\displaystyle \frac{a(1-r^n)}{1-r}}$ . Before solving such questions we should be well aware of the geometric progressions and their formulas. We should do the calculations very carefully especially while finding the sums of terms using the formula. It should be noted that the sum of n terms of arithmetic progression is given by ${\displaystyle \frac{1}{2}}\left(2a+(n-1)d\right)$ .