Question

Find G.P for which the sum of the first two terms is $ - 4 $ and the fifth term is 4 times the third term.

Answer 1

Hint: Here in this question G.P means geometric progression is of the form $ a,ar,a{r^2},a{r^3},... $ and the nth term is defined as $ {a_n} = a{r^{n - 1}} $ where r is the common ratio. By using data and using the geometric progression we obtain the solution for the question.

Complete step-by-step answer:
By the given data we have the fifth term is 4 times the third term. The general form of the G.P is given by $ a,ar,a{r^2},a{r^3},... $ where the nth term is defined as $ {a_n} = a{r^{n - 1}} $ . Therefore, the fifth term is expressed as $ {a_5} = a{r^{5 - 1}} \Rightarrow {a_5} = a{r^4} $ and the third term is expressed as $ {a_3} = a{r^{3 - 1}} \Rightarrow {a_3} = a{r^2} $ . By the given data we have the fifth term is 4 times the third term so we have,
 $ {a_5} = 4{a_3} $
 $ \Rightarrow a{r^4} = 4a{r^2} $
On simplification we have,
 $ \Rightarrow \dfrac{{a{r^4}}}{{a{r^2}}} = 4 $
By cancelling the same quantity, we have,
 $ \Rightarrow {r^2} = 4 $
Applying the square root, we have
 $ r = \pm 2 $
We have found the common ratio and we have to find the first element of the progression.
By the given data we have the sum of the first two terms is -4.
We have formula for the sum of the G.P we have two kind of depending on the common ratio and that is defined as
If the common ratio greater than 1 we have $ {S_n} = \dfrac{{a({r^n} - 1)}}{{r - 1}} $ and if the common ratio is less than 1 we have $ {S_n} = \dfrac{{a(1 - {r^n})}}{{1 - r}} $
Suppose $ r > 1 $ that is $ r = 2 $ we have
 $ {S_n} = \dfrac{{a({r^n} - 1)}}{{r - 1}} $
 $ \Rightarrow {S_2} = \dfrac{{a({2^2} - 1)}}{{2 - 1}} $
Since $ {S_2} = - 4 $ we have
 $
   \Rightarrow - 4 = \dfrac{{a(4 - 1)}}{1} \\
   \Rightarrow - 4 = 3a \\
   \Rightarrow a = \dfrac{{ - 4}}{3} \;
  $
So we have the first term $ a = \dfrac{{ - 4}}{3} $ and $ r = 2 $ then the geometric progression is $ \dfrac{{ - 4}}{3},\dfrac{{ - 8}}{3},\dfrac{{ - 16}}{3},.... $
So, the correct answer is “ $ \dfrac{{ - 4}}{3},\dfrac{{ - 8}}{3},\dfrac{{ - 16}}{3},.... $ ”.

Suppose $ r < 1 $ that is $ r = - 2 $ we have
 $ {S_n} = \dfrac{{a(1 - {r^n})}}{{1 - r}} $
 $ \Rightarrow {S_2} = \dfrac{{a(1 - {{( - 2)}^2})}}{{1 - ( - 2)}} $
Since $ {S_2} = - 4 $ we have
 $
   \Rightarrow - 4 = \dfrac{{a(1 - 4)}}{3} \\
   \Rightarrow - 4 = \dfrac{{ - 3a}}{3} \\
   \Rightarrow a = 4 \;
  $
So we have the first term $ a = 4 $ and $ r = - 2 $ then the geometric progression is $ 4, - 8,16, - 32,.... $
So, the correct answer is “$ 4, - 8,16, - 32,.... $ ”.

Note: We must know about the geometric progression arrangement and it is based on the first term and common ratio. The common ratio of the geometric progression is defined as $ \dfrac{{{a_2}}}{{{a_1}}} $ where $ {a_2} $ represents the second term and $ {a_1} $ represents the first term. The sum of n terms is defined on the basis of common ratio.