Question

In an infinite G.P., the sum of the first three terms is 70. If the extreme terms are multiplied by 4 and the middle term is multiplied by 5, the resulting terms form an A.P. then the sum of infinite terms of G.P.
(a) 120
(b) $-40$
(c) 160
(d) 80

Answer 1

Hint:Assume first term and common ratio as any variable. Write the three terms in terms of the assumed variables. By given conditions, first one is their sum being 70 you will arrive at an equation. By the second condition being that this geometric progression turns into arithmetic progression you will arrive at another equation. From 2 equations, find the 2 variables, first term, common ratio. Then apply the formula of sum of infinite geometric progression being S. “a” is first term of this sequence and “r” is common ratio of it is given by:
$S=\dfrac{a}{1-r}$

Complete step-by-step answer:
Let us assume the first term of geometric progression as “a” and let us assume the common ratio of given geometric progression to be “r”. From this we will get the first three terms: first term $=$ a ; second term $=ar$ ; third term $=a{{r}^{2}}$.
Given in question: sum of first three terms, value is 70.
By substituting the values into this condition, we get
$a+ar+a{{r}^{2}}=70$
By taking “a” common on left side, we get:
$a\left( 1+r+{{r}^{2}} \right)=70$ …………………………….(i)
Given in question: when 4 is multiplied to extremes and 5 is multiplied to the middle term, it turns into A.P.
The terms at extreme positions are called extreme terms. For any the first and last positions of the sequence are called extreme positions. Here we have 3 terms so we can say first and third are the extreme terms.
Middle term is defined as the term at middle position.
By using above definitions to this sequence, we get:
Extreme terms = first term, third term.
Middle term $=$ second term
(extreme terms) 4 and (middle terms) 5, by this we get sequence : - $4a,5ar$ and $4a{{r}^{2}}$
By basic knowledge of progression if a, b, c are in A.P:
$2b=a+c$ By applying this here, we turn this terms to:
$2\left( 5ar \right)=4a+4a{{r}^{2}}$
By cancelling ‘a’ and 2 on both sides of the equation, we get
$5r=2\left( 1+{{r}^{2}} \right)$
By subtracting 5r on both sides of equation, we get
$2{{r}^{2}}-5r+2=0$
We can write this equation as $\left( r-2 \right)\left( 2r-1 \right)=0$
So, r has 2 values which are $2,\dfrac{1}{2}$ .
Take the values of ‘r’ separately, to get both values possible for ‘a’.
If $r=2$, substitute this into equation (i), we get:
a (1+2+4) = 70
By simplifying above equation, we get
$a=10$
If $r=\dfrac{1}{2}$ , substitute this into equation (i), we get:
$a\left( 1+\dfrac{1}{2}+\dfrac{1}{4} \right)=70$
By taking least common multiple, we get:
$a\left( \dfrac{7}{4} \right)=70$
By simplifying the above equation, we get:
$a=40$.
So, by applying sum of infinite terms $\Rightarrow S=\dfrac{a}{1-r}$
We get 2 values for S because we have 2 values of a corresponding to 2 values of r.
$S=\dfrac{10}{1-2},\dfrac{40}{1-\dfrac{1}{2}}=-10,80$
By options we take 80. Option (d) is the answer.

Note: Be careful while taking A.P. condition as we have a quadratic equation, we get 2 values of “r”. Be careful while calculating values of ‘a’ because we need a pair of (a, r) correspondingly. While substituting into the equation of S carefully substitute the value of ‘a’ corresponding to which value of ‘r‘, you substituted before.Students should remember the definition of geometric progression and its general term i.e $a ,ar,ar^2...ar^{n-1}$ and formula of sum of infinite geometric progression being S. “a” is first term of this sequence and “r” is common ratio of it is given by:
$S=\dfrac{a}{1-r}$.