Question

Let $a_1, a_2, a_3 \ldots \ldots \ldots . .$. be an arithmetic progression and $b_1, b_2, b_3, \ldots \ldots$ be a geometric progression. The sequence $c _1, c _2, c _3, \ldots \ldots \ldots .$. is such that $c _{ n }= a _{ n }+ b _{ n } \forall n \in N$. Suppose $c _1=1, c _2=4$, $c_3=15$ and $c_4=2$.
The common ratio of geometric progression is equal to

• A. - 2
• B. - 3
• C. 2
• D. 3

Answer 1

Answer: (B)

We have
$c_1=a+b=1 $ ....(1)
$c_2=(a+d)+b r=4$....(2)
$c_3=(a+2 d)+b r^2=15 $....(3)
$c_4=(a+3 d)+b r^3=2$....(4)
Now,
(2) $-(1) \Rightarrow d+b(r-1)=3$....(5)
(3) $-(2) \Rightarrow d+b\left(r^2-r\right)=11$....(6)
(4) $-(3) \Rightarrow d+b\left(r^3-r^2\right)=-13$....(7)
Also,
$(6) -(5) \Rightarrow b\left(r^2-2 r+1\right)=8$....(8)
$(7)-(6) \Rightarrow b\left(r^3-2 r^2+r\right)=-24$....(9)
So, $\frac{(8)}{(9)} \Rightarrow \frac{( r -1)^2}{ r ( r -1)^2}=\frac{1}{ r }=\frac{-1}{3} \Rightarrow r =-3 ($ As $r \neq 1)$
$\Rightarrow a =\frac{1}{2}, b =\frac{1}{2}$ and $d =5$
Clearly common ratio of geometric progresssion is $r = - 3$