Sequences and Series 3 Question 3
3. The product of three consecutive terms of a GP is 512 . If 4 is added to each of the first and the second of these terms, the three terms now form an AP. Then, the sum of the original three terms of the given GP is
(2019 Main, 12 Jan I)
(a) 36
(b) 28
(c) 32
(d) 24
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Answer:
Correct Answer: 3. (b)
Solution:
- Let the three consecutive terms of a GP are $\frac{a}{r}, a$ and $a r$.
Now, according to the question, we have
$$ \begin{aligned} & \frac{a}{r} \cdot a \cdot a r=512 \\ & \Rightarrow \quad a^{3}=512 \\ & \Rightarrow \quad a=8 \end{aligned} $$
Also, after adding 4 to first two terms, we get
$\frac{8}{r}+4,8+4,8 r$ are in $AP$
$$ \begin{array}{lr} \Rightarrow & 2(12)=\frac{8}{8}+4+8 r \\ \Rightarrow & 24=\frac{8}{r}+8 r+4 \\ \Rightarrow & 5=\frac{2}{r}+2 r \\ \Rightarrow & 2 r^{2}-4 r-r+2=0 \\ \Rightarrow & 2 r(r-2)-1(r-2)=0 \\ \Rightarrow & (r-2)(2 r-1)=0 \\ \Rightarrow & r=2, \frac{1}{2} \end{array} $$
Thus, the terms are either $16,8,4$ or $4,8,16$. Hence, required sum $=28$.
