Sequences And Series Question 1
Question 1 - 24 January - Shift 1
For three positive integers $p, q, r, x^{p^{2}}=y^{q r}=z^{p^{2} r}$ and $r=p q+1$ such that $3,3 \log _{y} x, 3 \log _{z} y, 7 \log _{x} z$ are in A.P. with common difference $\frac{1}{2}$. Then $r-p-q$ is equal to
(1) 2
(2) 6
(3) 12
(4) -6
Show Answer
Answer: (1)
Solution:
Formula: Condition for three no. in AP
$pq^{2}=\log _{x} \lambda$
$qr=\log _{y} \lambda$
$p^{2} r=\log _{z} \lambda$
$\log _{y} x=\frac{qr}{pq^{2}}=\frac{r}{pq}$
$\log _{x} z=\frac{pq^{2}}{p^{2} r}=\frac{q^{2}}{pr}$
$\log _{z} y=\frac{p^{2} r}{q r}=\frac{p^{2}}{q}$
$3, \frac{3 r}{p q}, \frac{3 p^{2}}{q}, \frac{7 q^{2}}{p r}$ in A.P
$\frac{3 r}{pq}-3=\frac{1}{2}$
$r=\frac{7}{6} pq$
$r=pq+1$
$pq=6 \ldots \ldots \ldots \ldots \ldots \ldots(1)$
$\begin{aligned} & \mathrm{r}=7 \ldots \ldots \ldots \ldots \ldots \ldots . .(2) \\ & \frac{3 \mathrm{p}^2}{\mathrm{q}}=4 \\ & \left(\frac{3 p^2}{q}\right)(p q)=4 \times 6 \\ & \Rightarrow 3 p^3=24 \\ & \Rightarrow p^3=8=2^3 \\ & \Rightarrow p=2 \\ & \therefore q=\frac{6}{p}=\frac{6}{2}=3 \\ & \therefore r-p-q=7-2-3=7-5=2 . \end{aligned}$
