Question 13 Marks
If $p, q, r, s$ are in G.P., show that $\left(p^n+q^n\right),\left(q^n+r^n\right),\left(r^n+s^n\right)$ are also in G.P.
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verify that $p^2-2 q b a+r a^2=0$.
neither.
respectively, then show that $9 S _2{ }^2= S _3\left(1+8 S _1\right)$.
$1, \frac{-4}{3}, \frac{7}{9}, \frac{-10}{27} \ldots$
$3, \frac{6}{5}, \frac{9}{25}, \frac{12}{125}, \frac{15}{625}, \ldots$
$1, \frac{2}{4}, \frac{3}{16}, \frac{4}{64}, \ldots$
G.P.
0.4 + 0.44 + 0.444 + ….
For a G.P.
For a G.P. sum of the first 3 terms is 125 and the sum of the next 3 terms is 27, find the value of r.$\therefore \quad 125=a\left(\frac{1-r^3}{1-r}\right)$.........(i)
Also, $S_6=a\left(\frac{1-r^6}{1-r}\right)$
$\therefore \quad 152=a\left(\frac{1-r^6}{1-r}\right)$.......(ii)
Dividing (ii) by (i), we get
$\frac{152}{125}=\frac{1-r^6}{1-r^3}$
$\begin{array}{ll}\therefore & \frac{152}{125}=\frac{\left(1+r^3\right)\left(1-r^3\right)}{\left(1-r^3\right)} \\ \therefore & 1+r^3=\frac{152}{125} \\ \therefore & r^3=\frac{152}{125}-1 \\ \therefore & r^3=\frac{27}{125} \\ \therefore & r=\frac{3}{5}\end{array}$
$\therefore \quad r^3=\left(\frac{3}{5}\right)^3$
common ratio.