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SEQUENCES AND SERIES — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSSEQUENCES AND SERIES4 Marks
Question
If a, b, c are in A.P.; b, c, d are in G.P. and $\frac { 1 } { c } , \frac { 1 } { d } , \frac { 1 } { e }$ are in A.P., prove that a, c, e are in G.P.
✓

Answer

Since, $a, b, c$ are in A.P.
$\therefore b-a=c-b$
$\Rightarrow 2 b=a+c$
$\Rightarrow b=\frac{a+c}{2}$
Since, b, c, $d$ are in G.P.
$\therefore \frac{c}{b}=\frac{d}{c}$
$\Rightarrow c^2=b d .$……….(i)
Also $\frac{1}{c}, \frac{1}{d}, \frac{1}{e}$ are in A.P.
$\therefore \frac{1}{d}-\frac{1}{c}=\frac{1}{e}-\frac{1}{d}$
$\Rightarrow \frac{2}{d}=\frac{1}{c}+\frac{1}{e}$
$\Rightarrow \frac{2}{d}=\frac{c+e}{c e}$
$\Rightarrow d=\frac{2 c e}{c+e}$
Putting values of $b$ and $d$ in eq. (i), $c^2=\left(\frac{c+a}{2}\right)\left(\frac{2 c e}{c+e}\right)$
$\Rightarrow c^2=\frac{c e(c+a)}{c+e}$
$\Rightarrow c^2(c+e)=e c(c+a)$
$\Rightarrow c^2+c e=c e+a e$
$\Rightarrow c^2=a e$ which shows that $a, c$, e are in G.P.

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