8. Sequences and Series
medium

एक गुणोत्तर श्रेणी के प्रथम तीन पदों का योगफल $\frac{13}{12}$ है तथा उनका गुणानफल $1$ है, तो सार्व अनुपात तथा पदों को ज्ञात कीजिए ?

Option A
Option B
Option C
Option D

Solution

Let $\frac{a}{r}, a,$ ar be the first three terms of the $G.P.$ Then

$\frac{a}{r}+a r+a=\frac{13}{12}$       ……..$(1)$

and     $\left(\frac{a}{r}\right)(a)(a r)=-1$       ……..$(2)$

From $(2),$ we get $a^{3}=-1,$ i.e., $a=-1$ (considering only real roots)

Substituting $a=-1$ in $(1),$ we have

$-\frac{1}{r}-1-r=\frac{13}{12}$ or $12 r^{2}+25 r+12=0$

This is a quadratic in $r$, solving, we get $r=-\frac{3}{4}$ or $-\frac{4}{3}$

Thus, the three terms of $G.P.$ are $: \frac{4}{3},-1, \frac{3}{4}$ for $r=\frac{-3}{4}$ and $\frac{3}{4},-1, \frac{4}{3}$ for $r=\frac{-4}{3}$

Standard 11
Mathematics

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