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8. Sequences and Series
hard
અહી $a_1, a_2, a_3 \ldots$ એ ધન વધતી સમગુણોતર શ્રેણીમાં છે . જો $a_1 a_5=28$ અને $a_2+a_4=29$ તો $a_6$ ની કિમંત મેળવો.
A$628$
B$526$
C$784$
D$812$
(JEE MAIN-2025)
Solution
$a_1 \cdot a_5=28 \Rightarrow a \cdot a r^4=28 \Rightarrow a^2 r^4=28…..(1) $
$a_2+a_4=29 \Rightarrow a r+a r^3=29 $
$\Rightarrow \operatorname{ar}\left(1+r^2\right)=29 $
$\Rightarrow a^2 r^2\left(1+r^2\right)^2=(29)^2……(2)$
By Eq. $(1) \& (2)$
$\frac{r^2}{\left(1+r^2\right)^2}=\frac{28}{29 \times 29} $
$\Rightarrow \frac{r}{1+r^2}=\frac{\sqrt{28}}{29} \Rightarrow r=\sqrt{28} $
$\because a^2 r^4=28 \Rightarrow a^2 \times(28)^2=28 $
$\Rightarrow a=\frac{1}{\sqrt{28}} $
$\therefore a_6=a r^5=\frac{1}{\sqrt{28}} \times(28)^2 \sqrt{28}=784$
$a_2+a_4=29 \Rightarrow a r+a r^3=29 $
$\Rightarrow \operatorname{ar}\left(1+r^2\right)=29 $
$\Rightarrow a^2 r^2\left(1+r^2\right)^2=(29)^2……(2)$
By Eq. $(1) \& (2)$
$\frac{r^2}{\left(1+r^2\right)^2}=\frac{28}{29 \times 29} $
$\Rightarrow \frac{r}{1+r^2}=\frac{\sqrt{28}}{29} \Rightarrow r=\sqrt{28} $
$\because a^2 r^4=28 \Rightarrow a^2 \times(28)^2=28 $
$\Rightarrow a=\frac{1}{\sqrt{28}} $
$\therefore a_6=a r^5=\frac{1}{\sqrt{28}} \times(28)^2 \sqrt{28}=784$
Standard 11
Mathematics