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જો $a _{1}(>0), a _{2}, a _{3}, a _{4}, a _{5}$ સમગુણોતર શ્રેણીમાં હોય, $a _{2}+ a _{4}=2 a _{3}+1$ અને $3 a _{2}+ a _{3}=2 a _{4}$,હોય તો,$a _{2}+ a _{4}+2 a _{5}=\dots\dots\dots$
$30$
$20$
$30$
$40$
Solution
$a _{1}>0, a _{2}, a _{3}, a _{4}, a _{5} \rightarrow G \cdot P .$
$3 a _{2}+ a _{3}=2 a _{4}$
$3 ar + ar ^{2}=2 ar ^{3}$
$3+ r =2 r ^{2}$
$2 r ^{2}- r -3=0$
$r =-1$ and $r =\frac{3}{2}$
$a _{2}+ a _{4}=2 a _{3}+1$
$ar + ar ^{3}=2 ar ^{2}+1$
$a \left( r + r ^{3}-2 r ^{2}\right)=1$
$a\left(\frac{3}{2}+\frac{27}{8}-\frac{18}{4}\right)=1$
$a=\frac{8}{3}$
$When \;r =-1, a =-\frac{1}{4}\; (rejected, a _{1} > 0)$
$r =\frac{2}{3}, a =\frac{8}{3}(\text { selected })$
Now
$a_{2}+a_{4}+2 a_{5}$
$=\frac{8}{3} \times \frac{3}{2}+\frac{8}{3} \times \frac{27}{8}+2 \times \frac{8}{3} \times \frac{81}{16}$
$=4+9+27=40$
