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Q.
Three numbers are in $G.P$. whose sum is $70$ . If the extremes is each multiplied by $4$ and the mean by $5,$ they will be in
A.P. Then the sum of numbers is
Sequences and Series
Solution:
Let the numbers be $a, a r, a r^{2}$.
Sum $=70$
$\Rightarrow a\left(1+r+r^{2}\right)=70 \,\,\,\,\, (1)$
It is given that $4 a, 5 a r, 4 a r^{2}$ are in $A.P$.
$\Rightarrow 2(5 a r)=4 a+4 a r^{2}$
$\Rightarrow 5 r=2+2 r^{2}$
$\Rightarrow 2 r^{2}-5 r+2=0$
$\Rightarrow (2 r-1)(r-2)=0$
$\Rightarrow r=2$ or $r=1 / 2$
Putting $r=2$ in $(1),$ we get $a=10$
Putting $r=1 / 2$ in (1), we get $a=40$.
Hence, the numbers are $10,20,40$ or $40,20,10$ .
Also sum of the numbers is $70$ .