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In the four numbers first three are in G.P. and last three are in A.P. whose common difference is 6 . If the first and last numbers are same, then first will be
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The correct answer is:
$8$
Let the numbers be $\frac{a}{r}, a, a r, 2 a r-a$\ldots(i)
Where first three numbers are in G.P. and last three are in A.P. Given that the common difference of A.P. is 6 , so
$a r-a=6$\ldots(ii)
Also given, $\frac{a}{r}=2 a r-a \Rightarrow \frac{a}{r}=2(a r-a)+a$
$\Rightarrow \quad \frac{a}{r}=2(6)+a$
$\Rightarrow\left(\frac{a}{r}\right)-a=12 \Rightarrow$ $\Rightarrow a(1-r)=12 r \Rightarrow r=-\frac{1}{2}$
From (i) we get, $a\left[\left(-\frac{1}{2}\right)-1\right]=6$ or $a=-4$
Required numbers from (i) are $8,-4,2,8$.
Where first three numbers are in G.P. and last three are in A.P. Given that the common difference of A.P. is 6 , so
$a r-a=6$\ldots(ii)
Also given, $\frac{a}{r}=2 a r-a \Rightarrow \frac{a}{r}=2(a r-a)+a$
$\Rightarrow \quad \frac{a}{r}=2(6)+a$
$\Rightarrow\left(\frac{a}{r}\right)-a=12 \Rightarrow$ $\Rightarrow a(1-r)=12 r \Rightarrow r=-\frac{1}{2}$
From (i) we get, $a\left[\left(-\frac{1}{2}\right)-1\right]=6$ or $a=-4$
Required numbers from (i) are $8,-4,2,8$.
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