In a set of four numbers,the first three are | vedclass

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(D) Let the four numbers be $\frac{a}{r}, a, ar, x$. Since the last three numbers $a, ar, x$ are in $A.P.$ with common difference $d = 6$,we have $ar

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$8$

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(D) Let the four numbers be $\frac{a}{r}, a, ar, x$. Since the last three numbers $a, ar, x$ are in $A.P.$ with common difference $d = 6$,we have $ar - a = 6$ and $x - ar = 6$. Thus,$x = ar + 6$. Since $ar = a + 6$,we have $x = (a + 6) + 6 = a + 12$. Given that the first and last numbers are equal,$\frac{a}{r} = x = a + 12$. From $ar - a = 6$,we get $a(r - 1) = 6$,so $r - 1 = \frac{6}{a}$,which means $r = 1 + \frac{6}{a} = \frac{a + 6}{a}$. Substituting $r$ into $\frac{a}{r} = a + 12$: $\frac{a}{(a + 6)/a} = a + 12$ $\frac{a^2}{a + 6} = a + 12$ $a^2 = (a + 12)(a + 6)$ $a^2 = a^2 + 6a + 12a + 72$ $0 = 18a + 72$ $18a = -72$ $a = -4$. The first number is $\frac{a}{r} = a + 12 = -4 + 12 = 8$.

In a set of four numbers,the first three are in $G.P.$ and the last three are in $A.P.$ with a common difference of $6$. If the first and last numbers are equal,then the first number is:

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