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(D) Let the four numbers in arithmetic progression be $A_1, A_2, A_3, A_4$.Given that $A_1 + A_4 = 8$ $(i)$ and $A_2 	imes A_3 = 15$ $(ii)$.In an ari

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$1$

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(D) Let the four numbers in arithmetic progression be $A_1, A_2, A_3, A_4$.Given that $A_1 + A_4 = 8$ $(i)$ and $A_2 	imes A_3 = 15$ $(ii)$.In an arithmetic progression,the sum of terms equidistant from the beginning and the end is constant and equal to the sum of the first and last terms.Therefore,$A_2 + A_3 = A_1 + A_4 = 8$ $(iii)$.From $(ii)$ and $(iii)$,we have $A_2 + \frac{15}{A_2} = 8$,which simplifies to $A_2^2 - 8A_2 + 15 = 0$.Solving the quadratic equation,we get $(A_2 - 3)(A_2 - 5) = 0$,so $A_2 = 3$ or $A_2 = 5$.If $A_2 = 3$,then $A_3 = 5$. If $A_2 = 5$,then $A_3 = 3$.Using the property $A_2 = \frac{A_1 + A_3}{2}$,we find $A_1 = 2A_2 - A_3$.For $A_2 = 3$ and $A_3 = 5$,$A_1 = 2(3) - 5 = 1$. Then $A_4 = 8 - 1 = 7$.The series is $1, 3, 5, 7$. The least number is $1$.

Four numbers are in arithmetic progression. The sum of the first and last term is $8$ and the product of both middle terms is $15$. The least number of the series is

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