Four numbers are in an arithmetic progression | vedclass

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

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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 $A_1 + A_4 = 8$ $(i)$ and $A_2 	imes A_3 = 15$ $(ii)$.In an arithmetic progression,the sum of terms equidistant from the beginning and end is constant,so $A_1 + A_4 = A_2 + A_3 = 8$ $(iii)$.From $(ii)$ and $(iii)$,we have $A_2 + A_3 = 8$ and $A_2 	imes A_3 = 15$.Substituting $A_3 = 8 - A_2$ into the product equation: $A_2(8 - A_2) = 15 \Rightarrow A_2^2 - 8A_2 + 15 = 0$.Solving the quadratic equation: $(A_2 - 3)(A_2 - 5) = 0$,so $A_2 = 3$ or $A_2 = 5$.If $A_2 = 3$,then $A_3 = 5$. If $A_2 = 5$,then $A_3 = 3$.The common difference $d = A_3 - A_2 = 5 - 3 = 2$ or $3 - 5 = -2$.Case $1$: $A_2 = 3, A_3 = 5$. Then $d = 2$. $A_1 = A_2 - d = 3 - 2 = 1$ and $A_4 = A_3 + d = 5 + 2 = 7$.The sequence is $1, 3, 5, 7$.Case $2$: $A_2 = 5, A_3 = 3$. Then $d = -2$. $A_1 = A_2 - d = 5 - (-2) = 7$ and $A_4 = A_3 + d = 3 + (-2) = 1$.The sequence is $7, 5, 3, 1$.In both cases,the smallest number is $1$.

Four numbers are in an arithmetic progression. The sum of the first and last terms is $8$ and the product of the two middle terms is $15$. What is the smallest number in the sequence?

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