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Q.

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

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a

2

b

1

c

3

d

4

answer is D.

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Detailed Solution

$a - 3 d, a - d, a + d, a + 3 d are in A . P \begin{matrix} a - 3 d + a + 3 d = 8 \\ 2 a = 8 \\ a = 4 \end{matrix} \begin{matrix} (a - d) (a + d) = 15 \\ a^{2} - d^{2} = 15 \\ a = 4 \Rightarrow 16 - d^{2} = 15 \Rightarrow d = \pm 1 \end{matrix} ∴ nos are 1, 3, 5, 7$

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