Four numbers are in arithmetic progression. The sum of first and last term is $8$ and the product of both middle terms is $15$. The least number of the series is
Four numbers are in arithmetic progression. The sum of first and last term is $8$ and the product of both middle terms is $15$. The least number of the series is
- A
$4$
- B
$3$
- C
$2$
- D
$1$
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The number of terms of the $A.P. 3,7,11,15...$ to be taken so that the sum is $406$ is
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