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.

4

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1

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Solution

The correct option is D $1$
Let the terms be
$a + 3 d, a + d, a - d + a - 3 d$
It is given that
$T_{1} + T_{4} = 8$
Or
$a + 3 d + (a - 3 d) = 8$
$2 a = 8$
$a = 4$ ...(i)
And
$T_{2} \times T_{3} = 15$
$(a + d) (a - d) = 15$
$a^{2} - d^{2} = 15$
Now from i $a = 4$ .
Hence
$16 - d^{2} = 15$
$d^{2} = 1$
$d = 1$
Therefore the numbers are
$7, 5, 3, 1$
Hence the least number of the sequence is $1$ .

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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.

The correct option is D 1Let the terms bea+3d,a+d,a−d+a−3dIt is given that T1+T4=8Or a+3d+(a−3d)=82a=8a=4 ...(i)And T2×T3=15(a+d)(a−d)=15a2−d2=15Now from i a=4.Hence 16−d2=15d2=1d=1Therefore the numbers are 7,5,3,1Hence the least number of the sequence is 1.

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.