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Question

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
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B
3
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C
2
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D
1
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Solution

The correct option is D $$1$$
Let the terms be
$$a+3d,a+d,a-d+a-3d$$
It is given that
$$T_{1}+T_{4}=8$$
Or
$$a+3d+(a-3d)=8$$
$$2a=8$$
$$a=4$$ ...(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 $$1$$.

Mathematics

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