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
B
3
C
2
D
1

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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