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Question

For an increasing A.P. a1,a2,a3,,an, if a1+a3+a5=12
and a1a3a5=80, then which of the following is (are) TRUE?

A
a1=10
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B
a2=1
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C
a3=4
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D
a5=2
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Solution

The correct options are
A a1=10
C a3=4
D a5=2
Let a1=a and d be the common difference.
Given, a1+a3+a5=12
a+(a+2d)+(a+4d)=123a+6d=12a+2d=4 (1)

and a1a3a5=80
a(a+2d)(a+4d)=80
(42d)(4)(42d+4d)=80 [From (1)]
(42d)(4+2d)=20(2d)242=204d2=36d=±3

Since A.P. is increasing, so d=+3.
From eq(1), a=10
Hence, a1=10
a2=a+d=7a3=a+2d=4a5=a+4d=2

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