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Question

If first three terms of sequence 116,a,b,16 are in geometric series and last three terms are in harmonic series, then the value of a and b will be


A

a=-14,b=1

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B

a=112,b=19

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C

A and B both are true

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D

None of these

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Solution

The correct option is C

A and B both are true


Explanation for the correct option:

Step 1. Find the value of a and b:

Given, 116,a,b are in GP

a2=b16 ….(1)

Also, a,b,16 are in HP

b=2a6a+16

b=2a6a+1 ….(2)

Step 2. Substitute the value of b in equation (1), we get

a2=2a166a+1

a2=a86a+1

a=186a+1

48a2+8a=1

48a2+8a-1=0

(12a-1)(4a+1)=0

a=112,-14

Step 3. Find the value of b by putting the values of a:

When a=112,

b=21126112+1=218=19

When a=-14,

b=2-146-14+1=-2-2=1

b=19,1

Hence, Option ‘C’ is Correct.


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