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MathematicsSequence & SeriesA.P.Medium2 minPYQ_2020
MathematicsMediumstatement

If32sin2α-1,14and34-2sin2αare the first three terms of an A.P. for someα, then the sixth term of this A.P. is

Options:

Answer:
A
Solution:

If a, b, c are in AP, then b is A.M of a & c

 2 b=a+c

28=32sin2α-1+34-2sin2α

Putting, 32sin2α=x we get,

28=x3+81xx2-84x+243=0

(x-3)(x-81)=0

 32sin2α=3 or 34

sin2α=12sin2α2

Terms are 1,14,27, then

T6=1+5(13)=66

 

 

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:A.P.
2mℹ️ Source: PYQ_2020

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