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Question

If 32sin2α1, 14 and 342sin2α are the first three terms of an A.P. for some α, then the sixth term of this A.P. is:

A
65
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B
81
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C
78
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D
66
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Solution

The correct option is D 66
28=32sin2α1+342sin2α
28=9sin2α3+819sin2α
Let 9sin2α=t
28=t3+81t
t284t+243=0
t281t3t+243=0
t(t81)3(t81)=0
(t81)(t3)=0
t=81,3
9sin2α=92 or 3
sin2α=12,2 (not possible)
First term a=32sin2α1
a=1
Second term =14
Common difference d=13
T6=a+5d=1+5×13=66

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