If the common difference of an A.P is 1 and 6th term is \(\cfrac{(4c

1.	If the common difference of an A.P is 1 and 6th term is \(\cfrac{(4cos^2α+1)}{cos^2α}\) , then first term isA) cot2 α B) cosec2 α C) tan2 α D) sec2 α
Answer» Correct option is (C) tan² α We have common difference = d = 1 & \(6^{th}\) term \(=a_6=\frac{4\,cos^2\alpha+1}{cos^2\alpha}\) Let a be the first term of A.P. Then a+5d \(=\frac{4\,cos^2\alpha+1}{cos^2\alpha}\) \((\because a_6=a+5d)\) \(\Rightarrow\) \(a=\frac{4\,cos^2\alpha+1}{cos^2\alpha}-5d\) \(=\frac{4\,cos^2\alpha+1}{cos^2\alpha}-5\) \((\because d=1)\) \(=\frac{4\,cos^2\alpha+1-5\,cos^2\alpha}{cos^2\alpha}\) \(=\frac{1-cos^2\alpha}{cos^2\alpha}=\frac{sin^2\alpha}{cos^2\alpha}=tan^2\alpha\) \((\because1-cos^2\alpha=sin^2\alpha)\) Hence, first term of A.P. is \(tan^2 \alpha.\) Correct option is C) tan² α

If the common difference of an A.P is 1 and 6th term is \(\cfrac{(4cos^2α+1)}{cos^2α}\) , then first term isA) cot2 α B) cosec2 α C) tan2 α D) sec2 α

Answer»

Correct option is (C) tan² α

We have common difference = d = 1

& \(6^{th}\) term \(=a_6=\frac{4\,cos^2\alpha+1}{cos^2\alpha}\)

Let a be the first term of A.P.

Then a+5d \(=\frac{4\,cos^2\alpha+1}{cos^2\alpha}\) \((\because a_6=a+5d)\)

\(\Rightarrow\) \(a=\frac{4\,cos^2\alpha+1}{cos^2\alpha}-5d\)

\(=\frac{4\,cos^2\alpha+1}{cos^2\alpha}-5\) \((\because d=1)\)

\(=\frac{4\,cos^2\alpha+1-5\,cos^2\alpha}{cos^2\alpha}\)

\(=\frac{1-cos^2\alpha}{cos^2\alpha}=\frac{sin^2\alpha}{cos^2\alpha}=tan^2\alpha\) \((\because1-cos^2\alpha=sin^2\alpha)\)

Hence, first term of A.P. is \(tan^2 \alpha.\)

Correct option is C) tan² α