Question

If \[{3^{2\sin2\alpha - 1}}\] , \[14\] and \[{3^{4 – 2\sin2\alpha }}\] are the first three terms of an arithmetic progression for some \[\alpha \]. Then what is the sixth term of this arithmetic progression?
A. \[65\]
B. \[81\]
C. \[78\]
D. \[66\]

Answer 1

Hint: Since the given three terms are in the arithmetic progression. So, apply the condition for the three numbers in the arithmetic progression and create an equation. Simplify the equation using the substitution method and find the value of \[\sin2\alpha \]. Substitute the value of \[\sin2\alpha \] in the first term and calculate its value. After that, find the common difference of the given arithmetic progression. In the end, use the formula of the \[{n^{th}}\] term in the arithmetic progression to get the value of the sixth term.

Formula used:
If three numbers \[p, q\] and \[r\] are in the arithmetic progression, then \[2q = p + r\].
The formula of the \[{n^{th}}\] term in an arithmetic progression: \[{a_n} = a + \left( {n - 1} \right)d\], where \[a\] is the first term and \[d\] is the common difference.

Complete step by step solution:
Given:
\[{3^{2\sin2\alpha - 1}}\], \[14\] and \[{3^{4 – 2\sin2\alpha }}\] are the first three terms of an arithmetic progression.
Let’s apply the condition for the three numbers in an arithmetic progression.
\[2 \times 14 = {3^{2\sin2\alpha - 1}} + {3^{4 – 2\sin2\alpha }}\]
\[ \Rightarrow 28 = \dfrac{{{9^{\sin2\alpha }}}}{3} + \dfrac{{81}}{{{9^{\sin2\alpha }}}}\]
Substitute \[{9^{\sin2\alpha }} = x\] in the above equation.
\[28 = \dfrac{x}{3} + \dfrac{{81}}{x}\]
Now simplify the above equation.
\[28 = \dfrac{{{x^2} + 243}}{{3x}}\]
\[ \Rightarrow {x^2} + 243 - 84x = 0\]
Factorize the above equation.
\[{x^2} - 81x - 3x + 243 = 0\]
\[ \Rightarrow x\left( {x - 81} \right) - 3\left( {x - 81} \right) = 0\]
\[ \Rightarrow \left( {x - 81} \right)\left( {x - 3} \right) = 0\]
\[ \Rightarrow x - 81 = 0\] or \[x - 3 = 0\]
\[ \Rightarrow x = 81\] or \[x = 3\]

Resubstitute the value of \[x\].
\[{9^{sin2\alpha }} = 81\] or \[{9^{\sin2\alpha }} = 3\]
\[ \Rightarrow {9^{\sin2\alpha }} = {9^2}\] or \[{9^{\sin2\alpha }} = {9^{\dfrac{1}{2}}}\]
Now equate the exponents.
\[ \Rightarrow \sin2\alpha = 2\] or \[sin2\alpha = \dfrac{1}{2}\]
Since the range of the sine function is \[\left[ { - 1,1} \right]\] . So, \[\sin2\alpha = 2\] is rejected.
Thus, the only possible value is \[\sin2\alpha = \dfrac{1}{2}\].

Now substitute \[\sin2\alpha = \dfrac{1}{2}\] in the first term.
\[{3^{2\sin2\alpha - 1}} = {3^{2\left( {\dfrac{1}{2}} \right) - 1}}\]
\[ \Rightarrow {3^{2\sin2\alpha - 1}} = {3^{1 - 1}}\]
\[ \Rightarrow {3^{2\sin2\alpha - 1}} = {3^0}\]
\[ \Rightarrow {3^{2\sin2\alpha - 1}} = 1\]

Now calculate the common difference of the given arithmetic progression.
Let \[d\] be the common difference.
\[d = 14 - {3^{2\sin2\alpha - 1}}\]
\[ \Rightarrow d = 14 - 1\]
\[ \Rightarrow d = 13\]

Now apply the formula of the \[{n^{th}}\] term in an arithmetic progression.
\[{a_6} = 1 + \left( {6 - 1} \right)13\]
\[ \Rightarrow {a_6} = 1 + \left( 5 \right)13\]
\[ \Rightarrow {a_6} = 1 + 65\]
\[ \Rightarrow {a_6} = 66\]
The sixth term of the given arithmetic progression is \[66\].
Hence the correct option is D.

Note: Students often do a common mistake that is they are using the formula for \[{n^{th}}\] term is \[{a_n} = a + nd\] which is a wrong formula. The correct formula is \[{a_n} = a + \left( {n - 1} \right)d\]. We should also know that if three term a, b, and c are in GP then we can say that $b^2 = ac$.