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(a) Simplify; \(\frac{log_2 ^8 + log_2 ^{16} - 4 log_2 ^2}{log_4^{16}}\) (b) The first, ...

(a) Simplify; \(\frac{log_2 ^8 + log_2 ^{16} - 4 log_2 ^2}{log_4^{16}}\)
(b) The first, third, and seventh terms of an Arithmetic Progression (A.P) from three consecutive terms of a Geometric Progression (G.P). If the sum of the first two terms of the A.P is 6, find its:
(I) first term; (ii) common difference.
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    Correct Answer: Option
    Explanation:
    \(\frac{log_2 ^8 + log_2 ^ {16} - 4 log_2^ 2}{log_4^{16}}\)
    = \(\frac{3log_2 ^2 + 4 log_2^ 2 - 4 log_2 2}{2 log_4^4}\)
    = \(\frac{3 + 4 - 4}{2}\)
    = \(\frac{3}{2}\)

    (b)
    AP GP
    T\(_1\)T\(_3\)T\(_7\) T\(_1\)T\(_2\)T\(_3\)




    T\(_1\) + T\(_2\) = 6
    a + a + d = 6
    2a + d = 6.....(i)
    a = a
    a + 2d = ar, a + 6d = ar\(^2\)
    \(\frac{a + 2d}{a} = \frac{a + 6d}{a + 2d}\)
    (a + 2d)\(^2\) = \(\frac{a + 6d}{a + 2d}\)
    (a + 2d)\(^2\) = a(a + 6d)
    a\(^2\) + 4d\(^2\) + 4ad = a\(^2\) + 6ad
    4d\(^2\) = 6ad - 4ad
    4d\(^2\) = 2ad
    \(\frac{4d}{2} = \frac{2a}{2}\)
    a = 2d ......(ii)
    from;
    2a + d = 6
    2(2d) + d = 6
    4d + d = 6
    5d = 6
    d = \(\frac{6}{5}\)
    = 1\(\frac{1}{5}\)
    from; a = 2d
    a = \(\frac{2 \times 6}{5}\)
    = \(\frac{12}{5}\)
    = 2\(\frac{2}{5}\)
    First term = 2\(\frac{2}{5}\)
    Common difference = 1\(\frac{4}{5}\)

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