(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 thr...
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(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.
Answer Details
(a) Evaluate each logarithm to base 2:
\[\log_2 8 = 3,\quad \log_2 16 = 4,\quad 4\log_2 2 = 4,\quad \log_4 16 = 2\] \[\frac{\log_2 8 + \log_2 16 - 4\log_2 2}{\log_4 16} = \frac{3 + 4 - 4}{2} = \frac{3}{2}\](b) Let the A.P. have first term \(a\) and common difference \(d\). The 1st, 3rd and 7th terms are \(a,\ a + 2d,\ a + 6d\). These are consecutive G.P. terms, so:
\[(a + 2d)^2 = a(a + 6d)\] \[a^2 + 4ad + 4d^2 = a^2 + 6ad \ \Rightarrow\ 4d^2 - 2ad = 0 \ \Rightarrow\ 2d(2d - a) = 0\]Since \(d \neq 0\), \(a = 2d\). The sum of the first two A.P. terms is 6:
\[a + (a + d) = 2a + d = 6\]Substituting \(a = 2d\): \(4d + d = 6 \Rightarrow d = \tfrac{6}{5} = 1.2\), and \(a = 2d = \tfrac{12}{5} = 2.4\).
(i) First term \(= 2.4\). (ii) Common difference \(= 1.2\).
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