(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.
\(\frac{log_2 ^8 + log_2 ^ {16} - 4 log_2^ 2}{log_4^{16}}\)
= \(\frac{3log_2 ^2 + 4 log_2^ 2 - 4 log_2 2}{2 log_4^4}\)
= \(\frac{3 + 4 - 4}{2}\)
= \(\frac{3}{2}\)
(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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