Maths

22.
Let the first term be a and the common ratio be r.
a + ar = 54
a (1 + r) = 54
a = 54 / (1 + r) … (1)
a / (1 – r) = 72 … (2)

Substitute (1) into (2):
[54 / (1 + r)] / (1 – r) = 72
54 / (1 + r) = 72(1 – r)
54 = 72 (1 – r) (1 + r)
3 = 4 (1 – r^2)
4 – 4r^2 = 3
4r^2 – 1 = 0
r^2 = 1/4
r = ±1/2
When r = 1/2,
a = 54 / (1 + 1/2)
a = 54 / (3/2)
a = 36
When r = -1/2,
a = 54 / (1 – (1/2))
a = 54 / (1/2)
a = 108

∴ The first term is 108 and the common ratio is -1/2 or the first term is 36 and the common ratio is 1/2.

26.
(a)
Since logx + log2, logx2 + log4, logx3 + log8 are in arithmetic sequence.
The first term,
a = log(x) + log2
= log(2x)
The common difference,
d = (logx2 + log4) – (logx + log2)
= 2logx + 2log2 – logx – log2
= logx + log2
= log(2x)
The general term
= (log(2x)) + (n – 1) (logx + log2)
= log(2x) + n(logx + log2) – log(2x)
= n(logx + log 2)
= nlog(2x)

(b)
∵ the given sequence is an arithmetic sequence,
∴ the sum of the first ten terms
= (10/2)[2(log(2x)) + (10 – 1)(log(2x))]
= 5[2log(2x) + 9log(2x)]
= 5(11log(2x))
= 55log(2x)

logx+2log*2log+log4*3log+log8