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a) nth term = 3m + 1 - 2m
5th term = 35 + 1- 2(5)
= 36 - 10 = 729 - 10 = 719
b) (i) a = 580 000
nth term = 630, 400
580 000 + (n -1)d
= 630, 400
(n - 1)d
= 630 400 - 580 000
(n -1)d = 50 400
Sn = $\frac{n}{2}(2a \; +\;(n - 1)d)$
4 841 600 = $\frac{n}{2}(2a + 50 400)$
= $\frac{n}{2}(2a + 50 400)$
n = period of contract
= 8 years
ii) Annual increment
(n - 1)d = 50 400
(8 - 1)d = 50 400
7d = 50 400
d = 7 200
iii) Annual basic salary
in the third year
580 000 + (3 - 1)7 200
= 580 000 + 14 400
= sh 594 400
johnmulu answered the question on March 7, 2017 at 13:30
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