Question 1109475
(a) the formula for the nth term of an arithmetic progression is
:
x(n) = a + d(n-1), where a is the first term and d is the common difference
:
we are given
:
4 = a + d(3-1)
10 = a + d(7-1)
:
simplify both equations
:
(1) a + 2d = 4
(2) a + 6d = 10
:
we two equations in 2 unknowns
:
solve equation 1 for a
:
a = 4 - 2d
:
substitute for a in equation 2
:
(4 - 2d) + 6d = 10
:
4d = 6 
:
d = 6/4 = 1.5
:
using equation 1
:
a + 2(1.5) = 4
:
a + 3 = 4
:
a = 1
:
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the arithmetic progression for the pole heights is
:
1m, 2.5m, 4m, 5.5m, 7m, 8.5m, 10m, 11.5m, 13m, 14.5m, 16m, 17.5m
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(b) the formula for the sum of the first n terms of an arithmetic progression is
:
s(n) = (1/2) * n * (first term + the last term)
:
we want to sum the first 12 terms
:
s(12) = (1/2) * 12 * (1 + 17.5) = 111m
:
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cost of all the poles is 111m * 45.00naira per m = 4995.00naira 
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