Page 101 - Mathematics_Form_3

Sequences and series

Example 4.13 The formula for the sum of first n

The first term of an arithmetic terms of an AP is given by
progression is 2 and the common S = n n (2A + 1 (n − 1 d
) ).
difference is 5. If the sum of the first n 2
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terms is 245, find the number of terms Substituting the given values into the
of the progression. formula gives,

24
___
Solution S = 2 × 2860000 + (24 − 1 ) ×
24 2 (
Given A = 2, d = 5, S = 245.
1 n 445 250
)
) ).
From S = n n (2A + 1 (n − 1 d
2
Substituting the values into the formula = 12(5720000 + 10240750)
gives, = 191,529,000
n Therefore, in the first 24 years, Alinda
245 = 2 (2 2×+ ( 5 n − 1 )). will have saved Tshs 191,529,000 .
5 n − n − 490 = 0
2
Solving for values of n, it gives n = 10 (b) Recall the formula for the n term
th
or n = − 9.8 . of an AP. That is,
Since the number of terms cannot be A = A + (n − 1 ) d
negative, then n = 10 . n 1
Therefore, there are 10 terms in the Substituting the given values gives,
arithmetic progression. A = 2860000 + (13 − 1 ) × 445250
13
= 2860000 + 5343000
Example 4.14 Therefore, in the thirteenth year, she
Mathematics for Secondary Schools (a) How much money will she have 1. Find the sum of the following
= 8,203,000
Alinda saved Tshs 2,860,000 in the
first year of employment and each year
will save Tshs 8,203,000.
afterwards she saves Tshs 445,250
more than the previous year.
Exercise 4.4
saved in the first 24 years?
(b) How much money will she save in
arithmetic progression:
the thirteenth year?
Solution (a) − 10 − 7 − 4 − · · · + 50
(a) Given A = Tshs 2,860,000, (b) a + 3a + 5a + · · · + 21a
1
d = Tshs 445,250, n = 24 years. (c) 2.01 + 2.02 + 2.03 + · · · + 3.00

94 Student\s Book Form Three

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