Page 100 - Mathematics_Form_3

Sequences and series

Since there are n terms, each with Substituting the values gives,
( A + A ), then equation (7) becomes, 16
___
1 n S = 2 × 3 + (16 − 1 ) × 7
2S = ( nA + A n ) (8) 16 2 ( )
n
1

16
Dividing equation (8) by 2 both sides = 6 + 105
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(
)
2
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gives,
n = 8 × 111
S = ( A + A ) (9)
n
2 1 n = 888
Therefore, the formula S = n ( A + A n )
1
n
2
gives the sum of the first n terms of an Therefore, the sum of the first sixteen
arithmetic progression, where A is terms is 888.
1
the first term and A is the last term. In
n
other words, the sum equals the number Example 4.12
of terms times the average of the first
and the last terms. The sum of the first The first term of an arithmetic
n terms of an arithmetic progression progression is 2 and the last term
can be expressed by another formula is 46. If the arithmetic progression
obtained by replacing A in equation (9)
n consists of 23 terms, find the sum of
with A = A + (n − 1 ) d.
n 1 all the terms.
Thus, equation (9) becomes,
n Solution
S = ( A + ( A + (n − 1)d ))
n
2 1 1 Given A = 2, A = 46, n = 23. To find
1
23
n the sum of first 23 terms proceed as
= (2A + (n − 1)d )
2 1 follows.
n
Therefore, S = n 2 (2A + 1 (n − 1)d ). The formula for the sum of the first n
terms is given as
n
)
Example 4.11 S = (A + A.
n
2 1 n
Find the sum of the first sixteen terms Substituting the given values gives,
of the arithmetic series 23 Mathematics for Secondary Schools
+
3 + 10 + 17 + 24 + ... S = 2 (2 46 )
23
Solution 23
From the series given, d = 7, A = 3, = 2 × 48
1
and n = 16 = 552
Using the formula,
n Therefore, the sum of the first 23 terms
S = n 2 (2A + 1 (n − 1)d ). is 552.

Student\s Book Form Three 93

18/09/2025 09:59:22
MATHEMATIC F3 SB.indd 93 18/09/2025 09:59:22
MATHEMATIC F3 SB.indd 93