1.9 Progressions, SPM Practice (Paper 2)


Question 3:
An arithmetic progression has 16 terms. The sum of the 16 terms is 188, and the sum of the even terms is 96. Find
(a) the first term and the common difference,
(b) the last term.

Solution:
(a) Let the first term = a
Common difference = d

Given                   S 16 =188 Thus,   16 2 [ 2a+15d ]=188               8[ 2a+15d ]=188                    2a+15d= 188 8                    2a+15d=23.5( 1 )

Given the sum of the even terms = 96
T 2 + T 4 + T 6 +…..+ T 16 =96 ( a+d )+( a+3d )+( a+5d )+…..+( a+15d )=96 8 2 [ ( a+d )+( a+15d ) ]=96 4[ 2a+16d ]=96 2a+16d=24( 2 )

(2) – (1):
16d – 15d = 24 – 23.5
d = 0.5
Substitute d = 0.5 into (2):
2a + 16 (0.5) = 24
2a + 8 = 24
2a = 16
a = 8
Therefore, first term = 8 and common difference = 0.5.

(b) Last term
= T2
= 8 + 15 (0.5)
= 8 +7.5
= 15.5


Question 4:
The third term and the sixth term of a geometric progression are 24 and 7 1 9 respectively. Find
(a) the first term and the common ratio,
(b) the sum of the first five terms,
(c) the sum of the first n terms with n is very big approaching rn ≈ 0.

Solution:
(a)
Given  T 3 =24          a r 2 =24 ………..( 1 ) Given  T 6 =7 1 9          a r 5 = 64 9  ………..( 2 ) ( 2 ) ( 1 ) : a r 5 a r 2 = 64 9 24            r 3 = 8 27            r= 2 3

Substitute r= 2 3  into ( 1 )            a ( 2 3 ) 2 =24             a( 4 9 )=24                    a=24× 9 4                      =54  the first term 54 and the common ratio is  2 3 .

(b)
S 5 = 54[ 1 ( 2 3 ) 5 ] 1 2 3    =54× 211 243 × 3 1    =140 2 3  sum of the first five term is 140 2 3 .

(c)
When 1<r<1 and n becomes  very big approaching  r n 0,   S n = a 1r        = 54  1   2 3          =162
Therefore, sum of the first n terms with n is very big approaching rn ≈ 0 is 162.

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