SPM Practice Question 5


Question 5 (6 marks):
The sum of the first n terms of an arithmetic progression, Sn is given by S n = 3n( n33 ) 2 .  
Find
(a) the sum of the first 10 terms,
(b) the first term and the common difference,
(c) the value of q, given that qth term is the first positive term of the progression.

Solution:
(a)
S n = 3n( n33 ) 2 S 10 = 3( 10 )( 1033 ) 2 S 10 =345

(b)
S n = 3n( n33 ) 2 S 1 = 3( 1 )( 133 ) 2 S 1 =48 T 1 = S 1 =48 First term, a= T 1 =48 T n = S n S n1 T 2 = S 2 S 1 T 2 = 3( 2 )( 233 ) 2 ( 48 ) T 2 =45 Common difference, d = T 2 T 1 =45( 48 ) =3

(c)
First positive term,  T q >0 T q >0 a+( q1 )d>0 48+( q1 )3>0 48+3q3>0 3q>51 q>17 Thus, q=18.