Question 5 (6 marks):
The sum of the first n terms of an arithmetic progression, Sn is given by Sn=3n(n−33)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)
Sn=3n(n−33)2S10=3(10)(10−33)2S10=−345
(b)
Sn=3n(n−33)2S1=3(1)(1−33)2S1=−48T1=S1=−48First term, a=T1=−48Tn=Sn−Sn−1T2=S2−S1T2=3(2)(2−33)2−(−48)T2=−45Common difference, d=T2−T1=−45−(−48)=3
(c)
First positive term, Tq>0Tq>0a+(q−1)d>0−48+(q−1)3>0−48+3q−3>03q>51q>17Thus, q=18.
The sum of the first n terms of an arithmetic progression, Sn is given by Sn=3n(n−33)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)
Sn=3n(n−33)2S10=3(10)(10−33)2S10=−345
(b)
Sn=3n(n−33)2S1=3(1)(1−33)2S1=−48T1=S1=−48First term, a=T1=−48Tn=Sn−Sn−1T2=S2−S1T2=3(2)(2−33)2−(−48)T2=−45Common difference, d=T2−T1=−45−(−48)=3
(c)
First positive term, Tq>0Tq>0a+(q−1)d>0−48+(q−1)3>0−48+3q−3>03q>51q>17Thus, q=18.