Chapter 11 In a contest all
Chapter 1
1. In a contest ,all twenty finalists were given cash prize.The first winner was given RM860 , the second RM 800 ,the third RM740 and so on.Calculate the total amount of money awarded to all the finalists.
Solution:The prizes are RM860,RM800 ,RM740…..
It is an arithmetic progression .The first term a=rm860 , common difference d=-60 and n=20. The total amount given is
Sn = n/2[ 2a + (n-1)]
S26 =20/2[ 2(860) +(20-1)(-60)] =RM5800
The amount is RM 5800
2. Find the minimum number of terms that must be taken from the sequence : 4,16,6,256…so that the sum is more than 400.
Solution:
This is a geometric sequence with a=4 and r =4
Let the minimum number of terms be n.
From Sn= a(rn-1)/r-1 , we get 4(4n-1)/4-1 > 400 4(4n-10)> 1200 4n-1> 300 4n> 301 N log 4 > log 301 n> 4.117 n> 5
3. Given P , P+1 and P-2 are three consecutive terms of a geometric sequence , find P and the common ratio.
r = P+1/P= P-2/P+1 (P+1)2=P(P-2) P2+ 2P +1 = P2-2P P2-P2+2P + 2P =-1 4P=-1 P=-1/4 r= (-1/4+1)/(-1/4) =-3
Thus ,P= -1/4 and the common ratio r=-3
4. Azrul starts with a monthly salary of RM 1300 for the first year and receives an annual increment of RM 70 .How much is his monthly salary for the nth year of service .How much will he receive monthly for his tenth year of service?
Solution:
His monthly salaries for successive years are as follows
RM1300,RM 1370,RM 1440 ,RM1510..
This is an arithmetic progression with the first term RM13000 and the common difference RM70 .Therefore, the nth term is Tn= a+(n-d) = RM1300 +(n-1)(70) = RM1300+ 70n -70n =RM1230+70n
Thus , for the nth year , he receives a monthly salary of RM(70n+1230)
For the tenth year, he receives monthly salary of RM [(70x10)+1230]
=RM1930
5. Zara keep RM75 in her “Giant Fund” at the end of the first month and continues saving every month RM10 more than the previous month .Ahmad ,however keeps RM100 in
1. In a contest ,all twenty finalists were given cash prize.The first winner was given RM860 , the second RM 800 ,the third RM740 and so on.Calculate the total amount of money awarded to all the finalists.
Solution:The prizes are RM860,RM800 ,RM740…..
It is an arithmetic progression .The first term a=rm860 , common difference d=-60 and n=20. The total amount given is
Sn = n/2[ 2a + (n-1)]
S26 =20/2[ 2(860) +(20-1)(-60)] =RM5800
The amount is RM 5800
2. Find the minimum number of terms that must be taken from the sequence : 4,16,6,256…so that the sum is more than 400.
Solution:
This is a geometric sequence with a=4 and r =4
Let the minimum number of terms be n.
From Sn= a(rn-1)/r-1 , we get 4(4n-1)/4-1 > 400 4(4n-10)> 1200 4n-1> 300 4n> 301 N log 4 > log 301 n> 4.117 n> 5
3. Given P , P+1 and P-2 are three consecutive terms of a geometric sequence , find P and the common ratio.
r = P+1/P= P-2/P+1 (P+1)2=P(P-2) P2+ 2P +1 = P2-2P P2-P2+2P + 2P =-1 4P=-1 P=-1/4 r= (-1/4+1)/(-1/4) =-3
Thus ,P= -1/4 and the common ratio r=-3
4. Azrul starts with a monthly salary of RM 1300 for the first year and receives an annual increment of RM 70 .How much is his monthly salary for the nth year of service .How much will he receive monthly for his tenth year of service?
Solution:
His monthly salaries for successive years are as follows
RM1300,RM 1370,RM 1440 ,RM1510..
This is an arithmetic progression with the first term RM13000 and the common difference RM70 .Therefore, the nth term is Tn= a+(n-d) = RM1300 +(n-1)(70) = RM1300+ 70n -70n =RM1230+70n
Thus , for the nth year , he receives a monthly salary of RM(70n+1230)
For the tenth year, he receives monthly salary of RM [(70x10)+1230]
=RM1930
5. Zara keep RM75 in her “Giant Fund” at the end of the first month and continues saving every month RM10 more than the previous month .Ahmad ,however keeps RM100 in