CHAPTER 5. QUADRATIC SEQUENCES 5.2
SOLUTION
Step 1 : The next number in thesequence
The numbers go up in multiples of 6
1 + 6(1) = 7, then 7 + 6(2) = 19
19 + 6(3) = 37, then 37 + 6(4) = 61
Therefore 61 + 6(5) = 91
The next number in thesequence is 91.
Step 2 : Generalising the pattern
T ERMS 1 7 19 37 61
1 stdifference +6 +12 +18 +24
2 nddifference +6 + 6 +6
The pattern will yield aquadratic equation since the second differenceis
constant
Therefore Tn= an^2 + bn + c
For the first term: n = 1, then T 1 = 1
For the second term: n = 2, then T 2 = 7
For the third term: n = 3, then T 3 = 19
etc.
Step 3 : Setting up sets of equations
a + b + c = 1 ...eqn(1)
4 a + 2b + c = 7 ...eqn(2)
9 a + 3b + c = 19 ...eqn(3)
Step 4 : Solve the sets of equations
eqn(2)− eqn(1) : 3a + b = 6 ...eqn(4)
eqn(3)− eqn(2) : 5a + b = 12 ...eqn(5)
eqn(5)− eqn(4) : 2a = 6
∴ a = 3, b =− 3 and c = 1
Step 5 : Final answer
The general formula forthe pattern is Tn= 3n^2 − 3 n + 1
Step 6 : Term 100
Substitute n with 100 :
3(100)^2 − 3(100) + 1 = 29 701
The value for term 100 is 29 701.
