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The general term of the sequence is given by $T_{n} = an^{2} + bn + c$

$$T_{2} - T_{1} = 4 $$

$$T_{3} - T_{2} = 6 $$

$$T_{4} - T_{3} = 8 $$

 $$ T_{2} = 6$$

  • Calculate a,b and c
  • Determine the 12th term of the sequence. 
in Grade 12 Maths by Platinum (114k points) | 26 views

1 Answer

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T3 - 6=6

T3=12

then T2-T1=4

         6-T1=4

                T1=2

T4-T3=8

T4-12=8

      T4=20

Therefore terms= 2,6,12,20

1st difference=       4,6,8

2nd difference=        2,2,....

2a=2

 a=1

3a+b=4

3(1) + b=4

   b= 1

a+b+c= 2

c= 2-1- 1= 0

giving Tn= n^2 + n

T12 = 12^2 + 12 = 156
by Diamond (85.5k points)

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