SOLUTION:
Observe that:
- 10 - 1 = 9
- 25 - 10 = 14
- 46 - 25 = 19
- 73 - 46 = 24
And if we try to subtract their differences once again,
- 14 - 9 = 5
- 19 - 14 = 5
- 24 - 19 = 5
Since the terms have the same 2nd difference, this sequence is called quadratic sequence.
The formula in finding the nth term of a quadratic sequence is
[tex]a_n = an^2 + bn + c[/tex]
where a, b, c satisfy the conditions below:
- [tex]a = \frac{\textsf{2nd common difference}}{2}[/tex]
- [tex]3a + b = \textsf{2nd term} - \textsf{1st term}[/tex]
- [tex]a + b + c = \textsf{1st term}[/tex]
Solving for a,
Knowing that the 2nd common difference is 5, [tex]a = \frac{5}{2}[/tex]
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Solving for b,
[tex]3a + b = \textsf{2nd term} - \textsf{1st term}[/tex]
[tex]3(\frac{5}{2}) +b = 10 - 1[/tex]
[tex]\frac{15}{2} + b = 9[/tex]
[tex]b = 9 - \frac{15}{2}[/tex]
[tex]b = \frac{3}{2}[/tex]
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Solving for c,
[tex]a + b + c = \textsf{1st term}[/tex]
[tex]\frac{5}{2} + \frac{3}{2} + c = 1[/tex]
[tex]\frac{8}{2} + c = 1[/tex]
[tex]4 + c = 1[/tex]
[tex]c = 1- 4[/tex]
[tex]c = -3[/tex]
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Then plug in the value of a, b, and c to the formula to solve for 20th term
[tex]a_n = an^2 + bn + c[/tex]
[tex]a_{20} = \frac{5}{2}(20^2) + \frac{3}{2}(20) - 3[/tex]
[tex]a_{20} = \frac{5}{2}(400) + 3(10) - 3[/tex]
[tex]a_{20} = 1000 + 300 - 3[/tex]
[tex]a_{20} = \boxed{1297}[/tex]
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ANSWER:
1297
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