Question 8 (i)
Justify whether it is true to say that the following are the nth terms of an AP.
2n – 3

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Solution

Yes.
Here, $a_{n}$ = 2n – 3
Put n = 1, $a_{1}$ = 2(1) - 3 = -1
Put n = 2, $a_{2}$ = 2 (2) - 3 = 1
Put n = 3, $a_{3}$ = 2(3) - 3 = 3
Put n = 4, $a_{4}$ = 2(4) - 3 = 5
The numbers are;
- 1, 1, 3 ....
$H e r e, a_{2} - a_{1} = 1 - (- 1) = 1 + 1 = 2$
$a_{3} - a_{2} = 3 - 1 = 2$
$a_{4} - a_{3} = 5 - 3 = 2$
$∵ a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = \dots$
Hence, 2n - 3 is the nth term of an AP.

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