Question

find the nth term of sequence 1,-4,9,-16,25​

Answer 1

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$given \\ \: \: \: \: \: \: n = 9 \\ \: \: \: \: \: \: a = 1 \\ \: \: \: \: \: \: d = - 5 \\ \\ therefore \\ \: \: \: \: \: \: a_{9} = a + (n - 1)d \\ \: \: \: \: \: \: a_{9} = 1 + (9 - 1) - 5 \\ \: \: \: \: \: \: a_{9} = 1 - 40 \\ \: \: \: \: \: \: a_{9} = - 39$

$Hope\: you\: got\: it$

Answer 2

Given: A sequence 1,-4,9,-16,25 is given

To find: nth term of the sequence

Solution: Here, looking at the numbers we can infer that the magnitude of the numbers are the squares of the natural numbers.

Square of 1= 1

Square of 2 = 4

Square of 3 = 9

Square of 4 = 16

Square of 5 = 25

Here, when the numbers which are even are squared, these are written with a negative sign.

For example- 2 is an even number and 4 is written with a negative sign in the sequence.

Therefore, the nth term can be written as:

$- {( - 1)}^{n} {n}^{2}$

Here, if n is even, (-1)^n becomes 1 and the extra negative sign gives an overall negative sign to the number in the sequence.

If n is odd, (-1)^n remains -1 and the extra negative sign neutralises -1 and gives an overall positive sign to the number in the sequence.

Therefore, the nth term is

$- {( - 1)}^{n} {n}^{2}$