Question

Find the smallest positive integer n л such that ta of the arithmetic sequence 20, 19 1/4 18 1/2 is negative?​

Answer 1

Answer:

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Answer 2

Answer:

Here we have a=20,d=19

4

1

−20=−

4

3

.

We want to find the first positive integer n such that t

n

<0.

This is same as solving a+(n−1)d<0 for smallest nϵN.

That is solving 20+(n−1)(−

4

3

)<0 for smallest nϵN.

Now, (n−1)(−

4

3

)<−20

⇒(n−1)×

4

3

>20 (The inequality is reversed on multiplying both sides by −1)

⇒n−1>20×

3

4

=

3

80

=26

3

2

.

This implies n>26

3

2

+1. That is, n>27

3

2

=27.66

Thus, the smallest positive integer nϵN satisfying the inequality is n=28.

Hence, the 28

th

term, t

28

is the first negative term of the A.P.