Which of the following form an AP? Justify your answer.
$ -1,-1,-1,-1, \ldots $

To do:

We have to check whether the given sequences are in AP. 

Solution:

(i) In the given sequence,

$a_1=-1, a_2=-1, a_3=-1, a_4=-1$

$a_2-a_1=-1-(-1)=-1+1=0$

$a_3-a_2=-1-(-1)=-1+1=0$

$a_4-a_3=-1-(-1)=-1+1=0$

Here,

$a_2 - a_1 = a_3 - a_2=a_4-a_3$

Therefore, the given sequence is an AP.

(ii) In the given sequence,

$a_1=0, a_2=2, a_3=0, a_4=2$

$a_2-a_1=2-0=2$

$a_3-a_2=0-2=-2$

$a_4-a_3=2-0=2$

Here,

$a_2 - a_1 ≠ a_3 - a_2$

Therefore, the given sequence is not an AP.

(iii) In the given sequence,

$a_1=1, a_2=1, a_3=2, a_4=2$

$a_2-a_1=1-1=0$

$a_3-a_2=2-1=1$

$a_4-a_3=2-2=0$

Here,

$a_2 - a_1 ≠ a_3 - a_2$

Therefore, the given sequence is not an AP.

(iv) In the given sequence,

$a_1=11, a_2=22, a_3=33$

$a_2-a_1=22-11=11$

$a_3-a_2=33-22=11$

Here,

$a_2 - a_1 = a_3 - a_2$

Therefore, the given sequence is an AP.

(v) In the given sequence,

$a_1=\frac{1}{2}, a_2=\frac{1}{3}, a_3=\frac{1}{4}$

$a_2-a_1=\frac{1}{3}-\frac{1}{2}=\frac{2-3}{6}=\frac{-1}{6}$

$a_3-a_2=\frac{1}{4}-\frac{1}{3}=\frac{3-4}{12}=\frac{-1}{12}$

Here,

$a_2 - a_1 ≠ a_3 - a_2$

Therefore, the given sequence is not an AP. 

(vi) In the given sequence,

$a_1=2, a_2=2^2, a_3=2^3$

$a_2-a_1=2^2-2=4-2=2$

$a_3-a_2=2^3-2^2=8-4=4$

Here,

$a_2 - a_1 ≠ a_3 - a_2$

Therefore, the given sequence is not an AP. 

(vii) In the given sequence,

$a_1=\sqrt3, a_2=\sqrt{12}=\sqrt{4\times3}=2\sqrt3, a_3=\sqrt{27}=\sqrt{9\times3}=3\sqrt{3}, a_4=\sqrt{48}=\sqrt{16\times3}=4\sqrt{3}$

$a_2-a_1=2\sqrt3-\sqrt3=\sqrt3$

$a_3-a_2=3\sqrt3-2\sqrt3=\sqrt3$

$a_4-a_3=4\sqrt3-3\sqrt3=\sqrt3$

Here,

$a_2 - a_1 = a_3 - a_2=a_4 - a_3$

Therefore, the given sequence is an AP. 

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Updated on: 10-Oct-2022

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