Which of the following form of an AP ?

Solution:

(i) Here, $t_{1}=-1, t_{2}=-1, t_{3}=-1$ and $t_{4}=-1$

$t_{2}-t_{1}=-1+1=0$

$t_{3}-t_{2}=-1+1=0$

$t_{4}-t_{3}=-1+1=0$

Clearly, the difference of successive terms is same, therefore given list of numbers form an AP.

(ii) Here, $t_{1}=0, t_{2}=2, t_{3}=0$ and $t_{4}=2$

$t_{2}-t_{1}=2-0=2$

$t_{3}-t_{2}=0-2=-2$

$t_{4}-t_{3}=2-0=2$

Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

(iii) Here, $t_{1}=1, t_{2}=1, t_{3}=2$ and $t_{4}=2$

$t_{2}-t_{1}=1-1=0$

$t_{3}-t_{2}=2-1=1$

$t_{4}-t_{2}=2-2=0$

Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

(iv) Here, $t_{1}=11, t_{2}=22$ and $t_{3}=33$

$t_{2}-t_{1}=22-11=11$

$t_{3}-t_{2}=33-22=11$

$t_{4}-t_{3}=33-22=11$

Clearly, the difference of successive terms is same, therefore given list of numbers form an $A P$.

(v) $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$

Here, $t_{1}=\frac{1}{2}, t_{2}=\frac{1}{3}$ and $t_{3}=\frac{1}{4}$

$t_{2}-t_{1}=\frac{1}{3}-\frac{1}{2}=\frac{2-3}{6}=-\frac{1}{6}$

$t_{3}-t_{2}=\frac{1}{4}-\frac{1}{3}=\frac{3-4}{12}=-\frac{1}{12}$

Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

(vi) $2,2^{2}, 2^{3}, 2^{4}, \ldots$ i.e., $2,4,8,16, \ldots$

Here, $\quad t_{1}=2, t_{2}=4, t_{3}=8$ and $t_{4}=16$

$t_{2}-t_{1}=4-2=2$

$t_{3}-t_{2}=8-4=4$

$t_{4}-t_{3}=16-8=8$

Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

(vii) $\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, \ldots$ i.e., $\sqrt{3}, 2 \sqrt{3}, 3 \sqrt{3}, 4 \sqrt{3}, \ldots$

Here, $\quad t_{1}=\sqrt{3}, t_{2}=2 \sqrt{3}, t_{3}=3 \sqrt{3}$ and $t_{4}=4 \sqrt{3}$

$t_{2}-t_{1}=2 \sqrt{3}-\sqrt{3}=\sqrt{3}$

$t_{3}-t_{2}=3 \sqrt{3}-2 \sqrt{3}=\sqrt{3}$

$t_{4}-t_{3}=4 \sqrt{3}-3 \sqrt{3}=\sqrt{3}$

Clearly, the difference of successive terms is same, therefore given list of numbers form an AP.