Which of the following are APs ?

Solution:

(i) Not an $\mathrm{AP}$ because $\mathrm{t}_{2}-\mathrm{t}_{1}=2$

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

i.e., $\mathrm{t}_{2}-\mathrm{t}_{1} \neq \mathrm{t}_{3}-\mathrm{t}_{2}$

(ii) It is an AP. $a=2, d=1 / 2$

$\left\lceil\because t_{2}-t_{1}=t_{3}-t_{2}=t_{4}-t_{3}=1 / 2\right\rceil$

$\mathrm{t}_{5}=\frac{7}{2}+\frac{1}{2}=4, \mathrm{t}_{6}=4+\frac{1}{2}=\frac{9}{2}$

$t_{7}=\frac{9}{2}+\frac{1}{2}=5$

(iii) We have : $-1.2,-3.2,-5.2,-7.2$,

$\therefore \mathrm{t}_{1}=-1.2, \mathrm{t}_{2}=-3.2, \mathrm{t}_{3}=-5.2, \mathrm{t}_{4}=-7.2$

$\mathrm{t}_{2}-\mathrm{t}_{1}=-3.2+1.2=-2$

$\mathrm{t}_{3}-\mathrm{t}_{2}=-5.2+3.2=-2$

$\mathrm{t}_{4}-\mathrm{t}_{3}=-7.2+5.2=-2$

$\because t_{2}-t_{1}=t_{3}-t_{2}=t_{4}-t_{3}=-2$

$\Rightarrow d=-2$

$\therefore$ The given numbers from an A.P. such that $d=-2 .$

Now, $\mathrm{t}_{5}=\mathrm{t}_{4}+(-2)=-7.2+(-2)=-9.2$,

$\mathrm{t}_{6}=\mathrm{t}_{5}+(-2)=-9.2+(-2)=-11.2$

and $\mathrm{t}_{7}=\mathrm{t}_{6}+(-2)$

$=-11.2+(-2)=-13.2$

Thus, $\mathrm{d}=-2$ and $\mathrm{t}_{5}=-9.2, \mathrm{t}_{6}=-11.2$ and $\quad \mathrm{t}_{7}=-13.2$

(iv) It is an AP.

$a=-10, d=4, t_{5}=6, t_{6}=10, t_{7}=14$

(v) It is an AP.

$a=3, d=\sqrt{2}$

$\mathrm{t}_{5}=3+3 \sqrt{2}+\sqrt{2}=3+4 \sqrt{2}$

$\mathrm{t}_{6}=3+5 \sqrt{2}, \mathrm{t}_{7}=3+6 \sqrt{2} .$

(vi) It is not AP.

$\mathrm{t}_{2}-\mathrm{t}_{1}=0.22-0.2=0.02$

$\mathrm{t}_{3}-\mathrm{t}_{3}=0.222-0.22=0.002, \ldots$

i.e., $\mathrm{t}_{2}-\mathrm{t}_{1} \neq \mathrm{t}_{3}-\mathrm{t}_{2}$

(vii) We have : $0,-4,-8,-12, \ldots \ldots \ldots$

$\therefore \mathrm{t}_{1}=0, \mathrm{t}_{2}=-4, \mathrm{t}_{3}=-8, \mathrm{t}_{4}=-12$

$t_{2}-t_{1}=-4-0=-4$

$\mathrm{t}_{3}-\mathrm{t}_{2}=-8+4=-4$

$t_{4}-t_{3}=-12+8=-4$

$\because \mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{t}_{3}-\mathrm{t}_{2}=\mathrm{t}_{4}-\mathrm{t}_{3}=-4 \Rightarrow \mathrm{d}=-4$

$\therefore$ The given numbers from an A.P.

Now, $\mathrm{t}_{5}=\mathrm{t}_{4}+(-4)=-12+(-4)=-16$

$\mathrm{t}_{6}=\mathrm{t}_{5}+(-4)=-16+(-4)=-20$

$\mathrm{t}_{7}=\mathrm{t}_{6}+(-4)=-20+(-4)=-24$

Thus, $\mathrm{d}=-4$ and $\mathrm{t}_{5}=-16, \mathrm{t}_{6}=-20, \quad \mathrm{t}_{7}=-24$.

(viii) We have : $-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots$.

$\therefore \mathrm{t}_{1}=\mathrm{t}_{2}=\mathrm{t}_{3}=\mathrm{t}_{4}=-\frac{1}{2}$

$\mathrm{t}_{2}-\mathrm{t}_{1}=0, \mathrm{t}_{3}-\mathrm{t}_{3}=0, \mathrm{t}_{4}-\mathrm{t}_{3}=0 \Rightarrow \mathrm{d}=0$

$\therefore$ The given numbers from an A.P.

Now, $t_{5}=-\frac{1}{2}+0=-\frac{1}{2}$

$\mathrm{t}_{6}=-\frac{1}{2}+0=-\frac{1}{2}, \mathrm{t}_{7}=-\frac{1}{2}+0=-\frac{1}{2}$

Thus, $\mathrm{d}=0$ and $\mathrm{t}_{5}=-\frac{1}{2}, \mathrm{t}_{6}=-\frac{1}{2}, \mathrm{t}_{7}=-\frac{1}{2}$

(ix) Not an A.P. Here, $t_{2}-t_{1} \neq t_{3}-t_{2}$.

(x) We have : $\mathrm{a}, 2 \mathrm{a}, 3 \mathrm{a}, 4 \mathrm{a}, \ldots \ldots \ldots$

$\therefore \quad \mathrm{t}_{1}=\mathrm{a}, \mathrm{t}_{2}=2 \mathrm{a}, \mathrm{t}_{3}=3 \mathrm{a}, \mathrm{t}_{4}=4 \mathrm{a}$

$\mathrm{t}_{2}-\mathrm{t}_{1}=2 \mathrm{a}-\mathrm{a}=\mathrm{a}$

$t_{3}-t_{2}=3 a-2 a=a$ and

$\mathrm{t}_{4}-\mathrm{t}_{3}=4 \mathrm{a}-3 \mathrm{a}=\mathrm{a}$

$\because \quad \mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{t}_{3}-\mathrm{t}_{2}=\mathrm{t}_{4}-\mathrm{t}_{3}=\mathrm{a}$

$\Rightarrow d=a$

$\therefore$ The given numbers from an A.P.

Now, $\mathrm{t}_{5}=\mathrm{t}_{4}+\mathrm{a}=4 \mathrm{a}+\mathrm{a}=5 \mathrm{a}, \mathrm{t}_{6}=\mathrm{t}_{5}+\mathrm{a}$

$=5 a+a=6 a$ and $t_{7}=t_{6}+a=6 a+a=7 a$

Thus, $\mathrm{d}=\mathrm{a}$ and $\mathrm{t}_{5}=5 \mathrm{a}, \mathrm{t}_{6}=6 \mathrm{a}, \mathrm{t}_{7}=7 \mathrm{a}$

(xi) Not an $\mathrm{AP}$ if $\mathrm{a} \neq 1$.

Here, $\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{a}^{2}-\mathrm{a}=\mathrm{a}(1-\mathrm{a})$

$t_{3}-t_{2}=a^{3}-a^{2}=a^{2}(1-a)$

$\mathrm{t}_{3}-\mathrm{t}_{2} \neq \mathrm{t}_{2}-\mathrm{t}_{1}$ when $\mathrm{a} \neq 1$

It will be an $\mathrm{AP}$ if $\mathrm{a}=1$.

Hence, the given sequence is an $\mathrm{AP}$ only when $\mathrm{a}=1$.

In this case, first term $=1$,

common difference $=0$

(xii) $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots$ can be rewritten as

$\sqrt{2}, 2 \sqrt{2}, 3 \sqrt{2}, 4 \sqrt{2}, \ldots .$

a=\sqrt{2}, d=\sqrt{2}

$\mathrm{t}_{5}=5 \sqrt{2}, \mathrm{t}_{6}=6 \sqrt{2}, \mathrm{t}_{7}=7 \sqrt{2}$

i.e., $\mathrm{t}_{5}=\sqrt{50}, \mathrm{t}_{6}=\sqrt{72}, \mathrm{t}_{7}=\sqrt{98}$

(xiii) $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots$ can be rewritten as

$\sqrt{3}, \sqrt{2} \times \sqrt{3}, 3,2 \sqrt{3}, \ldots$

Here, $\mathrm{t}_{2}-\mathrm{t}_{1} \neq \mathrm{t}_{3}-\mathrm{t}_{2}$

Therefore, the given list is not an AP.

(xiv) We have $1^{2}, 3^{2}, 5^{2}, 7^{2}, \ldots \ldots \ldots$

$\left.\therefore \begin{array}{l}t_{1}=1^{2}=1 \\ t_{2}=3^{2}=9\end{array}\right\} \Rightarrow t_{2}-t_{1}=9-1=8$

$\left.\begin{array}{l}t_{3}=5^{2}=25 \\ t_{4}=7^{2}=49\end{array}\right\} \Rightarrow t_{4}-t_{3}=49-25=24$

$\because \mathrm{t}_{2}-\mathrm{t}_{1} \neq \mathrm{t}_{4}-\mathrm{t}_{3}$

$\therefore$ The given numbers do not form an A.P.

(xv) $1^{2}, 5^{2}, 7^{2}, 73, \ldots$ can be rewritten as $1,25,49,73, \ldots$

Here, $t_{2}-t_{1}=t_{3}-t_{2}=t_{4}-t_{3}=\ldots .=24$

Hence, it is an AP.

$\therefore \mathrm{t}_{5}=97, \mathrm{t}_{6}=121, \mathrm{t}_{7}=145$