Question.
Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
(i) $2,4,8,16, \ldots$
(ii) $2, \frac{5}{2}, 3, \frac{7}{2}, \ldots$
(iii) $-1.2,-3.2,-5.2,-7.2, \ldots$
(iv) $-10,-6,-2,2, \ldots \ldots$
(v) $3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2}, \ldots$
(vi) $0.2,0.22,0.222,0.2222, \ldots .$
(vii) $\quad 0,-4,-8,-12, \ldots .$
(viii) $-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots$
(ix) $1,3,9,27, \ldots \ldots$
(x) $a, 2 a, 3 a, 4 a, \ldots$
(xi) $a, a^{2}, a^{3}, a^{4}, \ldots$
(xii) $\quad \sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots$
(xiii) $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots$
(xiv) $1^{2}, 3^{2}, 5^{2}, 7^{2}, \ldots$
(xv) $\quad 1^{2}, 5^{2}, 7^{2}, 73, \ldots$
Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
(i) $2,4,8,16, \ldots$
(ii) $2, \frac{5}{2}, 3, \frac{7}{2}, \ldots$
(iii) $-1.2,-3.2,-5.2,-7.2, \ldots$
(iv) $-10,-6,-2,2, \ldots \ldots$
(v) $3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2}, \ldots$
(vi) $0.2,0.22,0.222,0.2222, \ldots .$
(vii) $\quad 0,-4,-8,-12, \ldots .$
(viii) $-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots$
(ix) $1,3,9,27, \ldots \ldots$
(x) $a, 2 a, 3 a, 4 a, \ldots$
(xi) $a, a^{2}, a^{3}, a^{4}, \ldots$
(xii) $\quad \sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots$
(xiii) $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots$
(xiv) $1^{2}, 3^{2}, 5^{2}, 7^{2}, \ldots$
(xv) $\quad 1^{2}, 5^{2}, 7^{2}, 73, \ldots$
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.
$\therefore \mathrm{t}_{5}=97, \mathrm{t}_{6}=121, \mathrm{t}_{7}=145$
(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.
$\therefore \mathrm{t}_{5}=97, \mathrm{t}_{6}=121, \mathrm{t}_{7}=145$