Solution

(i) It can be observed that

Taxi fare for $1^{\text{st }} km=15$

Taxi fare for first $2 km=15+8=23$

Taxi fare for first $3 km=23+8=31$

Taxi fare for first $4 km=31+8=39$

Clearly $15,23,31,39 \ldots$ forms an A.P. because every term is 8 more than the preceding term.

(ii) Let the initial volume of air in a cylinder be $V$ lit. In each stroke, the vacuum pump removes $\dfrac{1}{4}$ of air remaining in the cylinder at a time. In other words, after every stroke, only $1-\dfrac{1}{4}=\dfrac{3}{4}$ th part of air will remain.

Therefore, volumes will be $V, \dfrac{3}{4} V,(\dfrac{3}{4})^{2} V,(\dfrac{3}{4})^{3} V \ldots \ldots$.

Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

(iii) Cost of digging for first metre $=150$

Cost of digging for first 2 metres $=150+50=200$

Cost of digging for first 3 metres $=200+50=250$

Cost of digging for first 4 metres $=250+50=300$

Clearly, 150, 200, 250, $300 \ldots$ forms an A.P. because every term is 50 more than the preceding term.

(iv) We know that if Rs $P$ is deposited at $r %$ compound interest per annum for $n$ years, our money will

be $P(1+\dfrac{r}{100})^{n}$ aftern years.

Therefore, after every year, our money will be

$10000(1+\dfrac{8}{100}), 10000(1+\dfrac{8}{100})^{2}, 10000(1+\dfrac{8}{100})^{3}, 10000(1+\dfrac{8}{100})^{4}, \ldots$

Clearly, adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

Solution

(i) $a=10, d=10$

Let the series be $a_1, a_2, a_3, a_4, a_5 \ldots$

$a_1=a=10$

$a_2=a_1+d=10+10=20$

$a_3=a_2+d=20+10=30$

$a_4=a_3+d=30+10=40$

$a_5=a_4+d=40+10=50$

Therefore, the series will be $10,20,30,40,50 \ldots$

First four terms of this A.P. will be $10,20,30$, and 40.

(ii) $a=-2, d=0$

Let the series be $a_1, a_2, a_3, a_4 \ldots$

$a_1=a=-2$

$a_2=a_1+d=-2+0=-2$

$a_3=a_2+d=-2+0=-2$ $a_4=a_3+d=-2+0=-2$

Therefore, the series will be $-2,-2,-2,-2 \ldots$

First four terms of this A.P. will be $-2,-2,-2$ and -2 .

(iii) $a=4, d=-3$

Let the series be $a_1, a_2, a_3, a_4 \ldots$

$a_1=a=4$

$a_2=a_1+d=4-3=1$

$a_3=a_2+d=1-3=-2$

$a_4=a_3+d=-2-3=-5$

Therefore, the series will be $4,1,-2$ - $5 \ldots$

First four terms of this A.P. will be 4, 1, - 2 and -5 .

(iv) $a=-1, d=\dfrac{1}{2}$

Let the series be $a_1, a_2, a_3, a_4 \ldots$

$a_1=a=-1$

$a_2=a_1+d=-1+\dfrac{1}{2}=-\dfrac{1}{2}$

$a_3=a_2+d=-\dfrac{1}{2}+\dfrac{1}{2}=0$

$a_4=a_3+d=0+\dfrac{1}{2}=\dfrac{1}{2}$

Clearly, the series will be

$-1,-\dfrac{1}{2}, 0, \dfrac{1}{2}$

First four terms of this A.P. will be

$ -1,-\dfrac{1}{2}, 0 \text{ and } \dfrac{1}{2} $

(v) $a=-1.25, d=-0.25$

Let the series be $a_1, a_2, a_3, a_4 \ldots$

$a_1=a=-1.25$

$a_2=a_1+d=-1.25-0.25=-1.50$

$a_3=a_2+d=-1.50-0.25=-1.75$

$a_4=a_3+d=-1.75-0.25=-2.00$

Clearly, the series will be $1.25,-1.50,-1.75,-2.00 \ldots \ldots .$.

First four terms of this A.P. will be $-1.25,-1.50,-1.75$ and -2.00 .

Solution

(i) $3,1,-1,-3 \ldots$

Here, first term, $a=3$

Common difference, $d=$ Second term - First term

$=1-3=-2$

(ii) - 5, - 1, 3, $7 \ldots$

Here, first term, $a=-5$

Common difference, $d=$ Second term - First term

$=(-1)-(-5)=-1+5=4$

(iii) $\dfrac{1}{3}, \dfrac{5}{3}, \dfrac{9}{3}, \dfrac{13}{3} \ldots$

Here, first term, $a=\dfrac{1}{3}$

Common difference, $d=$ Second term - First term

$=\dfrac{5}{3}-\dfrac{1}{3}=\dfrac{4}{3}$

(iv) $0.6,1.7,2.8,3.9 \ldots$

Here, first term, $a=0.6$

Common difference, $d=$ Second term - First term

$=1.7-0.6$

$=1.1$

Solution

(i) $2,4,8,16 \ldots$

It can be observed that

$a_2-a_1=4-2=2$

$a_3-a_2=8-4=4$

$a_4-a_3=16-8=8$

i.e., $a _{k+1}-a_k$ is not the same every time. Therefore, the given numbers are not forming an A.P.

(ii) $2, \dfrac{5}{2}, 3, \dfrac{7}{2} \ldots$

It can be observed that $a_2-a=\dfrac{5}{2}-2=\dfrac{1}{2}$

$a_3-a_2=3-\dfrac{5}{2}=\dfrac{1}{2}$

$a_4-a_3=\dfrac{7}{2}-3=\dfrac{1}{2}$

i.e., $a _{k+1}-a_k$ is same every time.

Therefore, $d=\dfrac{1}{2}$ and the given numbers are in A.P.

Three more terms are

$a_5=\dfrac{7}{2}+\dfrac{1}{2}=4$

$a_6=4+\dfrac{1}{2}=\dfrac{9}{2}$

$a_7=\dfrac{9}{2}+\dfrac{1}{2}=5$

(iii) - 1.2, - 3.2, - 5.2, - $7.2 \ldots$

It can be observed that

$a_2-a_1=(-3.2)-(-1.2)=-2$

$a_3-a_2=(-5.2)-(-3.2)=-2$

$a_4-a_3=(-7.2)-(-5.2)=-2$

i.e., $a _{k+1}-a_k$ is same every time. Therefore, $d=-2$

The given numbers are in A.P.

Three more terms are

$a_5=-7.2-2=-9.2$

$a_6=-9.2-2=-11.2$

$a_7=-11.2-2=-13.2$

(iv) - 10, - 6, - 2, $2 \ldots$

It can be observed that

$a_2-a_1=(-6)-(-10)=4$

$a_3-a_2=(-2)-(-6)=4$

$a_4-a_3=(2)-(-2)=4$

i.e., $a _{k+1}-a_k$ is same every time. Therefore, $d=4$

The given numbers are in A.P.

Three more terms are $a_5=2+4=6$

$a_6=6+4=10$

$a_7=10+4=14$

(v) $3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2}, \ldots$

It can be observed that

$a_2-a_1=3+\sqrt{2}-3=\sqrt{2}$

$a_3-a_2=3+2 \sqrt{2}-3-\sqrt{2}=\sqrt{2}$

$a_4-a_3=3+3 \sqrt{2}-3-2 \sqrt{2}=\sqrt{2}$

i.e., $a _{k+1} . a_k$ is same every time. Therefore, $d=\sqrt{2}$

The given numbers are in A.P.

Three more terms are

$a_5=3+3 \sqrt{2}+\sqrt{2}=3+4 \sqrt{2}$

$a_6=3+4 \sqrt{2}+\sqrt{2}=3+5 \sqrt{2}$

$a_7=3+5 \sqrt{2}+\sqrt{2}=3+6 \sqrt{2}$

(vi) $0.2,0.22,0.222,0.2222 \ldots$

It can be observed that

$a_2-a_1=0.22-0.2=0.02$

$a_3-a_2=0.222-0.22=0.002$

$a_4-a_3=0.2222-0.222=0.0002$

i.e., $a _{k+1}-a_k$ is not the same every time.

Therefore, the given numbers are not in A.P.

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

It can be observed that

$a_2-a_1=(-4)-0=-4$

$a_3-a_2=(-8)-(-4)=-4$

$a_4-a_3=(-12)-(-8)=-4$

i.e., $a _{k+1}-a_k$ is same every time. Therefore, $d=-4$

The given numbers are in A.P.

Three more terms are

$a_5=-$