1. In which of the following situations, does the list of numbers involved make an arithmetic
progression, and why?
(i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
Answer:
Taxi fare for first km= ₹ 15 = $a_1$
Since the taxi fare increases by ₹ 8 for every additional km,
So, Taxi fare for second km= ₹(15+8)=₹ 23 =$a_2$
Taxi fare for third km= ₹(23+8)=₹ 31 = $a_3$
For any number to be in AP,
$a_2−a_1=a_3−a_2=a_4−a_3=....=a_n−a_{n−1}=d$
$a_2-a_1=23-15=8$
$a_3-a_2=31-23=8$
Since , the common difference between the sucessive term and preceding term is same,so the numbers are in AP.
(ii) The amount of air present in a cylinder when a vacuum pump removes $1/4$ of the air remaining in the
cylinder at a time.
Answer:
Let 1 be the amount of air present initially.
Air present after 1/4 of the air is pumped out=$a_1$=$(1-1/4)$=3/4
Air present after 1/4 of the air is pumped out=$a_2$=$3/4$$(1-1/4)$=9/16
Air present after 1/4 of the air is pumped out=$a_3$=$9/16$$(1-1/4)$=27/64
For any number to be in AP,
$a_2−a_1=a_3−a_2=a_4−a_3=....=a_n−a_{n−1}=d$
$a_2−a_1=9/16-3/4={9-12}/16=-3/16=-3/16$
$a_3−a_2=27/64-9/16={27-36}/64=-9/64$
Since $a_2−a_1 ≠ a_3−a_2$,
So, the given list of numbers does not form an AP.
(iii) The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre
and rises by ₹ 50 for each subsequent metre.
Answer:
The cost of digging for first meter = ₹ 150 and the cost rises by ₹ 50 for each subsequent depth,
so, $a_{k+1}-a_k = 50$ for each value of $k$
So, the given list of numbers form an AP.
(iv) The amount of money in the account every year, when ₹ 10000 is deposited at compound interest
at 8 % per annum.
Answer:
2. Write first four terms of the AP, when the first term $a$ and the common difference $d$ are given
as follows:
(i) $a = 10$, $d = 10$
Answer: Since the AP series is given by
$a, a+d,a+2d,a+3d.....$
Hence the first four terms are:
first term=$a$=10;
second term=$a+d$=10+10=20;
third term=$a+2d$=10+2(10)=30
fourth term=$a+3d$=10+3(10)=40
(ii) $a = –2$, $d = 0$
Answer
Since the AP series is given by
$a, a+d,a+2d,a+3d.....$
Hence the first four terms are:
first term=$a$=-2;
second term=$a+d$=-2+0=-2;
third term=$a+2d$=-2+2(0)=-2
fourth term=$a+3d$=-2+3(0)=-2
(iii) a = 4, d = – 3
Answer
Since the AP series is given by
$a, a+d,a+2d,a+3d.....$
Hence the first four terms are:
first term=$a$=4
second term=$a+d$=4+(-3)=1
third term=$a+2d$=4+2(-3)=-2
fourth term=$a+3d$=4+3(-3)=-5
So, the series are 4,1,-2,-5
(iv) a = – 1, d = $1/2$
Answer
Since the AP series is given by
$a, a+d,a+2d,a+3d.....$
Hence the first four terms are:
first term=$a$=-1
second term=$a+d$=$-1+1/2$=$-1/2$
third term=$a+2d$=$-1+2(1/2)$=$0$
fourth term=$a+3d$=$-1+3(1/2)$=$1/2$
So, the series are $-1,-1/2,0,1/2$
(v) a = – 1.25, d = – 0.25
Answer
Since the AP series is given by
$a, a+d,a+2d,a+3d.....$
Hence the first four terms are:
first term=$a$= -1.25
second term=$a+d$= -1.25+(-0.25)=-1.5
third term=$a+2d$= -1.25+2(-0.25)=-1.75
fourth term=$a+3d$= -1.25+3(-0.25)=-2.0
So, the series are -1.25,-1.5,-1.75,-2.0
3. For the following APs, write the first term and the common difference:
(i) 3, 1, – 1, – 3, . . .
Answer first term $a=3$
Now second term= $a+d=1$; $d=1-a$ ; i.e $d=1-3$ =$-2$
$a=3$ and $d=-2$
(ii) – 5, – 1, 3, 7, . . .
Answer first term $a=-5$
Now second term= $a+d=-1$;$d=-1-a$ i.e $d=-1-(-5)=4$
$a=-5$ and $d=4$
(iii)$1/3,5/3,9/3,13/3$
Answer first term $a=1/3$
Now second term= $a+d=5/3$;$1/3+d=5/3$;$d=5/3-1/3=4/3$
$a=1/3$ and $d=4/3$
(iv) 0.6, 1.7, 2.8, 3.9, . . .
Answer first term $a=0.6$
Now second term= $a+d=1.7$;$0.6+d=1.7$;$d=1.7-0.6=1.1$
$a=0.6$ and $d=1.1$
