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TOPICS
Exercise - 5.1

Question-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.
(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.
(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.
(iv) The amount of money in the account every year, when ₹10000 is deposited at compound interest at 8 % per annum.

Solution :-
```(i) Here, According to Question
Taxi fare for 1st km = ₹15
Taxi fare for 1st 2 km = 15 + 8 = ₹23
Taxi fare for 1st 3 km = 23 + 8 = ₹31
Taxi fare for 1st 4 km = 31 + 8 = ₹39
Therefore, 15, 23, 31, 39 .... forms an A.P. because every term is 8 more than the preceding term.
```
```(ii) Let the initial volume of air present in a cylinder be V litre.
In each stroke, the vacuum pump removes 1/4 of air remaining in the cylinder at a time.
In other words, after every stroke, only 1 - 1/4 = 3/4 th  part of air will remain.
Therefore, volumes will be V, (3V/4), (3V/4)2, (3V/4)3 ....
So, 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) Here, According to Question
Cost of digging for 1st metre = ₹150
Cost of digging for 1st 2 metres = 150 + 50 = ₹200
Cost of digging for 1st 3 metres = 200 + 50 = ₹250
Cost of digging for 1st 4 metres = 250 + 50 = ₹300
Therefore, 150, 200, 250, 300 .... 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+r/100)n after n years.
Therefore, after every year, our money will be 1000(1+0.08), 1000(1+0.08)2, 1000(1+0.08)3, 1000(1+0.08)4 ....
So, adjacent terms of this series do not have the same difference between them.
Therefore, this is not an A.P.
```

Question-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
(ii) a = –2, d = 0
(iii) a = 4, d = – 3
(iv) a = –1, d = 1/2
(v) a = –1.25, d = –0.25

Solution :-
```(i) Given : a = 10, d = 10
an = a + (n-1)d
n = 1, 2, 3, 4
a1 = 10 + (1-1)10 = 10 + 0 = 10
a2 = 10 + (2-1)10 = 10 + 10 = 20
a3 = 10 + (3-1)10 = 10 + 20 = 30
a4 = 10 + (4-1)10 = 10 + 30 = 40
So, Four terms of A.P. is 10, 20, 30 and 40
```
```(ii) Given : a = -2, d = 0
an = a + (n-1)d
n = 1, 2, 3, 4
a1 = -2 + (1-1)0 = -2 + 0 = -2
a2 = -2 + (2-1)0 = -2 + 0 = -2
a3 = -2 + (3-1)0 = -2 + 0 = -2
a4 = -2 + (4-1)0 = -2 + 0 = -2
So, Four terms of A.P. is -2, -2, -2 and -2
```
```(iii) Given : a = 4, d = -3
an = a + (n-1)d
n = 1, 2, 3, 4
a1 = 4 + (1-1)(-3) = 4 + 0 = 4
a2 = 4 + (2-1)(-3) = 4 - 3 = 1
a3 = 4 + (3-1)(-3) = 4 - 6 = -2
a4 = 4 + (4-1)(-3) = 4 - 9 = -5
So, Four terms of A.P. is 4, 1, -2 and -5
```
```(iv) Given : a = -1, d = 1/2
an = a + (n-1)d
n = 1, 2, 3, 4
a1 = -1 + (1-1)1/2 = -1 + 0 = -1
a2 = -1 + (2-1)1/2 = -1 + 10 = -1/2
a3 = -1 + (3-1)1/2 = -1 + 20 = 0
a4 = -1 + (4-1)1/2 = -1 + 30 = 1/2
So, Four terms of A.P. is -1, -1/2, 0 and 1/2
```
```(v) Given : a = -1.25, d = -0.25
an = a + (n-1)d
n = 1, 2, 3, 4
a1 = -1.25 + (1-1)(-0.25) = -1.25 + 0 = -1.25
a2 = -1.25 + (2-1)(-0.25) = -1.25 - 0.25 = -1.50
a3 = -1.25 + (3-1)(-0.25) = -1.25 - 0.50 = -1.75
a4 = -1.25 + (4-1)(-0.25) = -1.25 - 0.75 = -2.00
So, Four terms of A.P. is -1.25, -1.50, -1.75 and -2.00
```

Question-3 :-  For the following APs, write the first term and the common difference:
(i) 3, 1, –1, –3, ....
(ii) –5, –1, 3, 7, ....
(iii) 1/3, 5/3, 9/3, 13/3 ....
(iv) 0.6, 1.7, 2.8, 3.9, ....

Solution :-
```(i) 3, 1, –1, –3, ....
Here, First Term (a) = 3
and Common difference (d) = a2 - a1 = 1 - 3 = -2
```
```(ii) –5, –1, 3, 7, ....
Here, First Term (a) = -5
and Common difference (d) = a2 - a1 = -1 - (-5) = -1 + 5 = 4
```
```(iii) 1/3, 5/3, 9/3, 13/3 ....
Here, First Term (a) = 1/3
and Common difference (d) = a2 - a1 = 5/3 - 1/3 = 4/3
```
```(iv) 0.6, 1.7, 2.8, 3.9, ....
Here, First Term (a) = 0.6
and Common difference (d) = a2 - a1 = 1.7 - 0.6 = 1.1
```

Question-4 :-  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, ....
(ii) 2, 5/2, 3, 7/2, ....
(iii) -1.2, -3.2, -5.2, -7.2, ....
(iv) -10, -6, -2, 2, ....
(v) 3, 3+√2, 3+2√2, 3+3√2, ....
(vi) 0.2, 0.22, 0.222, 0.2222, ....
(vii) 0, -4, -8, -12, ....
(viii) -1/2, -1/2, -1/2, -1/2, ....
(ix) 1, 3, 9, 27, ....
(x) a, 2a, 3a, 4a, ....
(xi) a, a2, a3, a4, ....
(xii) √2, √8, √18, √32, ....
(xiii) √3, √6, √9, √12, ....
(xiv) 12, 32, 52, 72, ....
(xv) 12, 52, 72, 73, ....

Solution :-
```(i) 2, 4, 8, 16, ....
d1 = a2 - a1 = 4 - 2 = 2,
d2 = a3 - a2 = 8 - 4 = 4
Here, d1 ≠ d2
So, This series is not form in A.P.
```
```(ii) 2, 5/2, 3, 7/2, ....
d1 = a2 - a1 = 5/2 - 2 = 1/2,
d2 = a3 - a2 = 3 - 5/2 = 1/2
Here, d1 = d2
So, This series is form in A.P.
Now, a = 2, d = 1/2
According to question, 3 more terms are a5, a6 and a7
a5 = 2 + (5-1)1/2 = 2 + 2 = 4
a6 = 2 + (6-1)1/2 = 2 + 5/2 = 9/2
a7 = 2 + (7-1)1/2 = 2 + 3 = 5
```
```(iii) -1.2, -3.2, -5.2, -7.2, ....
d1 = a2 - a1 = -3.2 + 1.2 = -2.0,
d2 = a3 - a2 = -5.2 + 3.2 = -2.0
Here, d1 = d2
So, This series is form in A.P.
Now, a = -1.2, d = -2.0
According to question, 3 more terms are a5, a6 and a7
a5 = -1.2 + (5-1)(-2.0) = -1.2 - 8 = 9.2
a6 = -1.2 + (6-1)(-2.0) = -1.2 - 10 = 11.2
a7 = -1.2 + (7-1)(-2.0) = -1.2 - 12 = 13.2
```
```(iv) -10, -6, -2, 2, ....
d1 = a2 - a1 = -6 + 10 = 4,
d2 = a3 - a2 = -2 + 6 = 4
Here, d1 = d2
So, This series is form in A.P.
Now, a = -10, d = 4
According to question, 3 more terms are a5, a6 and a7
a5 = -10 + (5-1)(4) = -10 + 16 = 6
a6 = -10 + (6-1)(4) = -10 + 20 = 10
a7 = -10 + (7-1)(4) = -10 + 24 = 14
```
```(v) 3, 3+√2, 3+2√2, 3+3√2, ....
d1 = a2 - a1 = 3+√2 - 3 = √2
d2 = a3 - a2 = 3+2√2 - 3-√2 = √2
Here, d1 = d2
So, This series is form in A.P.
Now, a = 3, d = √2
According to question, 3 more terms are a5, a6 and a7
a5 = 3 + (5-1)(√2) = 3+4√2
a6 = 3 + (6-1)(√2) = 3+5√2
a7 = 3 + (7-1)(√2) = 3+6√2
```
```(vi) 0.2, 0.22, 0.222, 0.2222, ....
d1 = a2 - a1 = 0.22 - 0.2 = 0.02
d2 = a3 - a2 = 0.222 - 0.22 = 0.002
Here, d1 ≠ d2
So, This series is not form in A.P.
```
```(vii) 0, -4, -8, -12, ....
d1 = a2 - a1 = -4 - 0 = -4
d2 = a3 - a2 = -8 + 4 = -4
Here, d1 = d2
So, This series is form in A.P.
Now, a = 0, d = -4
According to question, 3 more terms are a5, a6 and a7
a5 = 0 + (5-1)(-4) = 0 - 16 = -6
a6 = 0 + (6-1)(-4) = 0 - 20 = -10
a7 = 0 + (7-1)(-4) = 0 - 24 = -14
```
```(viii) -1/2, -1/2, -1/2, -1/2, ....
d1 = a2 - a1 = -1/2 + 1/2 = 0
d2 = a3 - a2 = -1/2 + 1/2 = 0
Here, d1 = d2
So, This series is form in A.P.
Now, a = -1/2, d = 0
According to question, 3 more terms are a5, a6 and a7
a5 = -1/2 + (5-1)(0) = -1/2 - 0 = -1/2
a6 = -1/2 + (6-1)(0) = -1/2 - 0 = -1/2
a7 = -1/2 + (7-1)(0) = -1/2 - 0 = -1/2
```
```(xi) 1, 3, 9, 27, ....
d1 = a2 - a1 = 3 - 1 = 2
d2 = a3 - a2 = 9 - 3 = 6
Here, d1 ≠ d2
So, This series is not form in A.P.
```
```(x) a, 2a, 3a, 4a, ....
d1 = a2 - a1 = 2a - a = a
d2 = a3 - a2 = 3a - 2a = a
Here, d1 = d2
So, This series is form in A.P.
Now, a = a, d = a
According to question, 3 more terms are a5, a6 and a7
a5 = a + (5-1)(a) = a + 4a = 5a
a6 = a + (6-1)(a) = a + 5a = 6a
a7 = a + (7-1)(a) = a + 6a = 7a
```
```(xi) a, a2, a3, a4, ....
d1 = a2 - a1 = a2 - a = a(a-1)
d2 = a3 - a2 = a3 - a2 = a2(a-1)
Here, d1 ≠ d2
So, This series is not form in A.P.
```
```(xii) √2, √8, √18, √32, ....
d1 = a2 - a1 = √8 - √2 = 2√2 - √2 = √2
d2 = a3 - a2 = √18 - √8 = 3√2 - 2√2 = √2
Here, d1 = d2
So, This series is form in A.P.
Now, a = √2, d = √2
According to question, 3 more terms are a5, a6 and a7
a5 = √2 + (5-1)(√2) = √2 + 4√2 = 5√2 = √50
a6 = √2 + (6-1)(√2) = √2 + 5√2 = 6√2 = √72
a7 = √2 + (7-1)(√2) = √2 + 6√2 = 7√2 = √98
```
```(xiii) √3, √6, √9, √12, ....
d1 = a2 - a1 = √6 - √3
d2 = a3 - a2 = √9 - √6
Here, d1 ≠ d2
So, This series is not form in A.P.
```
```(xiv) 12, 32, 52, 72, ....
d1 = a2 - a1 = 32 - 12 = 9 - 1 = 8
d2 = a3 - a2 = 52 - 32 = 25 - 9 = 16
Here, d1 ≠ d2
So, This series is not form in A.P.
```
```(xv) 12, 52, 72, 73, ....
d1 = a2 - a1 = 52 - 12 = 25 - 1 = 24
d2 = a3 - a2 = 72 - 52 = 49 - 25 = 24
Here, d1 = d2
So, This series is form in A.P.
Now, a = 1, d = 24
According to question, 3 more terms are a5, a6 and a7
a5 = 1 + (5-1)(24) = 1 + 96 = 97
a6 = 1 + (6-1)(24) = 1 + 120 = 121
a7 = 1 + (7-1)(24) = 1 + 144 = 145
```
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