CBSE Class 10 Mathematics chapter 5 - ARITHMETIC PROGRESSIONS - Exercise 5.1
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Q1. |
In which of the following situations, does
the list of numbers involved make an arithmetic progression, and why? |
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(i) |
The taxi fare after each km when the fare
is Rs. 15 for the first km and Rs. 8 for each additional km. |
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Solution. |
Given – Taxi fare for 1 km = Rs. 15 For each additional km. = Rs. 8 So, Taxi fare for 2 km = Rs. 15 + 8 = 23 Taxi fare for 3 km = Rs. 23 + 8 = 31 Therefore, series makes 15, 23, 31 ……. Difference between 2nd and 1st
term (a2—a1) = 23 —15 = 8 Difference between 3rd and 2nd
term (a3—a2) = 31—23 = 8 Because difference is same so it is an AP. |
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(ii) |
The amount of air present in a cylinder
when a vacuum pump removes ¼ of the air remaining in the cylinder at a time. |
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Solution. |
Given – Let quantity of air in cylinder = 16 litres Air removed by pump in each step = ¼ So, Air removed by pump in 1st step
= ¼ of 16 = ¼×16 = 4 litres After first step balanced air in cylinder =
16—4 = 12 Air removed by pump in 2nd step
= ¼ of 12 = ¼×12 = 3 litres After second step balanced air in cylinder
= 12—3 = 9 Air removed by pump in 3rd step
= ¼ of 9 = ¼×9 = 2.25 litres After third step balanced air in cylinder =
9—2.25 = 6.75 Therefore, series makes 16, 12, 9, 6.25 ……. Difference between 2nd and 1st
term (a2—a1) = 16 —12 = 4 Difference between 3rd and 2nd
term (a3—a2) = 12—9 = 3 Difference between 4th and 3rd
term (a4—a3) = 9—6.75 = 2.25 Because difference is not same so it is not
an AP. |
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(iii) |
The cost of digging a well after every
metre of digging, when it costs Rs. 150 for the first metre and rises by Rs.
50 for each subsequent metre. |
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Solution. |
Given – Digging cost for 1 m = Rs. 150 For each additional m = Rs. 50 So, Digging cost for 2 m = Rs. 150 + 50 = 200 Digging cost for 3 m = Rs. 200 + 50 = 250 Therefore, series makes 150, 200, 250 ……. Difference between 2nd and 1st
term (a2—a1) = 200 —150 = 50 Difference between 3rd and 2nd
term (a3—a2) = 250—200 = 50 Because difference is same so it is an AP. |
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(iv) |
The amount of money in the account every
year, when Rs. 10000 is deposited at compound interest at 8 % per annum. |
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Solution. |
Given – Amount deposited in account = Rs. 10000 Rate of compound interest = 8% We know that compound interest = P(1+r/100)T
or P(1+r)T So, Amount after 1 year = 10000(1+8/100)1
= 10000(1+0.08)1 = 10000×1.08 = 10800 As we can see, every year amount increase
at 1.08 times, so Amount after 2 year = 10800×1.08 = 11664 Amount after 3 year = 11664×1.08 = 12597.12 Therefore, series makes 10800, 11664, 12597.12 ……. Difference between 2nd and 1st
term (a2—a1) = 11664 —10800 = 864 Difference between 3rd and 2nd
term (a3—a2) = 12697.12—11664 = 933.12 Because difference is not same so it is not
an AP. |
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Q2. |
Write first four terms of the AP, when the
first term a and the common difference d are given as follows: |
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(i) |
a = 10, d = 10 |
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Solution. |
Given First term (a) = 10 Common difference = 10 We know that a, a + d, a + 2d, a + 3d, . . . By putting value Second term (a+d) = 10 + 10 = 20 Third term (a+2d) = 10 + (2×10) = 30 Fourth term (a+3d) = 10 + (3×10) = 40 Therefore, first four terms of AP are 10, 20, 30, 40, ……… |
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(ii) |
a = –2, d = 0 |
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Solution. |
Given First term (a) = –2 Common difference = 0 We know that a, a + d, a + 2d, a + 3d, . . . By putting value Second term (a+d) = –2 + 0 = –2 Third term (a+2d) = –2 + (2×0) = –2 Fourth term (a+3d) = –2 + (3×0) = –2 Therefore, first four terms of AP are –2, –2, –2, –2, ……… |
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(iii) |
a = 4, d = – 3 |
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Solution. |
Given First term (a) = 4 Common difference = —3 We know that a, a + d, a + 2d, a + 3d, . . . By putting value Second term (a+d) = 4 + —3 = 1 Third term (a+2d) = 4 + (2×(—3)) = —2 Fourth term (a+3d) = 4 + (3×(—3)) = —5 Therefore, first four terms of AP are 4, 1, —2, —5, ……… |
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(iv) |
a = – 1, d = ½ |
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Solution. |
Given First term (a) = —1 Common difference = ½ We know that a, a + d, a + 2d, a + 3d, . . . By putting value Second term (a+d) = —1 + ½ = —½ Third term (a+2d) = —1 + (2×½) = —1+1 = 0 Fourth term (a+3d) = —1 + (3×½) = —1+3/2 =
½ Therefore, first four terms of AP are —1, —½, 0, ½, ……… |
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(v) |
a = — 1.25, d = — 0.25 |
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Solution. |
Given First term (a) = —1.25 Common difference = — 0.25 We know that a, a + d, a + 2d, a + 3d, . . . By putting value Second term (a+d) = —1.25 + —0.25 = —1.50 Third term (a+2d) = —1.25 + (2×(—0.25)) = —1.75 Fourth term (a+3d) = —1.25 + (3×(—0.25)) = —2.00 Therefore, first four terms of AP are —1.25, —1.50, —1.75, —2.00 ……… |
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Q3. |
For the following APs, write the first term
and the common difference: |
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(i) |
3, 1, – 1, – 3, . . . |
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Solution. |
In this AP First term (a) = 3 We know that d = an+1
— an d = Second term — First term d = 1 — 3 = —2 Hence, a = 3 and d = —2 |
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(ii) |
– 5, – 1, 3, 7, . . . |
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Solution. |
In this AP First term (a) = —5 We know that d = an+1
— an d = Second term — First term d = (—5) — (—1) = —5 +1 = 4 Hence, a = —5 and d = 4 |
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(iii) |
 |
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Solution. |
 |
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(iv) |
0.6, 1.7, 2.8, 3.9, . . . |
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Solution. |
In this AP First term (a) = 0.6 We know that d = an+1
— an d = Second term — First term d = 1.7 — 0.6 = 1.1 Hence, a = 0.6 and d = 1.1 |
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Q4. |
Which of the following are APs ? If they
form an AP, find the common difference d and write three more terms. |
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(i) |
2, 4, 8, 16, . . . |
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Solution. |
Difference between 2nd and 1st
term (a2—a1) = 4 —2 = 2 Difference between 3rd and 2nd
term (a3—a2) = 8—4 = 4 Difference between 4th and 3rd
term (a4—a3) = 16—8 = 8 Because difference is not same so it is not
an AP. |
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(ii) |
 |
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Solution. |
 |
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(iii) |
– 1.2, – 3.2, – 5.2, – 7.2, . . . |
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Solution. |
Difference between 2nd and 1st term (a2—a1)
= (—3.2)—(—1.2) = —3.2+1.2 = —2.0 Difference between 3rd and 2nd term (a3—a2)
= (—5.2)—(—3.2) = —5.2+3.2 = —2.0 Difference between 4th and 3rd term (a4—a3)
= (—7.2)—(—5.2) = —7.2+5.2 = —2.0 Because difference is same so it is an AP. Here common difference = —2.0 So, three more term 5th, 6th,
7th is: a4 + d = (—7.2) + (—2.0) = —7.2 + 2.0
=—9.2 a5 + d = (—9.2) + (—2.0) = —9.2 + 2.0
=—11.2 a6 + d = (—11.2) + (—2.0) = —11.2 +
2.0 =—13.2 |
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(iv) |
– 10, – 6, – 2, 2, . . . |
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Solution. |
Difference between 2nd and 1st term (a2—a1)
= (—6)—(—10) = —6+10 = 4 Difference between 3rd and 2nd term (a3—a2)
= (—2)—(—6) = —2+6 = 4 Difference between 4th and 3rd term (a4—a3)
= (2)—(—2) = 2+2 = 4 Because difference is same so it is an AP. Here common difference = 4 So, three more term 5th, 6th,
7th is: a4 + d = 2+ 4 = 6 a5 + d = 6 + 4 = 10 a6 + d = 10 + 4 = 14 |
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(v) |
3, 3 + √2
, 3+ 2√2
, 3+3√2
, . . . |
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Solution. |
Difference
between 2nd and 1st term (a2—a1)
= 3 + √2—3
= √2 Difference
between 3rd and 2nd term (a3—a2) = (3+ 2√2)—( 3 + √2) = 3+ 2√2— 3 — √2 = 3— 3 + 2√2 — √2 = 0+ √2 (2—1) = √2 Difference
between 4th and 3rd term (a4—a3) = = (3+3√2)—( 3+ 2√2) = 3+ 3√2— 3 — 2√2 = 3— 3 + 3√2 — 2√2 = 0+ 3√2 — 2√2 = √2 (3— 2) = √2 Because difference is same so it is an AP. Here common difference = √2 So, three more term 5th, 6th,
7th is: a4 + d = 3+3√2+ √2 = 3+√2(3+1) = 3+√2(4) = 3+4√2 a5 + d = 3+4√2 + √2 = 3+√2 (4+1) = 3+√2 (5) = 3+5√2 a6 + d = 3+5√2 + √2 = 3+√2 (5+1) = 3+√2 (6) = 3+6√2 |
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(vi) |
0.2, 0.22, 0.222, 0.2222, . . . |
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Solution. |
Difference
between 2nd and 1st term (a2—a1)
= (0.22)—(0.2) = 0.02 Difference
between 3rd and 2nd term (a3—a2)
= (0.222)—(0.22) = 0.002 Difference
between 4th and 3rd term (a4—a3)
= (0.2222)—(0.222) = 0.0002 Because difference is not same so it is not
an AP. |
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(vii) |
0, – 4, – 8, –12, . . . |
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Solution. |
Difference
between 2nd and 1st term (a2—a1)
= (—4)—(0) = —4 Difference
between 3rd and 2nd term (a3—a2)
= (—8)—(—4) = —8+4 = —4 Difference
between 4th and 3rd term (a4—a3)
= (—12)—(—8) = —12+8 = —4 Because difference is same so it is an AP. Here common difference = —4 So, three more term 5th, 6th,
7th is: a4 + d = (—12) + (—4) = —12— 4 = —16 a5 + d = (—16) + (—4) = —16— 4 = —20 a6 + d = (—20) + (—4) = —20— 4 = —24 |
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(viii) |
 |
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Solution. |
 |
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(ix) |
1, 3, 9, 27, . . . |
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Solution. |
Difference
between 2nd and 1st term (a2—a1)
= 3—1 = 2 Difference
between 3rd and 2nd term (a3—a2)
= 9—3 = 6 Difference
between 4th and 3rd term (a4—a3)
= 27—9 = 18 Because difference is not same so it is not
an AP. |
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(x) |
a, 2a, 3a, 4a, . . . |
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Solution. |
Difference
between 2nd and 1st term (a2—a1)
= 2a—a = a Difference
between 3rd and 2nd term (a3—a2)
= 3a—2a = a Difference
between 4th and 3rd term (a4—a3)
= 4a—3a = a Because difference is same so it is an AP. Here common difference = a So, three more term 5th, 6th,
7th is: a4 + d = 4a + a = 5a a5 + d = 5a + a = 6a a6 + d = 6a + a = 7a |
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(xi) |
a, a2, a3, a4,
. . . |
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Solution. |
Difference
between 2nd and 1st term (a2—a1)
= a2—a = a(a—1) Difference
between 3rd and 2nd term (a3—a2)
= a3—a2 = a2(a—1) Difference
between 4th and 3rd term (a4—a3)
= a4—a3 = a3(a—1) Because difference is not same so it is not
an AP. |
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(xii) |
√2, √8, √18, √32, ….. |
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Solution. |
 |
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(xiii) |
√3, √6, √9, √12, ….. |
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Solution. |
 |
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(xiv) |
12, 32, 52, 72,
. . . |
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Solution. |
Difference
between 2nd and 1st term (a2—a1)
= 32—12 = 9—1 = 8 Difference
between 3rd and 2nd term (a3—a2)
= 52—32 = 25—9 = 16 Because difference is not same so it is not
an AP. |
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(xv) |
12, 52, 72, 73,
. . . |
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Solution. |
Difference
between 2nd and 1st term (a2—a1)
= 52—12 = 25—1 = 24 Difference
between 3rd and 2nd term (a3—a2)
= 72—52 = 49—25 = 24 Difference
between 3rd and 2nd term (a4—a3)
= 73—72 = 73—49 = 24 Because difference is same so it is an AP. Here common difference = 24 So, three more term 5th, 6th,
7th is: a4 + d = 73 + 24 = 97 a5 + d = 97 + 24 = 121 a6 + d = 121 + 24 = 145 |
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