NCERT Solution for Arithmetic Progressions

(i) The taxi fare after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional km.

(ii) The amount of air Present in a cylinder when a vacuum pump removes 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 Rs. 50 for each subsequent metre.

(iv) The amount of money in the account every year, when Rs. 10,000 is deposited at compound interest at 8% per annum.

Sol. (i) Let us consider,

The first term (T_{1}) = Fare for the first km = Rs. 15 since, the taxi fare beyond the first km is Rs. 8 for each additional km. ⇒ T_{1} = 15

∴ Fare for 2 km = Rs. 15 + 1 x Rs. 8 ⇒ T_{2} = a + 8 [where a = 15]

Fare for 3km = Rs.15 + 2 × Rs. 8 ⇒ T_{3} = a + 16

Fare for 4km = Rs.15 + 3 × Rs.8 ⇒ T_{4} = a + 24

Fare for 5km = Rs.15 + 4 x Rs. 8 ⇒ T_{5} = a +32

Fare for n km = Rs. 15 + (n – 1)8 ⇒ T_{n} = a + (n – 1) 8

We see that above terms form an A.P.

(ii) Let the amount of air in the cylinder = x

(iii) Here, The cost of digging for 1st metre = Rs. 150

The cost of digging for first 2 metres = Rs. 150 + 50 = Rs. 200

The cost of digging for first 3 metres = Rs. 150 + (Rs. 50) × 2 = Rs. 250

The cost of digging for first 4 metres = Rs. 150 + (Rs. 50) × 3 = Rs. 300

∴ The terms are: 150, 200, 250, 300, ...

Since, 200 – 150 = 50

And 250 – 200 = 50

⇒ (200 – 150) = (250 – 200)

∴ The above terms form an A.P.

(i) a = 10, d = 10

(ii) a = –2, d = 0

(iii) a = 4, d = – 3

(iv)

(v) a = – 1.25, d = – 0.25

Sol. (i) ∵ T_{n} = a + (n – 1)d

∴ For a = 10 and d = 10, we have:

T_{1} = 10 + (1 – 1) × 10 = 10 + 10 = 10

T_{2} = 10 + (2 – 1) × 10 = 10 + 10 = 20

T_{3} = 10 + (3 – 1) × 10 = 10 + 20 = 30

T_{4} = 10 + (4 – 1) × 10 = 10 + 30 = 40

Thus, the first four terms of A.P. are:

10, 20, 30, 40.

(ii) T = a + (n – 1) d

∴ For a = – 2and d = 0, we have:

T_{1} = – 2 + (1 – 1) × 0 = – 2 + 0 = –2

T_{2} = – 2 + (2 – 1) × 0 = – 2 + 0 = –2

T_{3} = – 2 + (3 – 1) × 0 = – 2 + 0 = –2

T_{4} = – 2 + (4 – 1) × 0 = – 2 + 0 = –2

∴ The first four terms are:

– 2, – 2, – 2 – 2.

(iii) T_{n} = a + (n – 1)d

∴ For a = 4 and d = – 3, we have:

T_{1} = 4 + (1 – 1) × ( – 3) = 4 + 0 = 4

T_{2} = 4 + (2 – 1) × ( – 3) = 4 + (–3) = 1

T_{3} = 4 + (3 – 1) × ( – 3) = 4 + (–6) = –2

T_{4} = 4 + (4 – 1) × ( – 3) = 4 + (–9) = –5

Thus, the first four terms are:

4, 1, – 2, – 5.

(iv) T_{n} = a + (n – 1)d

(v) T_{n} = a + (n – 1)d

∴ For a = –1.25 and d = –0.25, we get

T_{1} = –1.25 + (1 – 1) × (–0.25) = –1.25 + 0 = – 1.25

T_{2} = –1.25 + (2 – 1) × (–0.25) = –1.25 + 0 = – 1.25

T_{3} = –1.25 + (3 – 1) × (–0.25) = –1.25 + 0 = – 1.25

T_{4} = –1.25 + (4 – 1) × (–0.25) = –1.25 + 0 = – 1.25

Thus, the four terms are:

–1.25 , –1.50, –1.75, –2.0

(i) 3, 1, –1, –3, ....

(ii) –5, –1, 3, 7, ....

(iii)

(iv) 0.6, 1.7, 2.8, 3.9, ...

Sol. (i) We have : 3, 1, –1, –3, ....

⇒ T_{1} = 3 ⇒ a = 3

T_{2} = 1

T_{3} = – 1

T_{4} = – 3

∴ T_{2} – T_{1} = 1 – 3 = –2

T_{4} – T_{3} = –3 – (–1) = –3 + 2 = –2

Thus, a = 3 and d = –2

(ii) We have: –5, –1, 3, 7....

⇒ T_{1} = –5 ⇒ a = –5

T_{2} = –1 ⇒ d = T_{2} – T_{1} = –1 – (–5) = –1 + 5 = 4

T_{3} = 3 ⇒ d = –1 + 5 = 4

T_{4} = 7 T_{4} – T_{3} = 7 – 3 = 4 ⇒ d = 4

Thus, a = –5 and d = 4

(iv) We have : 0.6, 1.7, 2.8, 3.9, ....

⇒ T_{1} = 0.6 ⇒ a = 0.6

T_{2} = 1.7 ⇒ d = T_{2} – T_{1} = 1.7 – 0.6 = 1.1

T_{3} = 2.8

T_{4} = 3.9 ⇒ d = T_{4} – T_{3} = 3.9 – 2.8 = 1.1

Thus, a = 0.6 and d = 1.1

∴ The numbers form an A.P.

Now, T_{5} = T_{4} + 24 = 73 + 24 = 97

T_{6} = T_{5} + 24 = 97 + 24 = 121

T_{7} = T_{6} + 24 = 121 + 24 = 145

Thus, d = 24 and T_{5} = 97, T_{6} = 121, T_{7} = 145

Sol. (i) a_{n} = a + (n – 1) d

⇒ a_{8} = 7 + (8 – 1) 3

= 7 + 7 × 3

= 7 + 21

⇒ a_{8} = 28

(ii) ⇒ a_{10} = – 18 + (10 – 1)d

⇒ 0 = – 18 + 9d

∴ d – 2

(iii) a_{n} = a + (n – 1) d

⇒ – 5 = a + (18 – 1) × ( – 3)

⇒ – 5 = a + 17 × ( – 3)

⇒ – 5 = a – 51

⇒ a = –5 + 51 = 46

Thus, a = 46

(iv) a_{n} = a + (n – 1)d

⇒ 3.6 = – 18.9 + (n – 1) × 2.5

⇒ (n – 1) × 2.5 = 3.6 + 18.9

⇒ (n – 1) × 2.5 = 22.5

⇒ n = 9 + 1 = 10 = 10

Thus, n = 10

(v) a_{n} = a + (n – 1)d

⇒ a_{n} = 3.5 + (105 – 1) × 0

⇒ a_{n} = 3.5 + 104 × 0

⇒ a_{n} = 3.5 + 0 = 3.5

Thus, a_{n} = 3.5

(i) 30th term of the A.P.: 10, 7, 4, ...., is

(A) 97 (B) 77 (C) –77 (D) –87

Sol. (i) Here, a = 10, n = 30

T_{n} = a + (n – 1)d and d = 7 – 10= – 3

∴ T_{30} = 10 + (30 – 1) × ( –3)

⇒ T_{30} = 10 + 29 × ( –3)

⇒ T_{30} = 10 – 87 = –77

Thus, the correct choice is (C) – 77.

Sol. (i) Here, a = 2, T_{3} – 26

Let common difference = d

∴ T_{n} = a + (n – 1) d

⇒ T_{3} = 2 + (3 – 1) d

⇒ 26 = 2 + 2d

⇒ 2d = 26 – 2 = 24

∴ The missing term = a + d

= 2 + 12 = 14

(ii) Let the first term = a and common difference = d

Here, T_{2} = 13 and T_{4} = 3

T_{2} = a + d = 13

T_{4} = a + 3d = 3

T_{4} – T_{2} = (a + 3d) – (a + d) = 3 – 13

⇒ 2d = – 10

Now, a + d = 13 ⇒ a + (–5) = 13

⇒ a = 13 + 5 =18

Thus, missing terms are a and a + 2d or 18 and 18 + (–10) = 8

i.e., T_{1} = 18 and T_{3} = 8

(iv) Here, a = – 4 and T_{6} = 6

∵ T_{n} = a + (n – 1) d

∴ T_{6} = – 4 + (6 – 1)d

⇒ 6 = – 4 + 5d

⇒ 5d = 6 + 4 = 10

⇒ d = 10 5=2

∴ T_{2} = a + d – 4 + 2 = – 2

T_{3} = a + 2d = –4 + 2(2) = 0

T_{4} = a + 3d = –4 + 3(2) = 2

T_{5} = a + 4d = – 4 + 4(2) = 4

∴ The missing terms are

(v) Here, T_{2} = 38 and T_{6} = – 22

∴ T_{2} = a + d = 38

T_{6} = a + 5d = –22

⇒ T_{6} – T_{2} = a + 5d – (a + d)= – 22 – 38

⇒ 4d = – 60

∴ a + d = 38 ⇒ a + ( – 15) = 38

⇒ a = 38 + 15 = 53

Now, T_{3} = a + 2d = 53 + 2 (–15) = 53 – 30 = 23

T_{4} = a + 3d = 53 + 3 (–15) = 53 – 45 = 8

T_{5} = a + 4d = 53 + 4(–15) = 53 – 60 = – 7

Thus, the missing terms are

Sol. Let the nth term = 78

Here, a = 3, ⇒ T_{1} = 3 and T_{2} = 8

∴ d = T_{2} – T_{1} = 8 – 3 = 5

Now, T_{n} = a + (n – 1) d

⇒ 78 = 3 + (n – 1) × 5

⇒ 78 – 3 = (n – 1) × 5

⇒ 75 = (n – 1) × 5

⇒ (n – 1) = 75 5 =15

⇒ n = 15 + 1 = 16

Thus, 78 is the 16th term of the given A.P.

(i) 7, 13, 19, ..., 205

(ii)

Sol. (i) Here, a = 7

d = 13 – 7 = 6

Let the number of terms be n

T_{n} = 205

Now, T_{n} = a + (n – 1) × d

⇒ 7 + (n – 1) × 6 = 205

⇒ (n – 1) × 6 = 205 – 7 = 198

∴ n = 33 + 1 = 34

Thus, the required number of terms is 34.

(ii) Here, a = 18

⇒ n – 1 = (13) × (–2) = 26

⇒ n = 26 + 1 = 27

Thus, the required number of terms is 27.

Sol. For the given A.P., we have

a = 11

d = 8 – 11 = – 3

Let – 150 is the nth term of the given A.P.

∴ T_{n} = a + (n – 1)d

⇒ –150 = 11 + (n – 1) × (–3)

⇒ –150 – 11 = (n – 1) × (–3)

⇒ –161 = (n – 1) × (–3)

But n should be a positive integer.

Thus, – 150 is not a term of the given A.P.

Sol. Here, T_{31} = ?

T_{11} = 38

T_{16} = 73

If the first term = a and the common difference = d.

Then,

a + (11 – 1) d = 38

⇒ a + 10 – d = 38

and a + (16 – 1) d = 73

⇒ a + 15d = 73

Subtracting (1) from (2), we get

(a + 15d) – (a + 10 – d) = 73 – 38

⇒ 5d = 35

From (1),

a + 10(7) = 38

⇒ a + 70 = 38

⇒ a = 38 – 70 = – 32

∴ T_{31} = – 32 + (31 – 1) × 7

⇒ T_{31} = – 32 + 30 × 7

⇒ T_{31} = – 32 + 210

⇒ T_{31} = 174

Thus, the 31st term is 178.

Sol. Here, n = 50

T_{3} = 12

T_{n} = 106 ⇒ T_{50} = 106

If first term = a and the common difference = d

∴ T_{3} = a + 2d = 12 ...(1)

T_{50} = a + 49d = 106 ...(2)

⇒ T_{50} – T_{3} = a + 49d – (a + 2d) = 106 – 12

⇒ 47d = 94

From (1), we have

a + 2d = 12 ⇒ a + 2(2) = 12

⇒ a = 12 – 4 = 8

Now, T_{29} = a + (29 – 1) d

= 8 + (28) × 2

= 8 + 56 = 64

Thus, the 29th term is 64.

Sol. Here, T_{3} = 4 and T_{9} = –8

∴ Using T_{n} = a + (n – 1) d

⇒ T_{3} = a + 2d = 4 ...(1)

T_{9} = n + 8d = – 8 ...(2)

Subtracting (1) from (2) we get

(a + 8d) – (a + 2d) = – 8 – 4

⇒ 6d = – 12

Now, from (1), we have:

a + 2d = 4

⇒ a + 2(–2) = 4

⇒ a – 4 = 4

⇒ a = 4 + 4 = 8

Let the nth term of the A.P. be 0.

∴ T_{n} = a + (n – 1)d = 0

⇒ 8 + (n – 1) × (–2) = 0

⇒ (n – 1) × – 2 = – 8

⇒ n = 4 + 1 = 5

Thus, the 5th term of the A.P. is 0.

Sol. Let 'a' be the first term and 'd' be the common difference of the given A.P.

Now, using T_{n} = a + (n – 1) d

T_{17} = a + 16d

T_{10} = a + 9d

According to the condition,

T_{n} + 7 = T_{17}

⇒ (a + 9d) + 7 = a + 16d

⇒ a + 9d – a – 16d = – 7

⇒ – 7d = – 7 ⇒ d = 1

Thus, the common difference is 1.

Sol. Here, a = 3

d = 15 – 3 = 12

Using T_{n} = a + (n – 1) d, we get

T_{54} = a + 53d

= 3 + 53 × 12

= 3 + 636 = 639

Let a_{n} be 132 more than its 54th term.

∴ a_{n}= T_{54} + 132

⇒ a_{n} = 639 + 132 = 771

Now a_{n} = a + (n – 1)d = 771

⇒ 3 + (n – 1) × 12 = 771

⇒ (n – 1) × 12 = 771 – 3 = 768

⇒ n = 64 + 1 = 65

Thus, 132 more than 54th term is the 65th term.

Sol. Let for the 1st A.P., the first term = a

∴ T_{100} = a + 99d

And for the 2nd A.P., the first term = a'

∴ T_{100} = a' + 99d

According to the condition, we have:

T_{100} – T'100 = 100

⇒ a + 99d – (a' + 99d) = 100

⇒ a – a' = 100

Let, T_{100} – T' = x

∴ a + 999d – (a' + 999d) = x

⇒ a – a' = x ⇒ x = 100

∴ The difference between the 1000th terms is 100.

Ans. The first three digit number divisible by 7 is 105.

The last such three digit number is 994.

∴ The A.P. is 105, 112, 119, ..., 994

Here, a = 105 and d = 7

Let n be the required number of terms.

∴ T_{n} = a + (n – 1)d

⇒ 994 = 105 + (n – 1) × 7

⇒ (n – 1) × 7 = 994 – 105 = 889

⇒ n = 127 + 1 = 128

Thus, 128 numbers of 3–digit are divisible by 7.

Sol. ∵ The first multiple of 4 beyond 10 is 12.

The multiple of 4 just below 250 is 248.

∴ The A.P. is given by:

12, 16, 20,.... 248

Here, a = 12 and d = 4

Let the number of terms = n

∴ Using T_{n} = a + (n – 1) d, we get

∴ T_{n} = 12 + (n – 1) × 4

⇒ 248 = 12 + (n – 1) × 4

⇒ (n – 1) × 4 = 248 – 12 = 236

⇒ n = 59 + 1 = 60

Thus, the required number of terms = 60.

Sol. For the 1st A.P.

∵ a = 63 and d = 65 – 63 = 2

∴ T_{n} = a + (n – 1)d

⇒ T_{n} = 63 + (n – 1) × 2

For the 2nd A.P.

a = 3 and d = 10 – 3 = 7

∴ T_{n} = a + (n – 1) d

⇒ T_{n} = 3 + (n – 1) × 7

Now, according to the condition,

3 + (n – 1) × 7 = 63 + (n – 1) × 2

⇒ (n – 1) × 7 – (n – 1) × 2 = 63 – 3

⇒ 7n – 7 – 2n + 2 = 60

⇒ 5n – 5 = 60

⇒ 5n = 60 + 5 = 65

Thus, the 13th terms of the two given A.Ps. are equal.

Sol. Let the first term = a and the common difference = d.

∴ Using T_{n} = a + (n – 1) d, we have:

T_{3} = a + 2d

⇒ a + 2d = 16 ...(1)

And T_{7} = a + 6d, T_{5} = a + 4d

According to the condition,

T_{7} – T_{5} = 12

⇒ (a + 6d) – (a + 4d) = 12

⇒ a + 6d – a – 4d = 12

⇒ 2d = 12

Now, from (1) and (2), we have:

a + 2 (6) = 16

⇒ a + 12 = 16

⇒ a = 16 – 12 = 4

∴ The required A.P. is

4, [4 + 61], [4 + 2 (6)], [4 + 3 (6)], ....

or 4,10,16,22, ....

Sol. We have, the last term I = 253

Here, d = 8 – 3 = 5

Since, the nth term before the last term is given by 1. – (n – 1) d,

∴ We have

20th term from the end = l – (20 – 1) × 5

= 253 – 19 × 5

= 253 – 95 = 158

Find the first three terms of the A.P.

Sol. Let the first term = a

And the common difference = d

∴ Using T_{n} = a + (n – 1) d,

T_{4} + T_{8} = 24

⇒ (a + 3d) + (a + 7d) = 24

⇒ 2a + 10d = 24

⇒ a + 5d = 12

And T_{6} + T_{10} = 44

⇒ (a + 5d) + (a + 9d) = 44

⇒ 2a + 14d = 44

⇒ a + 7d = 22

Now, subtracting (1) from (2), we get

(a + 7d) – (a + 5d) = 22 – 12

⇒ 2d = 10

∴ From (1), a + 5 × 5 = 12

⇒ a + 25 = 12

⇒ a = 12 – 25 = – 13

Now, the first three terms of the A.P. are given by:

a, (a + d), (a + 2d)

or –13, (–13 + 5), [–13 + 2 (5)]

or –13, –8, –3

Sol. Here, a = Rs. 5000 and d = Rs. 200

Say, in the nth year he gets Rs. 7000.

∴ Using T_{n} = a + (n – 1) d, we get

7000 = 5000 + (n – 1) × 200

⇒ (n – 1) × 200 = 7000 – 5000 = 2000

⇒ n = 10 + 1 = 11

Thus, his income becomes Rs. 7000 in 11 years.

If in the nth week, her weekly savings become Rs. 20.75, find n.

Sol. Here, a = Rs. 5 and d = Rs. 1.75

∵ In the nth week her savings become Rs. 20.75.

∴ T_{n} = Rs. 20.75

∴ Using T_{n} = a + (n – 1) d, we have

20.75 = 5 + (n – 1) × (1.75)

⇒ (n – 1) × 1.75 = 20.75 – 5

⇒ (n – 1) × 1.75 = 15.75

⇒ n – 9 + 1 = 10

Thus, the required number of years = 10.

(i) 2, 7, 12, ..., to 10 terms.

(ii) – 37, – 33, – 29, ..., to 12 terms.

(iii) 0.6, 1.7, 2.8, ..., to 100 terms.

(iv)

Sol. (i) Here, a = 2

d = 7 – 2 = 5

n = 10

= 50 [1.2 + 99 × 1.1]

= 50 [1.2 + 108.9]

= 50 [110.1]

= 5505

Thus, the required sum of first 100 terms is 5505.

(iii) Here, a = –5

d = – 8 – (–5) = –3

l = – 230

Let n be the number of terms.

∴ T_{n} = – 230

⇒ –230 = –5 + (n – 1) × (–3)

⇒ (n – 1) × (–3) = –230 + 5 = –225

= 38 × (– 235)

= – 8930

∴ The required sum = – 8930.

(i) given a = 5, d = 3, a_{n} = 50, find n and S_{n}.

(ii) given a = 7, a_{13} = 35, find d and S_{13}.

(iii) given a_{12} = 37, d = 3, find a and S_{12}.

(iv) given a_{3} = 15, S_{10} = 125, find d and a_{10}.

(v) given d = 5, S_{9} = 75, find a and a_{9}.

(vi) given a = 2, d = 8, S_{n} = 90, find n and an.

(vii) given a = 8, a_{n} = 62, S_{n} = 210, find n and d.

(viii) given a_{n} = 4, d = 2, S_{n} = – 14, find n and a.

(ix) given a = 3, n = 8, S = 192, find d.

(x) given l = 28, S = 144, and there are total 9 terms. Find a.

Sol. (i) Here, a = 5, d = 3 and a_{n} = 50 = l

a_{n} = a + (n – 1)d

∴ 50 = 5 + (n – 1) × 3

⇒ 50 – 5 = (n – 1) × 3

⇒ (n – 1) × 3 = 45

= 8 (55) = 440

Thus, n = 16 and S_{n} = 440

(ii) Here, a = 7 and a_{13} = 35 = l

∴ a_{n} = a + (n – 1) d

⇒ 35 = 7 + (13 – 1)d

⇒ 35 – 7 = 12d

⇒ 28 = 12d

(iii) Here, a_{12} = 37 = l and d = 3

Let the first term of the A.P. be 'a'.

Now a_{12} = a + (12 – 1) d

⇒ 37 = a + 11d

⇒ 37 = a + 11 × 3

⇒ 37 = a + 33

⇒ a = 37 – 33 = 4

(iv) a_{3} = 15 = l

S_{10} =125

Let first term of the A.P. be 'a' and the common difference = d

∴ a_{3} = a + 2d

⇒ a + 2d = 15 ...(1)

⇒ 125 = 5 [2a + 9d]

⇒ 2a + 9d = 25

Multiplying (1) by 2 and subtracting (2) from it, we get

2 [a + 2d =15] – [2a + 9d = 25]

⇒ 2a + 4d – 2a – 9d = 30 – 25

⇒ –5d = 5

∴ From (1), a + 2 (–1) = 15 ⇒ a = 15 + 2 ⇒ a = 15 + 2 ⇒ a = 17

Now, a_{10} = a + (10 – 1) d

= 17 + 9 × (– 1)

= 17 – 9 = 8

Thus, d = – 1 and a_{10} = 8

(v) Here, d = 5, S_{9} = 75

Let the first term of the A.P. is 'a'.

(vi) Here, a = 2, d = 8 and S_{n} = 90

⇒ 90 × 2 = 4n + n (n – 1) × 8

⇒ 180 = 4n + 8n^{2} – 8n

⇒ 180 = 8n^{2} – 4n

⇒ 45 = 2n^{2} – n

⇒ 2n^{2} – n – 45 = 0

⇒ 2n^{2} – 10n + 9n – 45 = 0

⇒ 2n (n – 5) + 9 (n – 5) = 0

⇒ (2n + 9) (n – 5) = 0

∴ Either 2n + 9 = 0 ⇒ n =

or n – 5 = 0 ⇒ n = 5

But n = is not required.

∴ n = 5

Now, a_{n} = a + (n – 1)d

⇒ a_{5} = 2 + (5 – 1) × 8

= 2 + 32 = 34

Thus, n = 5 and a_{5} = 34.

(vii) Here, a = 8, a_{n} = 62 = l and S_{n} = 210

Let the common difference = d

Now, S_{n} = 210

(viii) Here, a_{n} = 4, d = 2 and S_{n} = – 14

Let the first term be 'a'.

∵ a_{n} = 4

∴ a + (n –1)2 = 4

⇒ a + 2n –2 = 4

⇒ a = 4 – 2n + 2

⇒ a = 6 – 2n

Also S_{n} = –14

⇒ (a + l) = –14

⇒ (a + 4) = –14

⇒ n (a + 4) = – 28

Substituting the value of a from (1) into (2),

n [6 – 2n + 4] = – 28

⇒ n [10 – 2n] = –28

⇒ 2n [5 – n] = –28

⇒ n (5 – n) = –14

⇒ 5n – n^{2} + 14 = 0

⇒ n^{2} – 5n – 14 = 0

⇒ n^{2} – 7n + 2n – 14 = 0

⇒ n (n – 7) + 2 (n – 7) = 0

⇒ (n – 7) (n + 2) = 0

∴ Either n – 7 = 0 ⇒ n = 7

or n + 2 = 0 ⇒ n = –2

But n cannot be negative,

∴ n = 7

Now, from (1), we have

a = 6 – 2 × 7 ⇒ a = –8

Thus, a = – 8 and n = 7

(ix) Here, a = 3, n = 8 and S_{n} = 192

Let the common difference = d.

(x) Here, I = 28 and S_{9} = 144

Let the first term be 'a'.

Sol: Here, a = 9

Sol. Here, a = 5

l = 45 = T_{n}

S_{n} = 400

∵ T_{n} = a + (n –1) d

∴ 45 = 5+ (n – 1)d

⇒ (n – 1)d = 45 – 5

⇒ (n – 1)d = 40 ...(1)

Sol. We have,

First term a = 17

Last term l = 350 = T_{n}

Common difference d = 9

Let the number of terms be 'n'

T_{n} = a + (n – 1) d

∴ 350 = 17 + (n – 1) × 9

⇒ (n – 1) × 9 = 350 – 17 = 333

Sol. Here, n = 22, T_{22} = 149 = l

d = 7

Let the first term of the A.P. be 'a'.

∴ T_{n} = a + (n – 1) d

⇒ T_{n} = a + (22 – 1) × 7

⇒ a + 21 × 7 = 149

⇒ a + 147 = 149

⇒ a = 149 – 147 = 2

Sol. Here, n = 51, T_{2} = 14 and T_{3} = 18

Let the first term of the A.P. be 'a' and the common difference is d.

∴ We have:

T_{2} = a + d ⇒ a + d = 14 ...(1)

T_{3} = a + 2d ⇒ a + 2d = 18 ...(2)

Subtracting (1) from 2, we get

a + 2d – a – d = 18 – 14

⇒ d = 14

From (1), we get

a + d = 14 ⇒ a + 4 = 14

⇒ a = 14 – 4 = 10

Sol. Here, we have:

S_{7} = 49 and S_{17} = 289

Let the first term of the A.P, be 'a' and 'd be the common difference, then

(i) a_{n} = 3 + 4n (ii) a_{n} = 9 – 5n

Also find the sum of the first 15 terms in each case.

Sol. (i) Here, a_{n} = 3 + 4n

Putting n = 1, 2, 3, 4, ... n, we get:

a_{1} = 3 + 4(1) = 7

a_{2} = 3 + 4(2) = 11

a_{3} = 3 + 4(3) – 15

a_{4} = 3 + 4(4) = 19

.... .... ....

a_{n} = 3 + 4n

∴ The A.P. in which a = 7 and d = 11 – 7 = 4 1s:

7, 11, 15, 19, ...., (3 + 4n).

(ii) Here, a_{n} = 9 – 5n

Putting n = 1, 2, 3, 4,....., n, we get

a_{1} = 9 – 5 (1) = 4

a_{2} = 9 – 5 (2) = –1

a_{3} = 9 – 5 (3) = –6

a_{4} = 9 – 5 (4) = –11

∴ The A.P. is:

Sol. We have:

S_{n} = 4n – n^{2}

∴ S_{1} = 4 (1) – (1)2

= 4 – 1 = 3 ⇒ First term = 3

S_{2} = 4 (2) – (2)2

= 8 – 4 = 4 ⇒ Sum of first two terms = 4

∴ Second term (S_{2} – S_{1}) = 4 – 3 = 1

S_{3} = 4 (3) – (3)2

= 12 – 9 = 3 ⇒ Sum of first 3 terms = 3

∴Third term (S_{3} – S_{2}) = 3 – 4 = –1

S_{9} = 4 (9) – (9)2

= 36 – 81 = –45

S_{10} = 4 (10) – (10)2

= 40 – 100 = – 60

∴ Tenth term = S_{10} – S_{9} = [– 60] – [– 45] = – 15

Now, S_{n} = 4 (n) – (n)2 = 4n – n^{2}

Also S_{n} – 1 = 4 (n – 1) – (n – 1)2

= 4n – 4 – [n^{2} – 2n +1]

= 4n – 4 – n^{2} + 2n – 1

= 6n – n^{2} – 5

∴ nth term = S_{n} – S_{n} – 1

= [4n – n^{2}] – [6n – n^{2} – 5]

= 4n – n^{2} – 6n + n^{2} + 5 = 5 – 2n

Thus, S_{1} = 3 and a_{1} = 3

S_{2} = 4 and a_{2} = 1

S_{3} = 3 and a_{3}= –1

a_{10} = – 15 and an = 5 – 2n

Sol: ∵ The first 40 positive integers divisible by 6 are:

6,12,18, ...., (6 × 40).

And, these numbers are in A.P. such that

Thus, the sum of first 40 multiples of 6 is 4920.

Sol. The first 15 multiples of 8 are:

8, (8 × 2), (8 × 3), (8 × 4), ..., (8 × 15)

or 8, 16, 24, 32, ...., 120.

These numbers are in A.P., where

a = 8 and l = 120

= 15 × 64 = 960

Thus, the sum of first positive 15 multiples of 8 is 960.

Sol. Odd numbers between 0 and 50 are:

1, 3, 5, 7, ...., 49

The numbers are in A.P. such that

a = 1 and I = 49

Here, d = 3 – 1 = 2

∴ T_{n} = a + (n – 1)d

rArr; 49 = 1 + (n – 1)2

⇒ 49 – 1 = (n – 1)2

Thus, the sum of odd numbers between 0 and 50 is 625.

Sol. Here, penalty for delay on

1st day = Rs. 200

2nd day = Rs. 250

3rd day = Rs. 300

...............

...............

Now, 200, 250, 300, .... are in A.P. such that

a = 200, d = 250 – 200 = 50

∴ S_{30} is given by

= 15 [400 + 29 × 50]

= 15 [400 + 1450]

= 15 × 1850 = 27,750

Thus, penalty for the delay for 30 days is Rs. 27,750.

Sol. Sum of all the prizes = Rs. 700

Let the first prize = a

∴ 2nd prize = (a – 20)

3rd prize = (a – 40)

4th prize = (a – 60)

Thus, we have, first term = a

Common difference = – 20

Number of prizes, n = 7

Sum of 7 terms S_{n} = 700

Thus, the values of the seven prizes are:

Rs. 160, Rs. (160 – 20), Rs. (160 – 40), Rs. (160 – 60), Rs. (160 – 80), Rs. (160 – 100) and Rs. (160 – 120)

⇒ Rs. 160, Rs. 140, Rs. 120, Rs. 100, Rs. 80, Rs. 60 and Rs 40..

Sol. Number of classes = 12

∵ Each class has 3 sections.

Number of plants planted by class I = 1 × 3 = 3

Number of plants planted by class II = 2 × 3 = 6

Number of plants planted by class III = 3 × 3 = 9

Number of plants planted by class IV = 4 × 3 = 12

...................................................................................................

Number of plants planted by class XII = 12 × 3 = 36

The numbers 3, 6, 9, 12, ........., 36 are in A.P.

Here, a = 3 and d = 6 – 3 = 3

∵ Number of classes = 12

i.e., n = 12

∴ Sum of the n terms of the above A.P., is given by

= 6 [6 + 11 × 3]

= 6 [6 + 33]

= 6 × 39 = 234

Thus, the total number of trees = 234.

Sol. ∵ Length of a semi–circle = semi–circumference

Sol. We have:

The number of logs:

1st row = 20

2nd row = 19

3rd row = 18

obviously, the numbers

20, 19, 18, ...., are in A.P. such that

a = 20

d = 19 – 20 = –1

Let the numbers of rows be n.

∴ S_{n} = 200

Either

⇒ n – 16 = 0 ⇒ n = 16

or n – 25 = 0 ⇒ n = 25

T_{n} = 0 ⇒ a + (n – 1) d = 0 ⇒ 20 + (n – 1) × (–1) = 0

⇒ n – 1 = 20 ⇒ n = 21

∴ i.e., 21st term becomes 0

∴ n = 25 is not required.

Thus, n = 16

∴ Number of rows = 16

Now, T_{16} = a + (16 – 1) d

= 20 + 15 × (–1)

= 20 – 15 = 5

∴ Number of logs in the 16th (top) row is 15.

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

Sol. Here, number of potatoes = 10

The up–down distance of the bucket:

From the 1st potato = [5 m] × 2 = 10 m

From the 2nd potato = [(5 + 3) m] × 2 = 16 m

From the 3rd potato [(5 + 3 + 3) m] × 2 = 22 m

From the 4th potato = [(5 + 3 + 3 + 3) m] × 2 = 28 m

..................... ......................

∵ 10, 16, 22, 28, .... are in A.P. such that

a = 10 and d = 16 – 10 = 6

Thus, the sum of above distances = 370 m.

The competitor has to run a total distance of 370 m.

Sol. We have the A.P. having a = 121 and d = 117 – 121 = – 4

∴ a_{n} = a + (n – 1) d

= 121 + (n – 1) × (–4)

= 121 – 4n + 4

= 125 – 4n

For the first negative term, we have

a_{n} < 0

⇒ (125 – 4n) < 0

⇒ 125 < 4n

Thus, the first negative term is 32nd term.

Sol. Here, T_{3} + T_{7} = 6 and T_{3} × T_{7} = 8

Let the first term = a and the common difference = d

∴ T_{3} = a + 2d and T_{7} = a + 6d

∵ T_{3} + T_{7} = 6

(a + 2d) + (a + 6d) = 6

⇒ 2a + 8d = 6

⇒ a + 4d = 3 ...(1)

Again T_{3} × T_{7} = 8

⇒ (a + 2d) × (a + 6d) = 8

⇒ (a + 4d – 2d) × (a + 4d + 2d) = 8

⇒ [(a + 4d) – 2d] × [(a + 4d) + 2d] = 8

⇒ [(3) – 2d] × [(3) + 2d] = 8 [From (1)]

⇒ 32 – (2d)2 = 8

⇒ 9 – 4d2 = 8

⇒ –4d2 = 8 – 9 = –1

Sol. Total distance between bottom to top rungs

Distance between two consecutive rungs = 25 cm

Length of the 1st rung (bottom rung) = 45 cm

Length of the 11th rung (top rung) = 25 cm

Let the length of each successive rung decrease by x cm

∴ Total length of the rungs

= 45 cm + (45 – x) cm + (45 – 2x) cm + .... + 25 cm

Here, the numbers 45, (45 – x), (45 – 2x), ...., 25 are in an A.P. such that

First term 'a' = 45

Last term 'l' = 25

Number of terms 'n' = 11

∴ Total length of 11 rungs = 385 cm

i.e., Length of wood required for the rungs is 385 cm.

Sol: We have, the following consecutive numbers on the houses of a row ;

1, 2, 3, 4, 5, ...., 49.

These numbers are in an A.P., such that

a = 1

d = 2 – 1 = 1

n = 49

Let one of the houses be numbered as x

∴ Number of houses preceding it = x – 1

Number of houses following it = 49 – x

Now, the sum of the house–numbers preceding xis given by:

The houses beyond x are numbered as

(x + 1), (x + 2), (x + 3), ...., 49

∴ For these house numbers (which are in an A.P.),

First term (a) = x + 1

Last term (l) = 49

According to the question,

[Sum of house numbers preceding x] = [Sum of house numbers following x]

i.e., Sx – 1 = S_{49} – x

But x cannot be taken as –ve

∴ x = 35

Each step has a rise of m and a tread of m . (see Fig.). Calculate the total volume of concrete required to build the terrace.

∴ Volume of concrete required to build the 1st step

= Volume of the cuboidal step

= Length × Breadth × height

∴ Volume of concrete required to build the 3rd step

Thus, the volumes (in m3) of concrete required to build the various steps are:

obviously, these numbers form an A.P. such that

Here, total number of steps n = 15

Total volume of concrete required to build 15 steps is given by the sum of their individual volumes.

Thus, the required volume of concrete is 750 m3.