CBSE Class X

Question 19:

200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

Answer:

It can be observed that the numbers of logs in rows are in an A.P. 20, 19, 18…

For this A.P.,

a = 20

d = a₂ − a₁ = 19 − 20 = −1

Let a total of 200 logs be placed in n rows.

S_n = 200

400 = n (40 − n + 1)

400 = n (41 − n)

400 = 41n − n²

n² − 41n + 400 = 0

n² − 16n − 25n + 400 = 0

n (n − 16) −25 (n − 16) = 0

(n − 16) (n − 25) = 0

Either (n − 16) = 0 or n − 25 = 0

n = 16 or n = 25

a_n = a + (n − 1)d

a₁₆ = 20 + (16 − 1) (−1)

a₁₆ = 20 − 15

a₁₆ = 5

Similarly,

a₂₅ = 20 + (25 − 1) (−1)

a₂₅ = 20 − 24

= −4

Clearly, the number of logs in 16^th row is 5. However, the number of logs in 25^th row is negative, which is not possible.

Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16^th row is 5.

Question 18:

A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, ……… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7)

[Hint : Length of successive semicircles is l₁, l₂, l₃, l₄, . . . with centres at A, B, A, B, . . ., respectively.]

Answer:

Semi-perimeter of circle = πr

I₁ = π(0.5) = π/2 cm

I₂ = π(1) = π cm

I₃ = π(1.5) = 3π/2 cm

Therefore, I₁, I₂, I₃ ,i.e. the lengths of the semi-circles are in an A.P., π/2, π, 3π/2, 2π, ………..

a = π/2

d = π − (π/2) = π/2

S₁₃ =?

We know that the sum of n terms of an a A.P. is given by

Therefore, the length of such spiral of thirteen consecutive semi-circleswill be 143 cm.

Question 17:

In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?

Answer:

It can be observed that the number of trees planted by the students is in an AP.

1, 2, 3, 4, 5………………..12

First term, a = 1

Common difference, d = 2 − 1 = 1

Therefore, number of trees planted by 1 section of the classes = 78

Number of trees planted by 3 sections of the classes = 3 × 78 = 234
Therefore, 234 trees will be planted by the students.

Question 16:

A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20
less than its preceding prize, find the value of each of the prizes.

Answer:

Let the cost of 1^st prize be P.

Cost of 2^nd prize = P − 20

And cost of 3^rd prize = P − 40

It can be observed that the cost of these prizes are in an A.P. having common difference as −20 and first term as P.

a = P

d = −20

Given that, S₇ = 700

a + 3(−20) = 100

a − 60 = 100

a = 160

Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.

Question 15:

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days.

Answer:

It can be observed that these penalties are in an A.P. having first term as 200 and common difference as 50.

a = 200

d = 50

Penalty that has to be paid if he has delayed the work by 30 days = S₃₀

= 30/2 [2(200) + (30 − 1) 50]

= 15 [400 + 1450]

= 15 (1850)

= 27750

Therefore, the contractor has to pay Rs 27750 as penalty.

Question 14:

Find the sum of the odd numbers between 0 and 50.

Answer:

The odd numbers between 0 and 50 are 1, 3, 5, 7, 9 … 49

Therefore, it can be observed that these odd numbers are in an A.P.

a = 1

d = 2

l = 49

l = a + (n − 1) d

49 = 1 + (n − 1)2

48 = 2(n − 1)

n − 1 = 24

n = 25

Question 13:

Find the sum of first 15 multiples of 8.

Answer:

The multiples of 8 are 8, 16, 24, 32…

These are in an A.P., having first term as 8 and common difference as 8.

Therefore, a = 8

d = 8

S₁₅ =?

Question 12:

Find the sum of first 40 positive integers divisible by 6.

Answer:

The positive integers that are divisible by 6 are 6, 12, 18, 24 …

It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.

a = 6

d = 6

S₄₀ =?

Question 11:

If the sum of the first n terms of an AP is 4n − n², what is the first term (that is S₁)? What is the sum of first two terms? What is the second term? Similarly find the 3^rd, the10^th and the n^th terms.

Answer:

Given that,

S_n = 4n − n²

First term, a = S₁ = 4(1) − (1)² = 4 − 1 = 3

Sum of first two terms = S2

= 4(2) − (2)² = 8 − 4 = 4

Second term, a₂ = S₂ − S₁ = 4 − 3 = 1

d = a₂ − a = 1 − 3 = −2

a_n = a + (n − 1)d

= 3 + (n − 1) (−2)

= 3 − 2n + 2

= 5 − 2n

Therefore, a₃ = 5 − 2(3) = 5 − 6 = −1

a₁₀ = 5 − 2(10) = 5 − 20 = −15

Hence, the sum of first two terms is 4. The second term is 1. 3^rd, 10^th, and n^th terms are −1, −15, and 5 − 2n respectively.

Question 10:

Show that a₁, a₂ … , a_n , … form an AP where an is defined as below

(i) a_n = 3 + 4n             (ii) a_n = 9 − 5n

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

Answer:

(i) a_n = 3 + 4n

a₁ = 3 + 4(1) = 7

a₂ = 3 + 4(2) = 3 + 8 = 11

a₃ = 3 + 4(3) = 3 + 12 = 15

a₄ = 3 + 4(4) = 3 + 16 = 19

It can be observed that

a₂ − a₁ = 11 − 7 = 4

a₃ − a₂ = 15 − 11 = 4

a₄ − a₃ = 19 − 15 = 4

i.e., a_{k + 1} − a_k is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.

(ii) a_n = 9 − 5n

a₁ = 9 − 5 × 1 = 9 − 5 = 4

a₂ = 9 − 5 × 2 = 9 − 10 = −1

a₃ = 9 − 5 × 3 = 9 − 15 = −6

a₄ = 9 − 5 × 4 = 9 − 20 = −11

It can be observed that

a₂ − a₁ = − 1 − 4 = −5

a₃ − a₂ = − 6 − (−1) = −5

a₄ − a₃ = − 11 − (−6) = −5

i.e., a_{k + 1} − a_k is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.

Question 9:

If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Answer:

Given that,

S₇ = 49

S₁₇ = 289

17 = (a + 8d)

a + 8d = 17 (ii)

Subtracting equation (i) from equation (ii),

5d = 10

d = 2

From equation (i),

a + 3(2) = 7

a + 6 = 7

a = 1

Question 8:

Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Answer:

Given that,

a₂ = 14

a₃ = 18

d = a₃ − a₂ = 18 − 14 = 4

a₂ = a + d

14 = a + 4

a = 10

Question 6:

The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Answer:

Given that,

a = 17

l = 350

d = 9

Let there be n terms in the A.P.

l = a + (n − 1) d

350 = 17 + (n − 1)9

333 = (n − 1)9

(n − 1) = 37

n = 38

Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.

Question 7:

Find the sum of first 22 terms of an AP in which d = 7 and 22^nd term is 149.

Answer:

d = 7

a₂₂ = 149

S₂₂ = ?

a_n = a + (n − 1)d

a₂₂ = a + (22 − 1)d

149 = a + 21 × 7

149 = a + 147

a = 2

Question 5:

The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Answer:

Given that,

a = 5

l = 45

S_n = 400

n = 16

l = a + (n − 1) d

45 = 5 + (16 − 1) d

40 = 15d

d = 40/15 = 8/3

Question 4:

How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?

Answer:

Let there be n terms of this A.P.

For this A.P., a = 9

d = a₂ − a₁ = 17 − 9 = 8

636 = n [9 + 4n − 4]

636 = n (4n + 5)

4n² + 5n − 636 = 0

4n² + 53n − 48n − 636 = 0

n (4n + 53) − 12 (4n + 53) = 0

(4n + 53) (n − 12) = 0

Either 4n + 53 = 0 or n − 12 = 0

n = −53/4 or n = 12

n cannot be −53/4. As the number of terms can neither be negative nor fractional, therefore, n = 12 only.

Question 3:

In an AP

(i) Given a = 5, d = 3, a_n = 50, find n and S_n.

(ii) Given a = 7, a₁₃ = 35, find d and S₁₃.

(iii) Given a₁₂ = 37, d = 3, find a and S₁₂.

(iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀.

(v) Given d = 5, S₉ = 75, find a and a₉.

(vi) Given a = 2, d = 8, S_n = 90, find n and a_n.

(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.

Answer:

(i) Given that, a = 5, d = 3, a_n = 50

As a_n = a + (n − 1)d,

∴ 50 = 5 + (n − 1)3

45 = (n − 1)3

15 = n − 1

n = 16

(ii) Given that, a = 7, a₁₃ = 35

As a_n = a + (n − 1) d,

∴ a₁₃ = a + (13 − 1) d

35 = 7 + 12 d

35 − 7 = 12d

28 = 12d

(iii)Given that, a₁₂ = 37, d = 3

As a_n = a + (n − 1)d,

a₁₂ = a + (12 − 1)3

37 = a + 33

a = 4

(iv) Given that, a₃ = 15, S₁₀ = 125

As a_n = a + (n − 1)d,

a₃ = a + (3 − 1)d

15 = a + 2d              (i)

On multiplying equation (1) by 2, we obtain

30 = 2a + 4d (iii)

On subtracting equation (iii) from (ii), we obtain

−5 = 5d

d = −1

From equation (i),

15 = a + 2(−1)

15 = a − 2

a = 17

a₁₀ = a + (10 − 1)d

a₁₀ = 17 + (9) (−1)

a₁₀ = 17 − 9 = 8

(v)Given that, d = 5, S₉ = 75

(vi) Given that, a = 2, d = 8, S_n = 90

90 = n [2 + (n − 1)4]

90 = n [2 + 4n − 4]

90 = n (4n − 2) = 4n² − 2n

4n² − 2n − 90 = 0

4n² − 20n + 18n − 90 = 0

4n (n − 5) + 18 (n − 5) = 0

(n − 5) (4n + 18) = 0

Either n − 5 = 0 or 4n + 18 = 0

n = 5 or n =−(18/4) = −9/2

However, n can neither be negative nor fractional.

Therefore, n = 5

a_n = a + (n − 1)d

a₅ = 2 + (5 − 1)8

= 2 + (4) (8)

= 2 + 32 = 34

(vii) Given that, a = 8, a_n = 62, S_n = 210

n = 6

a_n = a + (n − 1)d

62 = 8 + (6 − 1)d

62 − 8 = 5d

54 = 5d

d = 54/5

(viii) Given that, a_n = 4, d = 2, S_n = −14

a_n = a + (n − 1)d

4 = a + (n − 1)2

4 = a + 2n − 2

a + 2n = 6

a = 6 − 2n (i)

−28 = n (a + 4)

−28 = n (6 − 2n + 4) {From equation (i)}

−28 = n (− 2n + 10)

−28 = − 2n² + 10n

2n² − 10n − 28 = 0

n² − 5n −14 = 0

n² − 7n + 2n − 14 = 0

n (n − 7) + 2(n − 7) = 0

(n − 7) (n + 2) = 0

Either n − 7 = 0 or n + 2 = 0

n = 7 or n = −2

However, n can neither be negative nor fractional.

Therefore, n = 7

From equation (i), we obtain

a = 6 − 2n

a = 6 − 2(7)

= 6 − 14

= −8

(ix)Given that, a = 3, n = 8, S = 192

192 = 4 [6 + 7d]

48 = 6 + 7d

42 = 7d

d = 6

(x)Given that, l = 28, S = 144 and there are total of 9 terms.

Question 2:

Find the sums given below :

(ii) 34 + 32 + 30 + ……….. + 10

(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)

Answer:

For this A.P.,

a = 7

l = 84

(ii)34 + 32 + 30 + ……….. + 10

For this A.P.,

a = 34

d = a₂ − a₁ = 32 − 34 = −2

l = 10

Let 10 be the n^th term of this A.P.

l = a + (n − 1) d

10 = 34 + (n − 1) (−2)

−24 = (n − 1) (−2)

12 = n − 1

n = 13

(iii)(−5) + (−8) + (−11) + ………… + (−230)

For this A.P.,

a = −5

l = −230

d = a₂ − a₁ = (−8) − (−5)

= − 8 + 5 = −3

Let −230 be the n^th term of this A.P.

l = a + (n − 1)d

−230 = − 5 + (n − 1) (−3)

−225 = (n − 1) (−3)

(n − 1) = 75

n = 76

Question 1:

Find the sum of the following APs.

(i) 2, 7, 12 ,…., to 10 terms.                                  (ii) − 37, − 33, − 29 ,…, to 12 terms

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

Answer:

(i)2, 7, 12 ,…, to 10 terms

For this A.P.,

a = 2

d = a₂ − a₁ = 7 − 2 = 5

n = 10

We know that,

(ii)−37, −33, −29 ,…, to 12 terms

For this A.P.,

a = −37

d = a₂ − a₁ = (−33) − (−37)

= − 33 + 37 = 4

n = 12

We know that,

(iii) 0.6, 1.7, 2.8 ,…, to 100 terms

For this A.P.,

a = 0.6

d = a₂ − a₁ = 1.7 − 0.6 = 1.1

n = 100

We know that,

For this A.P.,

Question 20:

Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the n^th week, her week, her weekly savings become Rs 20.75, find n.

Answer:

Given that,

a = 5

d = 1.75

a_n = 20.75

n = ?

a_n = a + (n − 1) d

n − 1 = 9

n = 10

Hence, n is 10.