Arithmetic Progressions Class 10 Mathematics

20 Questions | 76 Marks

The sum of the first 10 terms of an AP is 210 and the sum of its first 20 terms is 590. Find the first term and the common difference of the AP. 5marks

In an AP, the sum of first n terms is given by

S_{n} = 3 n^{2} + 2 n

. Find the common difference and the nth term of the AP. 3marks

The third term of an AP is 12 and the 10th term is 26. Find the first term and the common difference. Also, find the sum of the first 15 terms. 5marks

A sum of ₹1000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30 years making use of this fact. 3marks

How many terms of the AP: 24, 21, 18, ... must be taken so that their sum is 78? Explain the double answer. 5marks

Find the sum of the first 24 terms of the AP: 3, 8, 13, ... 3marks

Determine the AP whose 5th term is 19 and the 12th term is 40. 3marks

The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP. 5marks

A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes. 5marks

In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed? 3marks

Find the sum of the first 1000 positive integers. 2marks

Which term of the AP: 3, 15, 27, 39, ... will be 132 more than its 54th term? 3marks

How many three-digit numbers are divisible by 7? 3marks

If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero? 3marks

A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find: (i) the production in the 1st year (ii) the production in the 10th year. 5marks

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. 3marks

In an AP, the first term is 32 and the last term is -10. If the common difference is -2, find the number of terms. 2marks

Two APs have the same common difference. The difference between their 100th terms is 100. What is the difference between their 1000th terms? 2marks

Find the sum of the first 15 multiples of 8. 2marks

For what value of n, are the nth terms of two APs: 63, 65, 67, ... and 3, 10, 17, ... equal? 3marks

Answers

Let the first term be 'a' and the common difference be 'd'.

S_{10} = \frac{10}{2} [2 a + (10 - 1) d] = 5 (2 a + 9 d) = 210 ⟹ 2 a + 9 d = 42

(1)

S_{20} = \frac{20}{2} [2 a + (20 - 1) d] = 10 (2 a + 19 d) = 590 ⟹ 2 a + 19 d = 59

(2)Subtracting (1) from (2):

10 d = 17 ⟹ d = 1.7

.Substituting d in (1):

2 a + 9 (1.7) = 42 ⟹ 2 a + 15.3 = 42 ⟹ 2 a = 26.7 ⟹ a = 13.35

.First term = 13.35, Common difference = 1.7.

S_{n} = 3 n^{2} + 2 n

a_{1} = S_{1} = 3 (1)^{2} + 2 (1) = 5

S_{2} = 3 (2)^{2} + 2 (2) = 12 + 4 = 16

a_{2} = S_{2} - S_{1} = 16 - 5 = 11

.Common difference

d = a_{2} - a_{1} = 11 - 5 = 6

.The nth term

a_{n} = S_{n} - S_{n - 1} = (3 n^{2} + 2 n) - [3 (n - 1)^{2} + 2 (n - 1)] = (3 n^{2} + 2 n) - [3 (n^{2} - 2 n + 1) + 2 n - 2]

a_{n} = (3 n^{2} + 2 n) - (3 n^{2} - 6 n + 3 + 2 n - 2) = 3 n^{2} + 2 n - 3 n^{2} + 4 n - 1 = 6 n - 1

.Alternatively,

a_{n} = a_{1} + (n - 1) d = 5 + (n - 1) 6 = 5 + 6 n - 6 = 6 n - 1

.Common difference = 6, nth term =

6 n - 1

Given

a_{3} = 12

and

a_{10} = 26

a_{3} = a + 2 d = 12

(1)

a_{10} = a + 9 d = 26

(2)Subtracting (1) from (2):

7 d = 14 ⟹ d = 2

.Substituting d in (1):

a + 2 (2) = 12 ⟹ a + 4 = 12 ⟹ a = 8

.First term = 8, Common difference = 2.Sum of first 15 terms

S_{15} = \frac{15}{2} [2 a + (15 - 1) d] = \frac{15}{2} [2 (8) + 14 (2)] = \frac{15}{2} [16 + 28] = \frac{15}{2} (44) = 15 \times 22 = 330

The interest at the end of each year forms an AP.Principal (P) = ₹1000, Rate (R) = 8%Interest for 1st year =

\frac{1000 \times 8 \times 1}{100} = ₹ 80

.Interest for 2nd year =

\frac{1000 \times 8 \times 2}{100} = ₹ 160

.Interest for 3rd year =

\frac{1000 \times 8 \times 3}{100} = ₹ 240

.The sequence of interests is 80, 160, 240, ...This is an AP with first term

a = 80

and common difference

d = 160 - 80 = 80

.Interest at the end of 30 years is the 30th term of this AP.

a_{30} = a + (30 - 1) d = 80 + 29 \times 80 = 80 (1 + 29) = 80 \times 30 = 2400

.The interests form an AP. The interest at the end of 30 years is ₹2400.

The AP is 24, 21, 18, ...Here,

a = 24

d = 21 - 24 = - 3

. Let the sum be

S_{n} = 78

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

78 = \frac{n}{2} [2 (24) + (n - 1) (- 3)]

156 = n [48 - 3 n + 3]

156 = n [51 - 3 n]

156 = 51 n - 3 n^{2}

3 n^{2} - 51 n + 156 = 0

Divide by 3: $n^{2} - 17 n + 52 = 0$
Factoring the quadratic equation: $(n - 4) (n - 13) = 0$ .
So, $n = 4$ or $n = 13$ .
Both values of n are valid. This means that the sum of the first 4 terms is 78, and the sum of the first 13 terms is also 78. This happens because the terms from the 5th to the 13th sum up to zero.
The number of terms can be 4 or 13.

The AP is 3, 8, 13, ...Here,

a = 3

d = 8 - 3 = 5

. We need to find the sum of the first 24 terms,

S_{24}

S_{24} = \frac{24}{2} [2 a + (24 - 1) d]

S_{24} = 12 [2 (3) + 23 (5)]

S_{24} = 12 [6 + 115]

$S_{24} = 12 [121] = 1452$ .
The sum of the first 24 terms is 1452.

Let the AP be

a, a + d, a + 2 d, . . .

Given

a_{5} = 19

and

a_{12} = 40

a_{5} = a + 4 d = 19

(1)

a_{12} = a + 11 d = 40

(2)Subtracting (1) from (2):

7 d = 21 ⟹ d = 3

.Substituting d in (1):

a + 4 (3) = 19 ⟹ a + 12 = 19 ⟹ a = 7

.The AP is 7, 10, 13, ...

Let the AP be

a, a + d, a + 2 d, . . .

Given

a_{4} + a_{8} = 24

and

a_{6} + a_{10} = 44

a_{4} = a + 3 d

a_{8} = a + 7 d ⟹ a_{4} + a_{8} = 2 a + 10 d = 24 ⟹ a + 5 d = 12

(1)

a_{6} = a + 5 d

a_{10} = a + 9 d ⟹ a_{6} + a_{10} = 2 a + 14 d = 44 ⟹ a + 7 d = 22

(2)Subtracting (1) from (2):

2 d = 10 ⟹ d = 5

.Substituting d in (1):

a + 5 (5) = 12 ⟹ a + 25 = 12 ⟹ a = - 13

.The first three terms are:

a_{1} = a = - 13

a_{2} = a + d = - 13 + 5 = - 8

$a_{3} = a + 2 d = - 13 + 2 (5) = - 13 + 10 = - 3$ .
The first three terms are -13, -8, -3.

The cash prizes form an AP.Let the prizes be

a_{1}, a_{2}, . . ., a_{7}

a_{1}

is the first prize. The common difference is

d = - 20

.The sum of the seven prizes is

S_{7} = 700

S_{7} = \frac{7}{2} [2 a + (7 - 1) d]

700 = \frac{7}{2} [2 a + 6 (- 20)]

700 = \frac{7}{2} [2 a - 120]

200 = 2 a - 120

$320 = 2 a ⟹ a = 160$ .
The first prize is ₹160.
The prizes are:

a_{1} = 160

a_{2} = 160 - 20 = 140

a_{3} = 140 - 20 = 120

a_{4} = 120 - 20 = 100

a_{5} = 100 - 20 = 80

a_{6} = 80 - 20 = 60

$a_{7} = 60 - 20 = 40$ .
The values of the prizes are ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40.

The number of rose plants in the rows form an AP: 23, 21, 19, ..., 5.Here,

a = 23

d = 21 - 23 = - 2

, and the last term

a_{n} = 5

.We need to find the number of terms (rows), n.

a_{n} = a + (n - 1) d

5 = 23 + (n - 1) (- 2)

5 - 23 = - 2 (n - 1)

- 18 = - 2 (n - 1)

$9 = n - 1 ⟹ n = 10$ .
There are 10 rows in the flower bed.

The first 1000 positive integers are 1, 2, 3, ..., 1000.This is an AP with

a = 1

d = 1

, and

n = 1000

.The sum

S_{n} = \frac{n}{2} (a + l)

, where

l

is the last term.

S_{1000} = \frac{1000}{2} (1 + 1000) = 500 \times 1001 = 500500

.The sum of the first 1000 positive integers is 500500.

The AP is 3, 15, 27, 39, ...Here,

a = 3

and

d = 15 - 3 = 12

.The 54th term is

a_{54} = a + (54 - 1) d = 3 + 53 \times 12 = 3 + 636 = 639

.We need to find the term which is 132 more than

a_{54}

.Let this term be

a_{n}

. So,

a_{n} = a_{54} + 132 = 639 + 132 = 771

.Now we find which term this is:

a_{n} = a + (n - 1) d

771 = 3 + (n - 1) 12

768 = (n - 1) 12

n - 1 = \frac{768}{12} = 64

$n = 65$ .
The 65th term is 132 more than the 54th term.

Three-digit numbers range from 100 to 999.The first three-digit number divisible by 7 is 105 (

7 \times 15 = 105

).The last three-digit number divisible by 7 is 994 (

7 \times 142 = 994

).The sequence of three-digit numbers divisible by 7 is 105, 112, 119, ..., 994.This is an AP with

a = 105

d = 7

, and

a_{n} = 994

.We need to find the number of terms, n.

a_{n} = a + (n - 1) d

994 = 105 + (n - 1) 7

994 - 105 = 7 (n - 1)

889 = 7 (n - 1)

n - 1 = \frac{889}{7} = 127

$n = 128$ .
There are 128 three-digit numbers divisible by 7.

The AP is given, with

a_{3} = 4

and

a_{9} = - 8

a_{3} = a + 2 d = 4

(1)

a_{9} = a + 8 d = - 8

(2)Subtracting (1) from (2):

6 d = - 12 ⟹ d = - 2

.Substituting d in (1):

a + 2 (- 2) = 4 ⟹ a - 4 = 4 ⟹ a = 8

.We need to find which term is zero. Let the nth term be 0.

a_{n} = a + (n - 1) d

0 = 8 + (n - 1) (- 2)

0 = 8 - 2 n + 2

0 = 10 - 2 n

$2 n = 10 ⟹ n = 5$ .
The 5th term of the AP is zero.

Let the production in the nth year be

a_{n}

. This forms an AP.Given

a_{3} = 600

and

a_{7} = 700

a_{3} = a + 2 d = 600

(1)

a_{7} = a + 6 d = 700

(2)Subtracting (1) from (2):

4 d = 100 ⟹ d = 25

.Substituting d in (1):

a + 2 (25) = 600 ⟹ a + 50 = 600 ⟹ a = 550

.(i) The production in the 1st year is

a = ₹ 550

.(ii) The production in the 10th year is

a_{10} = a + (10 - 1) d = 550 + 9 (25) = 550 + 225 = 775

.The production in the 10th year is 775 sets.

The AP has first term

a = 5

, last term

l = 45

, and sum

S_{n} = 400

.We use the formula

S_{n} = \frac{n}{2} (a + l)

400 = \frac{n}{2} (5 + 45)

400 = \frac{n}{2} (50)

400 = 25 n

$n = \frac{400}{25} = 16$ .
The number of terms is 16.
Now we find the common difference using $l = a + (n - 1) d$ .

45 = 5 + (16 - 1) d

40 = 15 d

$d = \frac{40}{15} = \frac{8}{3}$ .
The number of terms is 16 and the common difference is $\frac{8}{3}$ .

In an AP,

a = 32

l = - 10

d = - 2

. We need to find the number of terms, n.Using the formula

l = a + (n - 1) d

- 10 = 32 + (n - 1) (- 2)

- 10 - 32 = - 2 (n - 1)

- 42 = - 2 (n - 1)

21 = n - 1

$n = 22$ .
There are 22 terms.

Let the two APs be AP1 and AP2. Let their first terms be

a_{1}, a_{2}

and their common difference be

d

.For AP1, the 100th term is

a_{1} + 99 d

.For AP2, the 100th term is

a_{2} + 99 d

.The difference between their 100th terms is

(a_{1} + 99 d) - (a_{2} + 99 d) = a_{1} - a_{2} = 100

.Now consider their 1000th terms.For AP1, the 1000th term is

a_{1} + 999 d

.For AP2, the 1000th term is

a_{2} + 999 d

.The difference between their 1000th terms is

(a_{1} + 999 d) - (a_{2} + 999 d) = a_{1} - a_{2}

.Since

a_{1} - a_{2} = 100

, the difference between their 1000th terms is also 100.

We need to find the sum of the first 15 multiples of 8.The multiples of 8 are 8, 16, 24, ...This is an AP with first term

a = 8

and common difference

d = 8

.We need to find the sum of the first 15 terms,

S_{15}

S_{15} = \frac{15}{2} [2 a + (15 - 1) d]

S_{15} = \frac{15}{2} [2 (8) + 14 (8)]

S_{15} = \frac{15}{2} [16 + 112]

S_{15} = \frac{15}{2} [128]

$S_{15} = 15 \times 64 = 960$ .
The sum of the first 15 multiples of 8 is 960.

Let the nth term of the first AP (63, 65, 67, ...) be

a_{n}

a = 63

d = 65 - 63 = 2

a_{n} = 63 + (n - 1) 2 = 63 + 2 n - 2 = 2 n + 61

.Let the nth term of the second AP (3, 10, 17, ...) be

b_{n}

a = 3

d = 10 - 3 = 7

b_{n} = 3 + (n - 1) 7 = 3 + 7 n - 7 = 7 n - 4

.We need to find the value of n for which

a_{n} = b_{n}

2 n + 61 = 7 n - 4

61 + 4 = 7 n - 2 n

65 = 5 n

$n = \frac{65}{5} = 13$ .
The 13th terms of both APs are equal.

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