# R S AGGARWAL AND V AGGARWAL Solutions Mathematics Class 10 Chapter 11 Arithmetic Progressions

Chapter 11 – Arithmetic Progressions Exercise Ex. 11A

Question 1

Show that each of the progressions given
below is an AP. Find the first term, common difference and next term of each.

9, 15, 21, 27, …..

Solution 1

Question 2

Show that each of the progressions given
below is an AP. Find the first term, common difference and next term of each.

11, 6, 1, –4, ….

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Find

The 20th term of the AP 9, 13,
17, 21, ….

Solution 6

Question 7

Find

The 35th term of the AP 20, 17,
14, 11, ….

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Find

The 15th term of the AP -40,
-15, 10, 35, ….

Solution 10

Question 11

Find the 37th term of the AP

Solution 11

The given AP is

First term = 6, common difference =

a = 6, d =

The nth term is given by

Hence, 37th term is 69

Question 12

Find the 25th term of the AP

Solution 12

The given AP is

The first term = 5,

common difference =

a = 5,

The nth term is given by

Hence the 25th term is – 7

Question 13

Find the nth term of each of the following
APs :

5, 11, 17, 23, …..

Solution 13

Question 14

Find the nth term of each of the following
APs :

16, 9, 2, –5, …..

Solution 14

Question 15

If the nth term of a progression is (4n – 10), show that it is an A.P. Find its (i) first term, (ii) common difference, and (iii) 16th term.

Solution 15

(i)First term = -6

(ii)Common difference

(iii)16th term = where a = -6 and d = 4

= (-6 + 15 4) = 54

Question 16

How many terms are there in the AP 6, 10, 14, 18, …, 174?

Solution 16

In the given AP, we have a = 6 and d = (10 – 6) = 4

Suppose there are n terms in the given AP, then

Hence there are 43 terms in the given AP

Question 17

How many terms are there in the AP 41, 38, 35, …8?

Solution 17

In the given AP we have a = 41 and d = 38 – 41 = – 3

Suppose there are n terms in AP, then

Hence there are 12 terms in the given AP

Question 18

Solution 18

Question 19

Which term of the AP 3, 8, 13, 18, … is 88?

Solution 19

In the given AP, we have a = 3 and d = 8 – 3 = 5

Suppose there are n terms in given AP, then

Hence, the 18th term of given AP is 88

Question 20

Which term of the AP 72, 68, 64, 60, … is 0?

Solution 20

In the given AP, we have a = 72 and d = 68 – 72 = – 4

Suppose there are n terms in given AP, we have

Hence, the 19th term in the given AP is 0

Question 21

Which term of the AP is 3?

Solution 21

In the given AP, we have

Suppose there are n terms in given AP, we have

Then,

Thus, 14th term in the given AP is 3

Question 22

Which term of the AP 21, 18, 15, … is -81?

Solution 22

Question 23

Which term of the AP 3, 8, 13, 18, … will be 55 more than its 20th term?

Solution 23

The given AP is 3, 8, 13, 18…..

First term a = 3, common difference a = 8 – 3 = 5

Let nth  term is 55 more than the 20th term

(5n – 2) – 98 = 55

Or 5n = 100 + 55 = 155

31st term is 55 more than the 20th term of given AP

Question 24

Which term of the AP 5, 15, 25, … will be 130 more than its 31st term?

Solution 24

The given AP is 5, 15, 25….

a = 5, d = 15 – 5 = 10

Thus, the required term is 44th

Question 25

If the 10th term of an AP is 52 and the 17th term is 20 more than the 13th term, find the AP.

Solution 25

In the given AP let the first term = a,

And common difference = d

So the required AP is 7, 12, 17, 22….

Question 26

Find the middle term of the AP 6, 13, 20, …, 216.

Solution 26

Question 27

Find the middle term of the AP 10, 7, 4, …, (-62).

Solution 27

Question 28

Solution 28

Question 29

Find the 8th term from the end of the AP 7, 10, 13, …, 184.

Solution 29

Here a = 7, d = (10 – 7) = 3, l = 184

And n = 8

Hence, the 8th term from the end is 163

Question 30

Find the 6th term from the end of the AP 17, 14, 11, …(-40).

Solution 30

Here a = 17, d = (14 – 17) = -3, l = -40

And n = 6

Now, nth term from the end = [l – (n – 1)d]

Hence, the 6th term from the end is – 25

Question 31

Is 184 a term of the AP 3, 7, 11, 15, …?

Solution 31

Question 32

Is-150 a term of the AP 11, 8, 5, 2, …?

Solution 32

Question 33

Which term of the AP 121, 117, 113, … is its first negative term?

Solution 33

Question 34

Solution 34

Question 35

The 7th term of an AP is -4 and its 13th term is -16. Find the AP.

Solution 35

In the given AP, let the first term = a common difference = d

So the required AP is 8, 6, 4, 2, 0……

Question 36

The 4th term of an AP is zero. Prove that its 25th term is triple its 11th term.

Solution 36

Let the first term of given AP = a and common difference = d

Hence 25th term is triple its 11th term

Question 37

The 8th term of an AP is zero.
Prove that its 38th term is triple its 18th term.

Solution 37

Question 38

The 4th term of an AP is 11. The
sum of the 5th and 7th terms of this AP is 34. Find its
common difference.

Solution 38

Question 39

The 9th term of an AP is -32 and
the sum of its 11th and 13th terms is -94. Find the
common difference of the AP.

Solution 39

Question 40

Determine the nth term of the AP whose 7th
term is -1 and 16th term is 17.

Solution 40

Question 41

If 4 times the 4th term of an AP is equal
to 18 times its 18th term then find its 22nd term.

Solution 41

Question 42

If 10 times the 10th term of an AP is equal to 15 times the 15th term, show that its 25th term is zero.

Solution 42

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

Question 43

Find the common difference of an AP whose
first term is 5 and the sum of its first four terms is half the sum of the
next four terms.

Solution 43

Question 44

The sum of the 2nd and the 7th terms of an
AP is 30. If its 15th term is 1 less than twice its
8th term, find the AP.

Solution 44

Question 45

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

Solution 45

First AP is 63, 65, 67….

First term = 63, common difference = 65 – 63 = 2

nthterm = 63 + (n – 1) 2 = 63 + 2n – 2 = 2n + 61

Second AP is 3, 10, 17 ….

First term = 3, common difference = 10 – 3 = 7

nth term = 3 + (n – 1) 7 = 3 + 7n – 7 = 7n – 4

The two nth terms are equal

2n + 61 = 7n – 4 or 5n = 61 + 4 = 65

Question 46

The 17th term of AP is 5 more than twice
its 8th term. If the 11th term of the AP is 43, find its nth term.

Solution 46

Question 47

The 24th term of an AP is twice its 10th
term. Show that its 72nd term is 4 times its 15th term.

Solution 47

Question 48

The 19th term of an AP is equal to 3 times
its 6th term. If its 9th term is 19, find the AP.

Solution 48

Question 49

If the pth term of an AP is q and its qth term is p, then show that its (p + q)th term is zero.

Solution 49

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

pth term = a +(p – 1)d = q(given)—–(1)

qth term = a +(q – 1) d = p(given)—–(2)

subtracting (2) from (1)

(p – q)d = q – p

(p – q)d = -(p – q)

d = -1

Putting d = -1 in (1)

Question 50

The first and last terms of an AP are a and l respectively. Show that the sum of the nth term from the beginning and the nth term from the end is (a + l).

Solution 50

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

nth term from the beginning = a + (n – 1)d—–(1)

nth term from end= l – (n – 1)d —-(2)

sum of the nth term from the beginning and nth term from the end = [a + (n – 1)d] + [l – (n – 1)d] = a + l

Question 51

How many two-digit numbers are divisible by
6?

Solution 51

Question 52

How many two-digit numbers are divisible by
3?

Solution 52

Question 53

How many three-digit numbers are divisible
by 9?

Solution 53

Question 54

How many numbers are there between 101 and
999, which are divisible by both 2 and 5?

Solution 54

Question 55

In a flower bed, there are 43 rose plants in the first row, 41 in the second, 39 in the third, and so on. There are 11 rose plants in the last row. How many rows are there in the flower bed?

Solution 55

Number of rose plants in first, second, third rows…. are 43, 41, 39 respectively.

There are 11 rose plants in the last row

So, it is an AP . viz. 43, 41, 39 …. 11

a = 43, d = 41 – 43 = -2, l = 11

Let nth term be the last term

Hence, there are 17 rows in the flower bed.

Question 56

A sum of Rs. 2800 is to be used to award
four prizes. If each prize after the first is
Rs. 200 less than the preceding prize, find
the value of each of the prizes.

Solution 56

## Chapter 11 – Arithmetic Progressions Exercise Ex. 11B

Question 1

Determine
k so that (3k – 2), (4k – 6) and (k + 2) are three consecutive terms of an
AP.

Solution 1

Question 2

Find
the value of x for which the numbers (5x + 2),(4x – 1) and (x + 2) are in AP.

Solution 2

Question 3

If
(3y – 1),(3y + 5) and (5y + 1) are three consecutive terms of an AP then find
the value of y.

Solution 3

Question 4

Find the value of x for which (x + 2), 2x, (2x + 3) are three consecutive terms of an AP.

Solution 4

If are consecutive terms of an AP, then

Question 5

Solution 5

Question 6

Find three numbers in AP whose sum is 15 and product is 80.

Solution 6

Let the required numbers be (a – d), a and (a + d)

Sum of these numbers = (a – d) + a + (a + d) = 3a

Product of these numbers = (a – d) × a × (a + d) =

But sum = 15 and product = 80

Hence, the required numbers are (2, 5, 8)

Question 7

The sum of three numbers in AP is 3 and their product is -35. Find the numbers.

Solution 7

Let the required numbers be (a – d), a, (a + d)

Sum of these number = (a – d) + a + (a + d) = 3a

Product of these numbers = (a – d) a (a + d)

But,sum = 3 and product = – 35

Thus, a = 1 and d = 6

Hence, the required numbers are (-5, 1, 7)

Question 8

Divide 24 into three parts such that they are in AP and their product is 440.

Solution 8

Let the required number be (a – d), a and (a + d)

Sum of these numbers = (a – d) + a + (a + d) = 3a

Product of these numbers = (a – d) x a x (a + d)

But sum = 24 and product = 440

Thus, a = 8 and d = 3

Hence the required numbers are (5, 8, 11)

Question 9

The sum of three consecutive terms of an AP is 21 and the sum of the squares of these terms is 165. Find these terms.

Solution 9

Let the required numbers be (a – d), a, (a + d)

Sum of these numbers = (a – d) + a + (a + d) = 3a

sum of these squares =

Sum of three numbers = 21, sum of squares of these numbers = 165

3a = 21

a = 7

Thus, a = 7 and d =

Hence, the required numbers are (4, 7, 10) or (10, 7, 4)

Question 10

The angles of a quadrilateral are in AP whose common difference is 10°. Find the angles.

Solution 10

Let the required angles be (a – 3d)°, (a – d) °, (a + d) ° and (a + 3d) °

Common difference = (a – d) – (a- 3d) = a – d – a + 3d = 2d

Common difference = 10°

2d = 10° = d = 5°

Sum of four angles of quadrilateral = 360°

First angle = (a – 3d)° = (90 – 3 × 5) ° = 75°

Second angle = (a – d)° = (90 – 5) ° = 85°

Third angle = (a + d)° = (90 + 5°) = 95°

Fourth angle = (a + 3d)° = (90 + 3 × 5)° = 105°

Question 11

Find four numbers in AP whose sum is 28 and the sum of whose squares is 216.

Solution 11

Let the required number be (a – 3d), (a – d), (a + d) and (a + 3d)

Sum of these numbers = (a – 3d) + (a – d)+ (a + d) + (a + 3d)

4a = 28 a = 7

Sum of the squares of these numbers

Hence, the required numbers (4, 6, 8, 10)

Question 12

Divide
32 into four parts which are the four terms of an AP such that the product of
the first and the fourth terms is to the product of the second and the third
terms as 7 : 15.

Solution 12

Question 13

The
sum of first three terms of an AP is 48. If the product of first and second
terms exceeds 4 times the third term by 12. Find the AP.

Solution 13

## Chapter 11 – Arithmetic Progressions Exercise Ex. 11C

Question 1

The first three terms of an AP are
respectively (3y – 1), (3y + 5) and (5y + 1), find the value of y.

Solution 1

Question 2

If k, (2k – 1) and (2k + 1) are the three
successive terms of an AP, find the value of k.

Solution 2

Question 3

If 18, a, (b – 3) are in AP, then find the
value of (2a – b).

Solution 3

Question 4

If the numbers a, 9, b, 25 form an AP, find
a and b.

Solution 4

Question 5

If the numbers (2n – 1), (3n + 2) and (6n –
1) are in AP, find the value of n and the numbers.

Solution 5

Question 6

How many three-digit natural numbers are
divisible by 7?

Solution 6

Question 7

How many three-digit natural numbers are
divisible by 9?

Solution 7

Question 8

If the sum of first m terms of an AP is (2m2
+ 3m) then what is its second term?

Solution 8

Question 9

What is the sum of first n terms of the AP
a, 3a, 5a, ….

Solution 9

Question 10

What is the 5th term from the end of the AP
2, 7, 12, …, 47?

Solution 10

Question 11

If an denotes the nth term of
the AP 2, 7, 12, 17, …, find the value of (a30
– a20).

Solution 11

Question 12

Find the sum of all
three-digit natural numbers which are divisible by 13.

Solution 12

All three digit
natural numbers divisible by 13 are 104, 117, 130, 143,…, 988

This is an AP in
which a = 104, d = (117 – 104) = 13, l = 988

Question 13

Find the sum of first 15 multiples of 8.

Solution 13

First 15 multiples of 8 are 8, 16, 24, … to 15th term

Question 14

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

Solution 14

Odd natural numbers between 0 and 50 are 1, 3, 5, … 49

a = 1, d = 3 – 1= 2, l = 49

Let the number of terms be n

Question 15

Find the sum of first hundred even natural numbers which are divisible by 5.

Solution 15

First 100 even natural numbers divisible by 5 are

10, 20, 30, … to 100 term

First term of AP = 10

Common difference d = 20 – 10 = 10

Number of terms = n = 100

Question 16

Which term of the AP 21, 18, 15, … is zero?

Solution 16

The given AP is 21, 18, 15, ….

First term = 21, common difference = 18 – 21= – 3

Let nth term be zero

a + (n – 1)d = 0or 21 + (n – 1)(-3) = 0

21 – 3n + 3 = 0

3n = 24

or

Hence, 8th term of given series is 0

Question 17

Find the sum of first n natural numbers.

Solution 17

Sum of n natural numbers = 1 + 2 + 3 + … + n

Here a = 1, d = 2 – 1 = 1

Question 18

Find the sum of first n even natural numbers.

Solution 18

Sum of even natural numbers = 2 + 4 + 6 + … to n terms

a = 2, d = 4 – 2 = 2

Question 19

The first term of an AP is p and its common difference is q. Find its 10th term.

Solution 19

First term of AP = a = p

Common difference = d = q

nth term = a + (n 1)d

10th term = p + (10 1)q

= p + 9q

Question 20

If are three consecutive terms of an AP, find the value of a.

Solution 20

Question 21

If (2p + 1), 13, (5p – 3) are in AP, find
the value of p.

Solution 21

Question 22

If (2p – 1), 7, 3p are in AP, find the
value of p.

Solution 22

Question 23

If the sum of first p terms of an AP is (ap2
+ bp), find its common difference.

Solution 23

Question 24

If the sum of first n terms is (3n2
+ 5n), find its common difference.

Solution 24

Question 25

Find an AP whose 4th term is 9 and the sum
of its 6th and 13th terms is 40.

Solution 25

## Chapter 11 – Arithmetic Progressions Exercise Ex. 11D

Question 1

Find
the sum of each of the following APs:

2,
7, 12, 17, ….. to 19 terms.

Solution 1

Question 2

Find
the sum of each of the following APs:

9,
7, 5, 3, ….. to 14 terms.

Solution 2

Question 3

Find
the sum of each of the following APs:

-37,
-33, –29, …. to 12 terms.

Solution 3

Question 4

Find
the sum of each of the following APs:

Solution 4

Question 5

Find
the sum of each of the following APs:

0.6,
1.7, 2.8, ….. to 100 terms.

Solution 5

Question 6

Find
the sum of each of the following arithmetic series:

Solution 6

Question 7

Find
the sum of each of the following arithmetic series:

Solution 7

Question 8

Find
the sum of each of the following arithmetic series:

Solution 8

Question 9

Find
the sum of first n terms of an AP whose nth term is (5 – 6n). Hence,
find the sum of its first 20 terms.

Solution 9

Question 10

Solution 10

Question 11

The sum of n terms of an AP is . Find its 20th term.

Solution 11

It is given that —–(1)

Now, 20th term

=(sum of first 20 term) – (sum of first 19 terms)

Putting = 20 in (1) we get

Putting n= 19 in (1), we get

Hence, the 20thterm is 99

Question 12

If the sum of the first n terms of an AP is given by Sn = (3n2 – n), find its (i) nth term, (ii) first term and (iii) common difference.

Solution 12

(i)The nth term is given by

(ii)Putting n = 1 in (1) , we get

(iii)Putting n = 2 in (1), we get = 8

Question 13

Solution 13

Question 14

How many terms of the AP 21, 18, 15, … must be added to get the sum 0?

Solution 14

Here a = 21, d = (18 – 21) = -3

Let the required number of terms be n, then

sum of first 15 terms = 0

Question 15

How
many terms of the AP 9, 17, 25, …. must be taken so that their sum is 636?

Solution 15

Question 16

How
many terms of the AP 63, 60, 57, 54, … must be taken
so that their sum is 693? Explain the double answer.

Solution 16

Question 17

Solution 17

Question 18

Write the next term of the AP

Solution 18

The term AP is

Question 19

Find
the sum of all natural numbers between 200 and 400 which are divisible by 7.

Solution 19

Question 20

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

Solution 20

Question 21

The nth term of an AP is (7 – 4n). Find its common difference.

Solution 21

Question 22

Find the sum of all multiples of 9 lying between 300 and 700.

Solution 22

All numbers between 300 and 700 that are multiples of 9 are 306, 315, 324, 333, …, 693

This is an AP in which a = 306, d = (315 – 306) = 9, l = 693

Let the number of these terms be n, then

Question 23

The nth term of an AP is (3n + 5). Find its common difference.

Solution 23

Thus, common difference = 3

Question 24

Write the next term of the AP

Solution 24

The given AP is

Common difference d =

Term next to

Question 25

Find the sum of the following:

Solution 25

The given AP is

First term

Common difference d =

Sum of n terms =

Question 26

Solution 26

Question 27

In an AP the first term is 2, the last term is 29 and sum of the terms is 155. Find the common difference of the AP.

Solution 27

First term ‘a’ of an AP = 2

The last term l = 29

common difference = 3

Question 28

In
an AP, the first term is -4, the last term is 29 and the sum of all its terms
is 150. Find its common difference.

Solution 28

Question 29

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

Solution 29

Question 30

The
first and the last terms of an AP are 5 and 45 respectively. If the sum of
all its terms is 400, find the common difference and the number of terms.

Solution 30

Question 31

In an AP the first term is 22, nth term is -11 and sum to first nth terms is 66. Find n and d, the common difference.

Solution 31

First term of an AP, a = 22

Last term = nth term = – 11

Thus, n = 12, d = -3

Question 32

The
12th term of an AP is -13 and the sum of its first four terms is
24. Find the sum of its first 10 terms.

Solution 32

Question 33

The
sum of the first 7 terms of an AP is 182. If its 4th and 17th
terms are in the ratio 1 : 5, find the AP.

Solution 33

Question 34

The
sum of the first 9 terms of an AP is 81 and that of its first 20 terms is
400. Find the first term and the common difference of the AP.

Solution 34

Question 35

The
sum of the first 7 terms of an AP is 49 and the sum of its first 17 terms is
289. Find the sum of its first n terms.

Solution 35

Question 36

Two
APs have the same common difference. If the first terms of these APs be 3 and
8 respectively, find the difference between the sums of their first 50 terms.

Solution 36

Question 37

The
sum of first 10 terms of an AP is -150 and the sum of its next 10 terms is
-550. Find the AP.

Solution 37

Question 38

The
13th term of an AP is 4 times its 3rd term. If its 5th
term is 16, find the sum of its first 10 terms.

Solution 38

Question 39

The
16th term of an AP is 5 times its 3rd term. If its 10th
term is 41, find the sum of its first 15 terms.

Solution 39

Question 40

An
AP 5, 12, 19, … has 50 terms. Find its last term.
Hence, find the sum of its last 15 terms.

Solution 40

Question 41

An
AP 8, 10, 12, … has 60 terms. Find its last term.
Hence, find the sum of its last 10 terms.

Solution 41

Question 42

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

Solution 42

Question 43

The
sum of first m terms of an AP is (4m2 – m).If its nth term is 107,
find the value of n. Also ,find the 21st term of this AP.

Solution 43

Question 44

The
sum of first q terms of an AP is (63q -3q2). If its pth term is -60, find the value of p. Also , find the 11th term of its AP.

Solution 44

Question 45

Find
the number of terms of the AP -12, -9, -6, …, 21.
If 1 is added to each term of this AP then find the sum of all terms of the
AP thus obtained.

Solution 45

Question 46

Sum
of the first 14 terms of an AP is 1505 and its first term is 10. Find its 25th
term.

Solution 46

Question 47

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

Solution 47

Question 48

In
a school, students decided to plant 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 double of the class in which they are studying.
If there are 1 to 12 classes in the school and each class has two sections,
find how many trees were planted by students. Which value is shown in the
question?

Solution 48

Question 49

In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are 10 potatoes in the line. 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 he continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

Solution 49

Question 50

There are 25 trees at equal distances of 5 m in a line with a water tank, the distance of the water tank from the nearest tree being 10 m. A gardener waters all the trees separately, starting from the water tank and returning back to the water tank after watering each tree to get water for the next. Find the total distance covered by the gardener in order to water all the trees.

Solution 50

Question 51

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 prize.

Solution 51

Question 52

A
man saved Rs. 33000 in 10 months. In each month after the first, he saved Rs.
100 more than he did in the preceding month. How much did he save in the
first month?

Solution 52

Question 53

A
man arranges to pay off a debt of Rs. 36000 by 40 monthly installments which
form an arithmetic series. When 30 of the installments are paid, he dies
leaving one-third of the debt unpaid. Find the value of the first installment.

Solution 53

Question 54

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 1 contractor has to pay as penalty, if he
has delayed the work by 30 days?

Solution 54

## Chapter 11 – Arithmetic Progressions Exercise MCQ

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

If 4, x1, x2, x3, 28 are in AP then x3= ?

(a) 19

(b) 23

(c) 22

(d) cannot be determined

Solution 4

Question 5

If the nth term of an AP is (2n + 1) then the sum of its first three terms is

(a) 6n + 3

(b) 15

(c) 12

(d) 21

Solution 5

Question 6

The sum of first n terms of an AP is (3n2 + 6n). The common difference of the AP is

(a) 6

(b) 9

(c) 15

(d) -3

Solution 6

Question 7

The sum of first n terms of an AP is (5n – n2). The nth term of the AP is

(a) (5 – 2n)

(b) (6 – 2n)

(c) (2n – 5)

(d) (2n – 6)

Solution 7

Question 8

The sum of first n terms of an AP is (4n2 + 2n). The nth term of this AP is

(a) (6n – 2)

(b) (7n – 3)

(c) (8n – 2)

(d) (8n + 2)

Solution 8

Question 9

The 7th term of an AP is -1 and its 16th term is 17. The nth term of AP is

(a) (3n + 8)

(b) (4n – 7)

(c) (15 – 2n)

(d) (2n -15)

Solution 9

Question 10

The 5th term of an AP is -3 and its common difference is -4. The sum of its first 10 terms is

(a) 50

(b) -50

(c) 30

(d) -30

Solution 10

Question 11

The 5th term of an AP is 20 and the sum of its 7th and 11th terms is 64. The common difference of the AP is

(a) 4

(b) 5

(c) 3

(d) 2

Solution 11

Question 12

The 13th term of an AP is 4 times its 3rd term. If its 5th term is 16 then the sum of its first ten terms is

(a) 150

(b) 175

(c) 160

(d) 135

Solution 12

Question 13

An AP 5, 12, 19, … has 50 terms. Its last term is

(a) 343

(b) 353

(c) 348

(d) 362

Solution 13

Question 14

The sum of first 20 odd natural numbers is

(a) 100

(b) 210

(c) 400

(d) 420

Solution 14

Question 15

The sum of first 40 positive integers divisible by 6 is

(a) 2460

(b) 3640

(c) 4920

(d) 4860

Solution 15

Question 16

How many two-digit numbers are divisible by 3?

(a) 25

(b) 30

(c) 32

(d) 36

Solution 16

Question 17

How many three-digit numbers are divisible by 9?

(a) 86

(b) 90

(c) 96

(d) 100

Solution 17

Question 18

What is the common difference of an AP in which a18 – a14 = 32?

(a) 8

(b) -8

(c) 4

(d) -4

Solution 18

Question 19

If an denotes the nth term of the AP 3, 8, 13, 18, … then what is the value of (a30 -a20)?

(a) 40

(b) 36

(c) 50

(d) 56

Solution 19

Question 20

Which term of the AP 72, 63, 54, … is 0?

(a) 8th

(b) 9th

(c) 10th

(d) 11th

Solution 20

Question 21

Which term of the AP 25, 20, 15, … is the first negative term?

(a) 10th

(b) 9th

(c) 8th

(d) 7th

Solution 21

Question 22

Which term of the AP 21, 42, 63, 84, … is 210?

(a) 9th

(b) 10th

(c) 11th

(d) 12th

Solution 22

Question 23

What is 20th term from the end of the AP 3, 8, 13, …, 253?

(a) 163

(b) 158

(c) 153

(d) 148

Solution 23

Question 24

(5 + 13 +21+… + 181) =?

(a) 2476

(b) 2337

(c) 2219

(d) 2139

Solution 24

Question 25

The sum of first 16 terms of the AP 10, 6, 2, … is

(a) 320

(b) -320

(c) -352

(d) -400

Solution 25

Question 26

How many terms of the AP 3, 7, 11, 15, … will make the sum 406?

(a) 10

(b) 12

(c) 14

(d) 20

Solution 26

Question 27

The 2nd term of an AP is 13 and its 5th term is 25. What is its 17th term?

(a) 69

(b) 73

(c) 77

(d) 81

Solution 27

Question 28

The 17th term of an AP exceeds its 10th term by 21. The common difference of the AP is

(a) 3

(b) 2

(c) -3

(d) -2

Solution 28

Question 29

The 8th term of an AP is 17 and its 14th term is 29. The common difference of the AP is

(a) 3

(b) 2

(c) 5

(d) -2

Solution 29

Question 30

The 7th term of an AP is 4 and its common difference is -4. What is its first term?

(a) 16

(b) 20

(c) 24

(d) 28

Solution 30

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