Students can Download Class 10 Maths Chapter 1 Arithmetic Progressions Additional Questions, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations

## Karnataka State Syllabus Class 10 Maths Chapter 1 Arithmetic Progressions Additional Questions

I. Multiple-choice Questions:

Question 1.

If a_{n} = n^{2} – 1 and a_{n} = 99 then the value of n is

a) 100

b) 10

c) 9

d) 99

Answer:

b) 10

Question 2.

If a_{n}= 2n^{2} – 1 then the value of 4^{th} term is

a) 32

b) 30

c) 31

d) 18

Answer:

c) 31

Question 3.

If a_{n}= 2n + 1 then the common difference of the AP is

a) 3

b) 5

c) 2

d) 1

Answer:

c) 2

Question 4.

The general form of an A.P is

a) a, ar, ar^{2},…

b) a, a + d, a + 2d,…

c) a, a-d,.a-2d,….

d) \(\frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2 d}\) …..

Answer:

b) a, a + d, a + 2d,…

Question 5.

In an A.P the common difference is 3 first term is 1 then the value of 10^{th} term is

a) 28

b) 27

c) 25

d) 40

Answer:

a) 28

Question 6.

The formula to find n^{th} term of an A.P is.

a) a + (n – 1)d

b) ar^{n-1}

c) \(\frac{1}{a+(n-1) d}\)

d) a – (n + 1)d

Answer:

a) a + (n – 1)d

Question 7.

If 20, x + 1, 4 are In AP then value of x is

a) 11

b) 12

e) 24

d) 13

Answer:

a) 11

Question 8.

In an arithmetic progression a_{n+5} = 35 & a_{n+1} = 23 then common difference is

a) 3n

b) 4n

c) 2

d) 3

Answer:

d) 3

Question 9.

In an A.P the relation between a_{5} and a_{7} and common difference is (d) is

a) a_{5} = a_{7} + 2d

b) a_{5} = a_{7} + d

c) a_{7} = a_{5} + 3d

d) a_{7} = a_{5} + 2d

Answer:

d) a7 = a5 + 2d

Question 10.

If a is constant then a + 2a + 3a +…. + na is

Answer:

Question 11.

The value of Σ(n – 1) s

a) \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)

b) \(\frac{n(n-1)}{2}\)

c) \(\frac{n}{2}\)

d) \(\frac{n^{2}}{2}\)

Answer:

b) \(\frac{n(n-1)}{2}\)

Question 12.

In an A.P, if S_{5} = 35 and S_{4} = 22 then 5^{th} term is.

a) 35

b) 10

c) 13

d) 22

Answer:

c) 13

Question 13.

The n^{th} term of 3, 7, 11, 15, is

a) 4n – 1

b) 4n + 1

c) 4n + 3

d) 3n + 4

Answer:

a) 4n – 1

Question 14.

In an A.P, if a_{4} = 8 & a = 2, then its common difference is

a) 6

b) 4

c) 2

d) 10

Answer:

c) 2

Question 15.

In an A.P a_{n+5} = 50 and a_{n+1} = 38 then common difference

a) 3

b) 2

c) 3n

d) 2n

Answer:

a) 3

Question 16.

Among the following arithmetic progression is

a) 1, 4, 6,…

b) 12, 10, 14,…

c) 35, 32, 25,….

d) 8, 13, 19,

Answer:

c) 35, 32, 25,….

Question 17.

In an AP, the correct relation is

a) a_{n-5} = a_{n-4} + d

b) a_{n-5} = a_{n-6} + d

C) a_{n-5} = a_{n} + d

d) a_{n+5} = a_{n-5} – d

Answer:

b) a_{n-5} = a_{n-6 }+ d

Question 18.

The value of \(\sum_{n=1}^{n}10\) is

a) 10

b) 11

c) 55

d) 110

Answer:

c) 55

Question 19.

If 2x + 1, 4x, 13-x are in Arithmetic progression then x is equal to

a) 2

b) 3

c) 4

d) 5

Answer:

a) 2

Question 20.

In a progression, if a_{n} = 2n^{2} + 1, then S_{2 }is

a) 9

b) 12

c) 10

d) 11

Answer:

b) 12

II. Short answer Questions:

Question 1.

Write the formula to find n^{th} term of an A.P?

Answer:

a_{n} = a + (n – 1)d

Question 2.

Write the formula to find arithmetic mean of 3 numbers

Answer:

AM = \(\frac{a+b+c}{3}\)

Question 3.

The first term of an A.P is q and its common difference is p. Find its 20th term.

Answer:

a_{20} = a + 19d

= q + 19p

Question 4.

If a_{n} = 5_{n} – 3, find the common difference

Answer:

a_{n} = 5n – 3

a_{1} = 5(1) – 3 = 5 – 3 = 2

a_{2} = 5(2) – 3 = 10 – 3 = 7

d = a_{2} – a_{1} = 7 – 2 = 5

common difference = 5

Question 5.

Find the 9th term from the end (towards the first term) of the A.P 5, 9, 13, ….185

Answer:

a = 185, d = – 4 & 1 = 5

a_{9} = a + (n – 1)d

a_{9}= 1 + (9 – 1) – 4

a_{9} = 185 + 8x – 4

= 185 – 32

a_{9} = 153

Question 6.

For what value of p are 2p + 1, 13, 5p – 3, three consecutive terms of a AP?

Answer:

a_{2} – a_{1}, = a_{3} – a_{2}

13 – (2p + 1) = 5p – 3 – 13

13 – 2p – 1 = 5p – 16

12 + 16 = 5p + 2p

7p = 28

p = 4

Question 7.

Find the common difference of an AP in which a_{18} – a_{14} = 32

Answer:

a_{18} – a_{14} = 32

(a + 17d) – (a + 13d) = 32

a + 17d – a – 13d = 32

4d = 32

d = 8

Question 8.

In an A.P a_{n} = 20 & S_{n} = 399 find S_{n-1}

Answer:

a_{n} = S_{n} – S_{n-1}

S_{n-1} = S_{n} – a_{n}

= 399 – 20

S_{n-1} = 379

Question 9.

In an AP, if d = – 4, n = 7, a_{n} = 4 then find a.

Answer:

a_{n} = a + (n – 1)d

4 = a + (7 – 1) – 4

4 = a + 6 x – 4

4 = a – 24

a = 4 + 24

a = 28

III. Long Answer Questions:

Question 1.

Find the number of terms in the A.P, 100, 96, 92,….., 12

Answer:

100, 96, 92,………, 12

a = 100 d = a_{2} – a_{1}, d = 96 – 100 = – 4

a_{n} = 12 & n = ?

a_{n} = a + (n – 1)d

12 = 100 + (n – 1)d

12 – 100 = (n – 1) x – 4

– 88 = (n – 1) x – 4

\(\frac{-88}{-4}\) = n – 1

n – 1 = 22

n = 22 + 1

n = 23

Question 2.

The angles of a triangle are in A.P if the smallest angle is 50°, find the other two angles.

Answer:

Let the smallest angle = 50°

Three angles of AP are a, a + d, a + 2d – Sum of the angles of a triangle = 180° a + a + d + a + 2d = 180°

50° + 50° + d + 50° + 2d = 180°

3d = 180°- 150°

3d = 30°

d = 10°

Required angles a = 50°

a + d = 50° + 10° = 60°

a + 2d = 50 + 2(10) = 50 + 20 = 70°

Question 3.

An A.P consists of 50 terms of which 3^{rd} term is 12 and last term is 106. Find the 29^{th} term.

Answer:

n = 50

a_{3} = 12

a + 2d = 12

a = 12 – 12d → (1)

a = 106

a_{29} = ?

a_{n} = a + (n – 1) d

106 = 12 – 2d + (50 – 1)d

106 – 12 = – 2d + 49d

94 = 47d

d = 2

Put d = 2 is eqn (1)

a = 12 – 2 (2)

a = 12 – 4

a = 8

Question 4.

The sum of 4^{th} and 8^{th} terms of an A.P is 24 and the sum of 6^{th} and 10^{th} terms of the same A.P is 44. Find the first three terms.

Answer:

Sum of 4^{th} and 8^{th} terms = 24

a_{4} + a_{8} = 24

a + 3d + a + 7d = 24

2a + 10d = 24

÷ by 2 on both the sides

a + 5d = 12 → (1)

Sum of 6^{th} and 10^{th} terms = 44

a_{6} + a_{10} = 44

a + 5d + a + 9d = 44

2a + 14d = 44 by 2

a + 7d = 22 → (2)

Subtract eqn (1) & (2)

a + 5d = 12

a + 7d = 22

(-) (-) (-)

________

– 2d = – 10

+ d = \(\frac{10}{2}\) = 5

Put d = 5 is eqn (1)

a = 12 – 5d

a = 12 – 5(5)

= 12 – 25 a = – 13

required terms a, a + d, a + 2d…….

– 13, – 13,+ 5, – 13 + 2(5)

– 13,- 8, – 3

Question 5.

The ratio of 7^{th} to 3^{rd} term of an A.P is 12 : 5 find the ratio of 13^{th} to 4^{th} term

Answer:

\(\frac{a_{7}}{a_{3}}=\frac{12}{5}\)

\(\frac{a+6 d}{a+2 d}=\frac{12}{5}\)

5(a + 6d) = 12 (a + 2d)

5a + 30d= 12a + 24d

30d – 24d = 12a – 5a

6d = 7a

d = \(\frac{7}{6}\) a

The ratio of 13^{th} to 4^{th} term

Question 6.

A company employed 400 persons in the year 2001 and each year increased by 35 persons. In which year the number of employees in the company will be 785?

Answer:

2001 + 2002 ………………..

400, 435, 470, ………….. 785

a = 400, d = a_{2} – a_{1} an = 785, n = ? d = 435 – 400

d = 35

a_{n} = a + (n – 1)d

785 = 400 + (n – 1) 35

785 – 400 = (n – 1) × 35

385 = (n – 1) x 35

n – 1 = \(\frac{385}{35}\)

n – 1 = 11

n = 11 + 1

n = 12

Employes in the company will be 785 reach in year 2012.

Question 7.

If the pth term of an A.P is q and the qth term is p. Prove that the n^{th} term equal to (p + q – n)

Answer:

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

a_{p} = a + (p – 1)d and a_{q} = a + (q – 1)d

Now a_{p} = q and a_{q} = p given

a + (p – 1)d = q → (1)

a + (q – 1)d = p → (2)

On subtracting (1) from (2) we get

Putting d = – 1 in the eqn (i) we get a + (p – 1 ) x – 1 = q

a – p + 1 = q a = p + q – 1

∴ n^{th} term = a + (n – 1)d

= p + q – 1 + (n – 1) x – 1

= p + q – 1 + n + 1

n^{th} term = p + q – n

Hence it is proved

Question 8.

Find four numbers in A.P such that the sum of 2^{nd} and 3^{rd} terms is 22 and the product of 1^{st} and 4^{th} terms is 85.

Answer:

Sum of 2^{nd} and 3^{rd} terms = 22

Question 9.

Find the sum of all natural numbers between 1 and 201 which are divisible by 5.

Answer:

5 + 10 +15 +………….+ 200

which are divisible by 5

a = 5

d = a_{2} – a_{1} = 10 – 5 = 5

l = a_{n} = 200 + (n – 1)d

200 = 5 + (n – 1) x 5

195 = (n – 1) x 5

n – 1 = \(\frac{195}{5}\)

Question 10.

Find three numbers in AP whose sum and products are respectively. 21 and 231

Answer:

Let the numbers be a – d, a, a + d

Sum = 21

Question 11.

The sum of 6 terms which form an A.P is 345. The difference b/w the first and last terms is 55 find the terms.

Answer:

Let the numbers be a, a + d, a + 2d, a + 3d

a + 4d, a + 5d, a + 6d

6d + 15d = 345 – 3

2a + 5d = 115 → (1)

the difference b/w first term and last term is 55

a_{6} – a_{1} = 55

a + 5d – a = 55

5d = 5

d = \(\frac{55}{5}\)

d = 11

Put d = 11 in eqn (1)

2a + 5d = 115

2a + 55 = 115

2a = 115 – 55

2a = 60

a = \(\frac{60}{2}\)

a = 30

required numbers

a, a + d, a + 2d, a + 3d, a + 4d, a + 5d

30, 30 + 11, 30 + 2(11), 30 + 3(11), 30 + 4(11), 30 + 5(11)

30, 41, 52, 63, 74, 85.

Question 12.

In an AP whose first term is 2, the sum of first five terms is one fourth the sum of the next five terms show that T_{20} = – 112, Find S_{20}.

Answer:

Let the five number a, a + d, a + 2d, a + 3d, a + 4d, next 5 numbers be

a + 5d, a + 6d, a + 7d, a + 8d, a + 9d a = 2

a + a + d + a + 2d + a + 3d + a + 4d

= \(\frac{1}{4}\) (a + 5d + a + 6d + a + 7d + a + 8d + a + 9d)

5a + 10d = \(\frac{1}{4}\) (5a + 35d)

5(2) + 10d = \(\frac{1}{4}\) [(5 x 2) + 35d]

4(10 + 10d) = 10 + 35d

40 + 40d = 10 + 35d

40d – 35d = 10 – 40

5d = – 30

d = \(\frac{-30}{5}\)

d = – 6

a_{20} = ?, n = 20

a_{n} = a + (n – 1)d

a_{20} = 2 + (20 – 1) x 6

= 2 – 114 a_{20} = – 112

a_{20} = l = – 112

S_{20} = ? n = 20

S = f{2+(-112)]

s_{20} =10(2-112) = 10(-110)

s_{20} = -1100

Question 13.

The third term of an A.P is 8 and the ninth term of the A.P exceeds three times the third term by 2 find the sum of its first 19 terms.

Answer:

a_{3} = 8 (given)

a + 2d = 8 → (1)

a_{9} = 3(a_{3}) + 2 (given)

a + 8d = 3(8) + 2 [bye qn (1)]

a + 8d = 26

a + 2d + 6d = 26

8 + 6d = 26 (a + 2d = 28)

6d = 26 – 8

6d = 18

d = \(\frac{18}{6}\)

d = 3

Put d = 3 in eqn (1)

a + 2(3) = 8

a + 6 = 8

a = 8 – 6

a = 2

n = 19

Question 14.

If the sum of m terms of an AP is the same as the sum of its n terms, show that the sum of its (m + n) terms is 120°. Find the number of sides of the zero.

Answer:

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

Then S_{m} = S_{n}

Question 15.

Find the sum of all the first ‘n’ odd natural numbers.

Answer:

1, 3, 5, 7, 9,…… up to n

a = 1, d = 3 – 1 = 2, n = n

Question 16.

The common difference between any two consecutive Interior angles of a polygon is 5°. If the smallest angle is 120°. Find the number of sides of the polygon?

Answer:

Let the AP be 120°, 125°…… with

a = 120°and d = 5

The number of sides in a polygon is the sum of its interior angles. Therefore (2n – 4)90°

= 180n – 360°

2 [180n- 360] = n [240 + 5n – 5]

360n – 720 = n (235 + 5n)

360n – 720 = 235n + 5n^{2}

5n^{2} + 235n – 360n + 720 = 0

5n^{2} – 125n + 720 = 0 + by 5

n^{2}-25n + 144 = 0

n^{2} – 16n – 9n + 144 0

n(n – 16) – 9(n – 16) = 0

(n – 16) (n – 9) = 0

n – 16 = 0, and n – 9 = 0

n = 16 and n = 9

∴ n = 9

∴ number of sides is 9 so it is a nanogon.

Question 17.

In an AP whose first term is 2, the sum of first five terms is one fourth the sum of the next five terms show that T_{20} = – 112

Answer:

a = 2, S_{5} =\(\frac{1}{4}\)[S_{10 }– S_{5}], T_{20} = – 112

Question 18.

In an arithmetic progression the sum of first 10 terms is 175 and the sum of the next 10 terms is 475 find the arithmetic progression.

Answer: