KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Additional Questions

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

I. Multiple-choice Questions:

Question 1.
If a_n = n² – 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= 2n² – 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, ar²,…
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 a₅ and a₇ and common difference is (d) is
a) a₅ = a₇ + 2d
b) a₅ = a₇ + d
c) a₇ = a₅ + 3d
d) a₇ = a₅ + 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 S₅ = 35 and S₄ = 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 a₄ = 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 = 2n² + 1, then S₂ 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:
a₂₀ = a + 19d
= q + 19p

Question 4.
If a_n = 5_n – 3, find the common difference
Answer:
a_n = 5n – 3
a₁ = 5(1) – 3 = 5 – 3 = 2
a₂ = 5(2) – 3 = 10 – 3 = 7
d = a₂ – a₁ = 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
a₉ = a + (n – 1)d
a₉= 1 + (9 – 1) – 4
a₉ = 185 + 8x – 4
= 185 – 32
a₉ = 153

Question 6.
For what value of p are 2p + 1, 13, 5p – 3, three consecutive terms of a AP?
Answer:
a₂ – a₁, = a₃ – a₂
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 a₁₈ – a₁₄ = 32
Answer:
a₁₈ – a₁₄ = 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 = a₂ – a₁, 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
a₃ = 12
a + 2d = 12
a = 12 – 12d → (1)
a = 106
a₂₉ = ?
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
a₄ + a₈ = 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
a₆ + a₁₀ = 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 = a₂ – a₁ 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 = a₂ – a₁ = 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
a₆ – a₁ = 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 T₂₀ = – 112, Find S₂₀.
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
a₂₀ = ?, n = 20
a_n = a + (n – 1)d
a₂₀ = 2 + (20 – 1) x 6
= 2 – 114 a₂₀ = – 112
a₂₀ = l = – 112
S₂₀ = ? n = 20
S = f{2+(-112)]
s₂₀ =10(2-112) = 10(-110)
s₂₀ = -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:
a₃ = 8 (given)
a + 2d = 8 → (1)
a₉ = 3(a₃) + 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 + 5n²
5n² + 235n – 360n + 720 = 0
5n² – 125n + 720 = 0 + by 5
n²-25n + 144 = 0
n² – 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 T₂₀ = – 112
Answer:
a = 2,  S₅ =\(\frac{1}{4}\)[S₁₀ – S₅], T₂₀ = – 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: