Arithmetic Progressions
Class 10 Maths
Arithmetic Progressions Ex 5.2
1. Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the A.P.
| a | d | n | ||
| (i) | 7 | 3 | 8 | ……… |
| (ii) | -18 | …….. | 10 | 0 |
| (iii) | …….. | -3 | 18 | -5 |
| (iv) | -18.9 | 2.5 | …….. | 3.6 |
| (v) | 3.5 | 0 | 105 | ……… |
Solutions:
(i) First term, a = 7
Common difference, d = 3
Number of terms, n = 8,
find the nth term, an = ?
an = a+(n−1)d
an =7+(8 −1) 3
an = 7+(7) 3
an = 7+21 = 28
Hence, an = 28
(ii) First term, a = -18
Common difference, d = ?
Number of terms, n = 10
Nth term, an = 0
an = a+(n−1)d
0 = − 18 +(10−1)d
18 = 9d
d =
d = 2
Hence, common difference, d = 2
(iii) Given,
First term, a = ?
Common difference, d = -3
Number of terms, n = 18
Nth term, an = -5
As we know, for an A.P.,
an = a+(n−1)d
Putting the values,
−5 = a+(18−1) (−3)
−5 = a+(17) (−3)
−5 = a−51
a = 51−5 = 46
Hence, a = 46
(iv) Given,
First term, a = -18.9
Common difference, d = 2.5
Number of terms, n = ?
Nth term, an = 3.6
As we know, for an A.P.,
an = a +(n −1)d
Putting the values,
3.6 = − 18.9+(n −1)2.5
3.6 + 18.9 = (n−1)2.5
22.5 = (n−1)2.5
(n – 1) = 22.5/2.5
n – 1 = 9
n = 10
Hence, n = 10
(v) Given,
First term, a = 3.5
Common difference, d = 0
Number of terms, n = 105
Nth term, an = ?
As we know, for an A.P.,
an = a+(n −1)d
Putting the values,
an = 3.5+(105−1) 0
an = 3.5+104×0
an = 3.5
Hence, an = 3.5
Q2. Choose the correct choice in the following and justify :
(i) 30th term of the AP: 10, 7, 4, . . . , is
(A) 97 (B) 77 (C) –77 (D) – 87
Solution:
a = 10, d = 7 – 10 = -3
30th term = ?
a30 = a + 29d
= 10 + 29(-3)
= 10 – 87
= – 77
Correct Answer: (C) – 77
(ii) 11th term of the AP: -3, − , 2, …, is
a = -3 , d = – + 3 =
=
11th term = ?
a = a + 10d
= – 3 + 10( )
= -3 + 25
= 22
Correct Answer: (B) 22
Q3. In the following APs, find the missing terms in the boxes:
(i) a = 2, b = ?, c = 26
Solution:
Solution:
(ii) a2 = 13,
∴ a + d = 13 ……………. (1)
a4 = 3
∴ a + 3d = 3 ……………..(2)
On subtracting (ii) from (i)
a + 3d – (a + d) = 3 – 13
a + 3d – a – d = -10
2d = – 10
d = -5
Substituting the value of (d) in (i)
a + d = 13
a + (- 5) = 13
a = 13 + 5
a = 18
a3 = a + 2d = 18 + 2 (-5)
= 18 – 10 = 8
now : 18, 13, 8,
Q 4. Which term of the AP : 3, 8, 13, 18, . . ., is 78?
Solution:
a = 3, d = 8 – 3 = 5, an = 78
an = a + (n – 1) d
78 = 3 + (n – 1) 5
78 – 3 = (n – 1) 5
75 = (n – 1) 5
n – 1 = 75/5
n – 1 = 15
n = 15 + 1
n = 16
Therefore, 16th term of the given AP is 78.
Q5. Find the number of terms in each of the following APs: ?
(i) 7, 13, 19, …………….. , 205
Solution:
a = 7, d = 13 – 7 = 6, an = 205
an = a + (n – 1) d
205 = 7 + (n – 1) 6
205 – 7 = (n – 1) 6
198 = (n – 1) 6
n – 1 = 33
n = 33 + 1
n = 34
Hence, this given AP has 34 terms in it
Hence, this given AP has 27 terms in it
Q6. Check whether – 150 is a term of the AP : 11, 8, 5, 2 . ..
Solution:
a = 11, d = 8 – 11 = – 3 and an = – 150
an = a + (n – 1) d
– 150 = 11 + (n – 1) – 3
– 150 – 11 = (n – 1) -3
– 161 = (n – 1) – 3
n – 1 = 53. 66
n = 53.66 + 1
n = 54.66
is not an integer.
Hence, −150 is not a term of this given AP.
Q7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Solution:
31st term = ?
a11 = 38
⇒a + 10d = 38 ………………… (1)
a16 = 73
⇒ a + 15d = 73 ………………… (2)
On subtracting (1) from (2)
a + 15d – (a + 10d ) = 73 – 38
a + 15d – a – 10d = 35
5d = 35
d = 7
Substituting the value of (d) in (i)
a + 10d = 38
a = 10 (7) = 38
a = 38 – 70
a = – 32
अब, a31 = a + 30d
⇒ a31 = – 32 + 30(7)
⇒ a31 = – 32 + 210
⇒ a31 = 178
Therefore, the 31st term in given AP is 178
Q8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:
According to the question it is given that
50term in AP
now : n = 50
a3 = 12
⇒ a + 2d = 12 ………………… (1)
an = 106
a50 = 106
⇒ a + 49d = 106 ………………… (2)
On subtracting (I) from (II)
a + 49d – (a + 2d ) = 106 – 12
a + 49d – a – 2d = 94
47d = 94
d = 2
Substituting the value of (d) in (i)
a + 2d = 12
a = 2(2) = 12
a = 12 – 4
a = 8
a29 = a + 28d
⇒ a29 = 8 + 28(2)
⇒ a29 = 8 + 56
⇒ a29 = 64
the 29th term in given AP is 64.
Q9. If the 3rd and the 9th terms of an AP are 4 and −8 respectively, which term of this AP is zero?
Solution:
a3 = 4
⇒ a + 2d = 4 ………………… (1)
a9 = – 8
⇒ a + 8d = – 8 ………………… (2)
On subtracting equation (I) from (II)
a + 8d – (a + 2d ) = – 8 – 4
a + 8d – a – 2d = – 12
6d = – 12
d = – 2
Substituting the value of eq… in (i)
a + 2d = 4
a = 2(-2) = 4
a = 4 + 4
a = 8
a = 8, and d = – 2
Let nth term of the given AP be zero
an = 0
an = a + (n – 1) d
⇒ 0 = 8 + (n – 1) -2
⇒ – 8 = (n – 1) -2
⇒ n – 1 = 4
⇒ n = 4 + 1 = 5
Therefore, the 5th term of the given AP is 0
Q10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution:
AP, we know that: an = a + (n − 1)d
∴ a17 – a10 = 7
⇒ a + 16d – (a + 9d) = 7
⇒a + 16d – a – 9d = 7
⇒ 7d = 7
⇒ d = 1
the common difference is 1.
Q11. Which term of the AP: 3, 15, 27, 39, . .. will be 132 more than its 54th term?
Solution:
a = 3, d = 15 – 3 = 12
a54 = a + 53d
= 3 + 53(12)
= 3 + 636
= 639
will be 132 more than
an = a54 + 132
= 639 + 132
= 771
an = a + (n – 1) d
⇒ 771 = 3 + (n – 1) 12
⇒ 771 – 3 = (n – 1) 12
⇒ 768 = (n – 1) 12
⇒ n – 1 = 64
⇒n = 64 + 1 = 65
the 65th term will be 132 more than 54th term.
Q12. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
Let the first term of the given two AP be a and a’
and the common difference of these AP be d
a100 – a’100 = 100
a + 99d – (a’ + 99d) = 100
a + 99d – a’ – 99d = 100
a – a’ = 100 ……………. (1)
a1000 – a’1000 = a + 999d – (a’ + 999d)
= a + 999d – a’ – 999d
= a + a’
because : a – a’ = 100 ( in eq…. 1)
difference between 1000th terms of these AP will be 100.
Q13. How many three-digit numbers are divisible by 7?
Solution:
First three-digit number that is divisible by 7 = 105
105, 112, 119, …………………… 994
∴ a = 105, d = 7 and an = 994
an = a + (n – 1) d
⇒ 994 = 105 + (n – 1) 7
⇒ 994 – 105 = (n – 1) 7
⇒ 889 = (n – 1) 7
⇒ n – 1 = 127
⇒ n = 127 + 1 = 128
there are 128 three-digit numbers that are divisible by 7.
Q14. How many multiples of 4 lie between 10 and 250?
Solution:
First multiple of 4 that is greater than 10 is 12. Next will be 16.
So, 12,16,20,24, . . . ..
when we divide 250 by 4, the remainder will be 2.
Hence, 250 − 2 = 248 is divisible by 4
∴ a = 12, d = 4 and an = 248
an = a + (n – 1) d
⇒ 248 = 12 + (n – 1) 4
⇒ 248 – 12 = (n – 1) 4
⇒ 236 = (n – 1) 4
⇒ n – 1 = 59
⇒ n = 59 + 1 = 60
So, there are 60 multiples of 4 lying between 10 and 250.
Q15. For what value of , are the th terms of two APs: 63, 65, 67, . .. and 3, 10, 17, . .. equal?
Solution:
consider the first AP:
A.P: 63, 65, 67, ……….
a = 63, d = 65 – 63 = 2
an = a + (n – 1) d
= 63 + (n – 1) 2
= 63 + 2n – 2
= 61 + 2n ……………….. (1)
consider the second AP:
A.P: 3, 10, 17, ………………
a = 3, d = 10 – 3 = 7
an = a + (n – 1) d
= 3 + (n – 1) 7
= 3 + 7n – 7
= – 4 + 7n ……………….. (2)
nth term of the two APs are equal to each other. Equating both (1) and (2),
61 + 2n = – 4 + 7n
61 + 4 = 7n – 2n
5n = 65
n = 65/5
n = 13
13th terms of two APs are equal.
Q16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution:
a3 = 16
an = a + (n − 1)d
Substituting the value of = 3, we get
a + (3 − 1)d = 16
⇒ a + 2d = 16 … (1)
a7 – a5 = 12
⇒ a + 6d – (a + 4d) = 12
⇒ a + 6d – a – 4d = 12
⇒ 2d = 12
⇒ d = 6
Substituting the value of (d) in (i)
a + 2d = 16
a + 2(6) = 16
a + 12 = 16
a = 16 – 12
a = 4
a, a + d, a + 2d, a + 3d ………………
⇒ 4, 4 + 6, 4 + 2(6), 4 + 3(6), ……………
A.P: ⇒ 4, 10, 16, 22 …………………………
Q17. Find the 20th term from the last term of the AP : 3, 8, 13, . . . , 253.
Solution:
given AP is 3, 8, 13, . . . . . . . . . . , 253
a = 3, d = 8 – 3 = 5
So, this AP can be written in reverse order as
253, 248, 243, . . . . . , 13, 8, 5
a = 253, n = 20,
Common difference, d = – 5,
a20 = a + 19d
= 253 + 19(-5)
= 253 – 95
= 158
20th term from the last term of the given AP is 158
Arithmetic Progressions Ex 5.2
Q18. 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
Solution:
an = a + (n − 1)d
a4 = a + (4 − 1)d
a4 = a + 3d
Similarly,
a8 = a + 7d
A6 = a + 5d
a10 = a + 9d
it is given that, a4 + a8 = 24
= (a + 3d) + (a + 7d) = 24
⇒ 2a + 10d = 24
⇒ a + 5d = 12 … (1)
It is also given in the question that, A6 + a10 = 44
⇒ a + 5d + a + 9d = 44
⇒ 2a + 14d = 44
⇒ a + 7d = 22 … (2)
On subtracting equation (1) from (2), we get
2d = 22 − 12
⇒ 2d = 10
⇒ d = 5
From equation (1), we get
a + 5d = 12
⇒ a + 5(5) = 12
⇒ a + 25 = 12
⇒ a = −13
now : , -13, -13 + 5, -13 + 2(5)
-13, – 8, – 3
the first three terms of the given AP are −13, −8, and −3
Arithmetic Progressions Ex 5.2
Q19. Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?
Solution:
the income Subba Rao obtained in various years are in AP as every year, his salary is increased by ₹ 200.
A.P: 5000, 5200, 5400, ………………….. 7000
a = 5000, d = 200, an = 7000
an = a + (n – 1)d
7000 = 5000 + (n – 1)200
7000 – 5000 = (n – 1)200
2000 = (n – 1)200
n – 1 = 10
n = 10 + 1
n = 11 वर्ष
, in 11th year, Subba Rao’s salary will become ₹ 7000.
Arithmetic Progressions Ex 5.2
Q20. Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the th week, her weekly savings become ₹ 20.75, find .
Solution:
A.P: 5, 6.75, 8.50, ………………………, 20.75
A = 5, d = 1.75, an = 20.75
an = a + (n – 1)d
20.75 = 5 + (n – 1)1.75
20.75 – 5 = (n – 1)1.75
15.75 = (n – 1)1.75
n – 1 = 9
n = 9 + 1
n = 10
Arithmetic Progressions Ex 5.2
Arithmetic Progressions Ex 5.2