NCERT Solutions for Class 10 Maths Chapter 5
Arithmetic Progressions Exercise 5.2
Page 105
1. Fill in the blanks in the following table, given that a is the first term, d the common difference and a_{n} the nth term of the AP:
a | d | n | a_{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 | … |
Answer
(i) Given,
First term, a = 7
Common difference, d = 3
Number of terms, n = 8,
We have to find the nth term, a_{n} = ?
As we know, for an Arithmetic Progression
a_{n} = a+(n−1)d
Putting the values,
⇒ 7+(8 −1) 3
⇒ 7+(7) 3
⇒ 7+21 = 28
Hence, a_{n} = 28
(ii) Given,
First term, a = -18
Common difference, d = ?
Number of terms, n = 10
Nth term, a_{n} = 0
As we know, for an Arithmetic Progression
a_{n} = a+(n−1)d
Putting the values,
0 = − 18 +(10−1)d
18 = 9d
d = 18/9 = 2
Hence, common difference, d = 2
(iii) Given,
First term, a = ?
Common difference, d = -3
Number of terms, n = 18
Nth term, a_{n} = -5
As we know, for an Arithmetic Progression
a_{n} = 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, a_{n} = 3.6
As we know, for an Arithmetic Progression
a_{n} = 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 =
Number of terms, n = 105
Nth term, a_{n} = ?
As we know, for an Arithmetic Progression
a_{n} = a+(n −1)d
Putting the values,
a_{n} = 3.5+(105−1) 0
a_{n} = 3.5+104×0
a_{n} = 3.5
Hence, a_{n} = 3.5
2. Choose the correct choice in the following and justify:
(i) 30^{th} term of the A.P: 10,7, 4, …, is
(A) 97 (B) 77 (C) −77 (D) −87
Answer
Given here,
Arithmetic Progression = 10, 7, 4, …
Therefore, we can find,
First term, a = 10
Common difference, d = a_{2} − a_{1 }= 7−10 = −3
As we know, for an Arithmetic Progression,
a_{n} = a +(n−1)d
Putting the values;
a_{30} = 10 + (30−1)(−3)
a_{30} = 10 + (29)(−3)
a_{30} = 10 − 87 = −77
Hence, the correct answer is option C.
(ii) 11^{th }term of the A.P. -3, -1/2, ,2 …. is
(A) 28 (B) 22 (C) -38 (D) -48½
Answer
Given here,
Arithmetic Progression = -3, -1/2, ,2 …
Therefore, we can find,
First term a = – 3
Common difference, d = a_{2} − a_{1} = (-1/2) -(-3)
⇒ (-1/2) + 3 = 5/2
As we know, for an Arithmetic Progression,
a_{n} = a+(n−1)d
Putting the values;
a_{11} = 3+(11-1)(5/2)
a_{11} = 3+(10)(5/2)
a_{11} = -3+25
a_{11} = 22
Hence, the answer is option B.
3. In the following APs, find the missing terms in the boxes :
(i) 2, ☐, 26
Answer
For the given Arithmetic Progression, 2, 2, 26
The first and third term are :-
a = 2
a_{3} = 26
As we know, for an Arithmetic Progression,
a_{n} = a+(n −1)d
Therefore, putting the values here,
a_{3} = 2+(3-1)d
26 = 2+2d
24 = 2d
d = 12
a_{2} = 2+(2-1)12
= 14
Therefore, 14 is the missing term.
(ii) ☐, 13, ☐, 3
For the given Arithmetic Progression, , 13, ,3
a_{2} = 13 and
a_{4} = 3
As we know, for an Arithmetic Progression,
a_{n} = a+(n−1) d
Therefore, putting the values here,
a_{2} = a +(2-1)d
13 = a+d ……(i)
a_{4} = a+(4-1)d
3 = a+3d ……(ii)
On subtracting equation (i) from (ii), we get,
– 10 = 2d
d = – 5
From equation (i), putting the value of d, we get
13 = a+(-5)
a = 18
a_{3} = 18+(3-1)(-5)
⇒ 18+2(-5) = 18-10 = 8
Therefore, the missing terms are 18 and 8, respectively.
(iii) 5, ☐, ☐, 9^{1}⁄_{2}
For the given Arithmetic Progression
a = 5 and
a_{4} = 19/2
As we know, for an Arithmetic Progression
a_{n} = a+(n−1)d
Therefore, putting the values here,
a_{4} = a+(4-1)d
19/2 = 5+3d
(19/2) – 5 = 3d
3d = 9/2
d = 3/2
a_{2} = a+(2-1)d
a_{2} = 5+3/2
a_{2} = 13/2
a_{3} = a+(3-1)d
a_{3} = 5+2×3/2
a_{3} = 8
Therefore, the missing terms are 13/2 and 8, respectively.
(iv) -4, ☐, ☐, ☐, ☐, 6
For the given Arithmetic Progression
a = −4 and
a_{6} = 6
As we know, for an Arithmetic Progression
a_{n} = a +(n−1) d
Therefore, putting the values here,
a_{6} = a+(6−1)d
6 = − 4+5d
10 = 5d
d = 2
a_{2} = a+d = − 4+2 = −2
a_{3} = a+2d = − 4+2(2) = 0
a_{4} = a+3d = − 4+ 3(2) = 2
a_{5} = a+4d = − 4+4(2) = 4
Therefore, the missing terms are −2, 0, 2, and 4 respectively.
(v) ☐, 38, ☐, ☐, ☐, -22
For the given Arithmetic Progression
a_{2} = 38
a_{6} = −22
As we know, for an Arithmetic Progression
a_{n} = a+(n −1)d
Therefore, putting the values here,
a_{2} = a+(2−1)d
38 = a+d ……(i)
a_{6} = a+(6−1)d
−22 = a+5d ……(ii)
On subtracting equation (i) from (ii), we get
− 22 − 38 = 4d
−60 = 4d
d = −15
a = a_{2} − d = 38 − (−15) = 53
a_{3} = a + 2d = 53 + 2 (−15) = 23
a_{4} = a + 3d = 53 + 3 (−15) = 8
a_{5} = a + 4d = 53 + 4 (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7, respectively.
4. Which term of the AP : 3, 8, 13, 18, . . . ,is 78?
Given the Arithmetic Progression series as3, 8, 13, 18, …
First term, a = 3
Common difference, d = a_{2} − a_{1} = 8 − 3 = 5
Let the n^{th} term of given Arithmetic Progression be 78. Now as we know,
a_{n} = a+(n−1)d
Therefore,
78 = 3+(n −1)5
75 = (n−1)5
(n−1) = 15
n = 16
Hence, 16^{th} term of this Arithmetic Progression is 78.
5. Find the number of terms in each of the following APs :
(i) 7, 13, 19, . . . , 205
Answer
Given, 7, 13, 19, …, 205 is the Arithmetic Progression
∴ First term, a = 7
Common difference, d = a_{2} − a_{1} = 13 − 7 = 6
Let there are n terms in this Arithmetic Progression
a_{n} = 205
As we know, for an Arithmetic Progression
a_{n} = a + (n − 1) d
Therefore, 205 = 7 + (n − 1) 6
198 = (n − 1) 6
33 = (n − 1)
n = 34
Therefore, this given series has 34 terms in it.
(ii) 18, 15^{1}⁄_{2} , 13, . . . , – 47
First term, a = 18
Common difference, d = a_{2}-a_{1 }=
15^{1}⁄_{2 }– 18
d = (31-36)/2 = -5/2
Let there are n terms in this Arithmetic Progression
a_{n} = -47
As we know, for an Arithmetic Progression
a_{n} = a+(n−1)d
-47 = 18+(n-1)(-5/2)
-47-18 = (n-1)(-5/2)
-65 = (n-1)(-5/2)
(n-1) = -130/-5
(n-1) = 26
n = 27
Therefore, this given Arithmetic Progression has 27 terms in it.
6. Check whether – 150 is a term of the AP : 11, 8, 5, 2 . . .
Answer
For the given series, Arithmetic Progression 11, 8, 5, 2..
First term, a = 11
Common difference, d = a_{2}−a_{1} = 8−11 = −3
Let −150 be the n^{th} term of this Arithmetic Progression
As we know, for an Arithmetic Progression,
a_{n} = a+(n−1)d
-150 = 11+(n -1)(-3)
-150 = 11-3n +3
-164 = -3n
n = 164/3
Clearly, n is not an integer but a fraction.
∴ – 150 is not a term of this Arithmetic Progression.
7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Answer
Given that, 11^{th} term, a_{11} = 38
and 16^{th} term, a_{16} = 73
We know that,
a_{n} = a+(n−1)d
a_{11} = a+(11−1)d
38 = a+10d ……(i)
In the same way,
a_{16} = a +(16−1)d
73 = a+15d ……(ii)
On subtracting equation (i) from (ii), we get
35 = 5d
d = 7
From equation (i), we can write,
38 = a+10×(7)
38 − 70 = a
a = −32
a_{31} = a +(31−1) d
= − 32 + 30 (7)
= − 32 + 210
= 178
Hence, 31^{st} term is 178.
8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Answer
Given that,
3^{rd} term, a_{3} = 12
50^{th} term, a_{50} = 106
We know that,
a_{n} = a+(n−1)d
a_{3} = a+(3−1)d
12 = a+2d ……(i)
In the same way,
a_{50 }= a+(50−1)d
106 = a+49d ……(ii)
On subtracting equation (i) from (ii), we get
94 = 47d
d = 2 = common difference
From equation (i), we can write now,
12 = a+2(2)
a = 12−4 = 8
a_{29} = a+(29−1) d
a_{29} = 8+(28)2
a_{29} = 8+56 = 64
Therefore, 29^{th} term is 64.
9. If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?
Answer
Given that,
3^{rd} term, a_{3} = 4
and 9^{th} term, a_{9} = −8
We know that,
a_{n} = a+(n−1)d
∴ a_{3} = a+(3−1)d
4 = a+2d ……(i)
a_{9} = a+(9−1)d
−8 = a+8d ……(ii)
On subtracting equation (i) from (ii), we will get here,
−12 = 6d
d = −2
From equation (i), we can write,
4 = a+2(−2)
4 = a−4
a = 8
Let n^{th} term of this Arithmetic Progression be zero.
a_{n }= a+(n−1)d
0 = 8+(n−1)(−2)
0 = 8−2n+2
2n = 10
n = 5
Hence, 5^{th} term of this Arithmetic Progression is 0.
10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Answer
We know that, for an Arithmetic Progression series;
a_{n} = a+(n−1)d
a_{17} = a+(17−1)d
a_{17} = a +16d
In the same way,
a_{10} = a+9d
As it is given in the question,
a_{17} − a_{10} = 7
∴ (a +16d)−(a+9d) = 7
7d = 7
d = 1
∴ the common difference is 1.
11. Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?
Answer
Given Arithmetic Progression is 3, 15, 27, 39, …
first term, a = 3
common difference, d = a_{2} − a_{1} = 15 − 3 = 12
We know that, a_{n} = a+(n−1)d
∴ a_{54} = a+(54−1)d
⇒3+(53)(12)
⇒3+636 = 639
a_{54 }= 639
We have to find the term of this Arithmetic Progression which is 132 more than a_{54, }i.e.771.
Let n^{th} term be 771.
a_{n} = a+(n−1)d
771 = 3+(n −1)12
768 = (n−1)12
(n −1) = 64
n = 65
∴ 65^{th} term was 132 more than 54^{th} term.
Or another method is :
Let n^{th} term be 132 more than 54^{th} term.
n = 54 + 132/2
= 54 + 11 = 65^{th} term
12. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Answer
Let, the first term of two Arithmetic Progressions be a_{1} and a_{2} respectively
And the common difference of these Arithmetic Progressions be d.
For the first Arithmetic Progression, we know,
a_{n} = a+(n−1)d
∴ a_{100} = a_{1}+(100−1)d
= a_{1} + 99d
a_{1000} = a_{1}+(1000−1)d
a_{1000} = a_{1}+999d
For second Arithmetic Progression, we know,
a_{n} = a+(n−1)d
Therefore,
a_{100} = a_{2}+(100−1)d
= a_{2}+99d
a_{1000} = a_{2}+(1000−1)d
= a_{2}+999d
Given that, difference between 100^{th} term of the two Arithmetic Progressions = 100
Therefore, (a_{1}+99d) − (a_{2}+99d) = 100
a_{1}−a_{2} = 100 …..(i)
Difference between 1000^{th} terms of the two Arithmetic Progressions
(a_{1}+999d) − (a_{2}+999d) = a_{1}−a_{2
}From equation (i),
This difference, a_{1}−a_{2 }= 100
Hence, the difference between 1000^{th} terms of the two Arithmetic Progression will be 100.
13. How many three-digit numbers are divisible by 7?
Answer
First three-digit number that is divisible by 7 are :-
First number = 105
Second number = 105+7 = 112
Third number = 112+7 =119
Therefore, 105, 112, 119, …
All are three digit numbers are divisible by 7 and thus, all these are terms of an Arithmetic Progression having first term as 105 and common difference as 7.
As we know, the largest possible three-digit number is 999.
When we divide 999 by 7, the remainder will be 5.
∴ 999-5 = 994 is the maximum possible three-digit number that is divisible by 7.
Now the series is as follows.
105, 112, 119, …, 994
Let 994 be the nth term of this Arithmetic Progression.
first term, a = 105
common difference, d = 7
a_{n} = 994
n = ?
As we know,
a_{n} = a+(n−1)d
994 = 105+(n−1)7
889 = (n−1)7
(n−1) = 127
n = 128
∴ 128 three-digit numbers are divisible by 7.
14. How many multiples of 4 lie between 10 and 250?
Answer
The first multiple of 4 that is greater than 10 is 12.
Next multiple will be 16.
Therefore, the series formed as :-
12, 16, 20, 24, …
All these are divisible by 4 and thus, all these are terms of an Arithmetic Progression with first term as 12 and common difference as 4.
When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4.
The series is as follows :-
12, 16, 20, 24, …, 248
Let 248 be the n^{th} term of this Arithmetic Progression
first term, a = 12
common difference, d = 4
a_{n} = 248
As we know,
a_{n} = a+(n−1)d
248 = 12+(n-1)×4
236/4 = n-1
59 = n-1
n = 60
∴ there are 60 multiples of 4 between 10 and 250.
15. For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?
Answer
Given two APs as; 63, 65, 67,… and 3, 10, 17,….
Taking first Arithmetic Progression
63, 65, 67, …
First term, a = 63
Common difference, d = a_{2}−a_{1} = 65−63 = 2
We know, n^{th} term of this Arithmetic Progression = a_{n} = a+(n−1)d
a_{n}= 63+(n−1)2 = 63+2n−2
a_{n} = 61+2n …..(i)
Taking second Arithmetic Progression,
3, 10, 17, …
First term, a = 3
Common difference, d = a_{2} − a_{1} = 10 − 3 = 7
We know that,
n^{th} term of this Arithmetic Progression = 3+(n−1)7
a_{n} = 3+7n−7
a_{n} = 7n−4 …..(ii)
Given, n^{th} term of these Arithmetic Progressions are equal to each other.
Equating both these equations, we get,
61+2n = 7n−4
61+4 = 5n
5n = 65
n = 13
Therefore, 13^{th} terms of both these Arithmetic Progressions are equal to each other.
16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Answer
Given, Third term, a_{3} = 16
As we know,
a +(3−1)d = 16
a+2d = 16 …..(i)
It is given that, 7^{th} term exceeds the 5^{th} term by 12.
a_{7} − a_{5} = 12
[a+(7−1)d]−[a +(5−1)d]= 12
(a+6d)−(a+4d) = 12
2d = 12
d = 6
From equation (i), we get,
a+2(6) = 16
a+12 = 16
a = 4
∴ Arithmetic Progression will be4, 10, 16, 22, ….
17. Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.
Answer
Given Arithmetic Progression is3, 8, 13, …, 253
Common difference, d= 5.
∴ we can write the given Arithmetic Progression in reverse order as;
253, 248, 243, …, 13, 8, 5
Now for the new Arithmetic Progression,
first term, a = 253
and common difference, d = 248 − 253 = −5
n = 20
Therefore, using nth term formula, we get,
a_{20} = a+(20−1)d
a_{20} = 253+(19)(−5)
a_{20} = 253−95
a = 158
∴ 20^{th} term from the last term of the Arithmetic Progression 3, 8, 13, …, 253 is 158.
18. 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.
Answer
We know that, the nth term of the Arithmetic Progression is;
a_{n} = a+(n−1)d
a_{4} = a+(4−1)d
a_{4} = a+3d
In the same way, we can write,
a_{8} = a+7d
a_{6} = a+5d
a_{10} = a+9d
Given that,
a_{4}+a_{8} = 24
a+3d+a+7d = 24
2a+10d = 24
a+5d = 12 …..(i)
a_{6}+a_{10} = 44
a +5d+a+9d = 44
2a+14d = 44
a+7d = 22 …..(ii)
On subtracting equation (i) from (ii), we get,
2d = 22 − 12
2d = 10
d = 5
From equation (i), we get,
a+5d = 12
a+5(5) = 12
a+25 = 12
a = −13
a_{2} = a+d = − 13+5 = −8
a_{3} = a_{2}+d = − 8+5 = −3
∴ the first three terms of this Arithmetic Progressions are −13, −8, and −3.
19. 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?
Answer
It can be seen from the given question, that the incomes of Subba Rao increases every year by ₹ 200 and hence, forms an Arithmetic Progression.
Therefore, after 1995, the salaries of each year are;
5000, 5200, 5400, …
Here, first term, a = 5000
and common difference, d = 200
Let after n^{th} year, his salary be ₹ 7000.
Therefore, by the n^{th} term formula of Arithmetic Progression,
a_{n} = a+(n−1) d
7000 = 5000+(n−1)200
200(n−1)= 2000
(n−1) = 10
n = 11
Therefore, in 11th year, his salary will be ₹ 7000.
20. Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, find n.
Answer
Given that, Ramkali saved ₹ 5 in first week and then started saving each week by ₹ 1.75.
Hence, First term, a = 5
and common difference, d = 1.75
Also given, a_{n }= 20.75
Find, n = ?
As we know, by the n^{th} term formula,
a_{n} = a+(n−1)d
Therefore,
20.75 = 5+(n -1)×1.75
15.75 = (n -1)×1.75
(n -1) = 15.75/1.75 = 1575/175
= 63/7 = 9
n -1 = 9
n = 10
Hence, n is 10.
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