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# Arithmetic progressions Notes | EduRev

## : Arithmetic progressions Notes | EduRev

``` Page 1

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Class 10 Arithmetic progression CBSE Test Paper- 2
Q.1. Prove that S
n
– S
n-1
= t
n

Solution:  S
n
=
?? 2
[2 a + ( n - 1) d ]
S
n-1
=
?? -1
2
[2 a + ( n – 1 - 1) d ] ?
?? -1
2
[2a + ( n – 2)d]
Now, S
n
– S
n-1
=
?? 2
[2 a + ( n - 1) d ] -
?? -1
2
[2a + ( n – 2)d]
=
?? [ 2?? + ?? -1 ?? -(?? -1)[2a + ( n – 2)d]
2

=
2 ?? ?? +?? ?? -?? ?? -2?? ?? -?? ?? -2 ?? +2?? +(?? -2)?? 2

?
2?? +2 ?? -1 ?? 2

= ?? + (?? - ?? )??
= ?? ??
Q.2. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29
th

term.
Solution: Given, n = 50; t
n
= 106
t
3
= a + (3-1)d ? 12 = a + 2d -------------(i)
t
n
= 106 ? 106 = a + 49d ------(ii)
Substract (i) from (ii) , 106 - 12 = 49d – 2d  ? 94 = 47d ? Or, d = 2
Put this value in (i), 12 = a + 2 x 2 ? a = 12 – 4 Hence, a = 8
And, T
29
= a + 28 d ? 8 + 28 x 2 = 64
Q. 3. If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is
zero?
Solution: the 3rd term of an AP = 4  ? 4 = a + 2d -----------(i)
The 9rd term of an AP = - 8   ? - 8 = a + 8d –(ii)
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Class 10 Arithmetic progression CBSE Test Paper- 2
Q.1. Prove that S
n
– S
n-1
= t
n

Solution:  S
n
=
?? 2
[2 a + ( n - 1) d ]
S
n-1
=
?? -1
2
[2 a + ( n – 1 - 1) d ] ?
?? -1
2
[2a + ( n – 2)d]
Now, S
n
– S
n-1
=
?? 2
[2 a + ( n - 1) d ] -
?? -1
2
[2a + ( n – 2)d]
=
?? [ 2?? + ?? -1 ?? -(?? -1)[2a + ( n – 2)d]
2

=
2 ?? ?? +?? ?? -?? ?? -2?? ?? -?? ?? -2 ?? +2?? +(?? -2)?? 2

?
2?? +2 ?? -1 ?? 2

= ?? + (?? - ?? )??
= ?? ??
Q.2. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29
th

term.
Solution: Given, n = 50; t
n
= 106
t
3
= a + (3-1)d ? 12 = a + 2d -------------(i)
t
n
= 106 ? 106 = a + 49d ------(ii)
Substract (i) from (ii) , 106 - 12 = 49d – 2d  ? 94 = 47d ? Or, d = 2
Put this value in (i), 12 = a + 2 x 2 ? a = 12 – 4 Hence, a = 8
And, T
29
= a + 28 d ? 8 + 28 x 2 = 64
Q. 3. If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is
zero?
Solution: the 3rd term of an AP = 4  ? 4 = a + 2d -----------(i)
The 9rd term of an AP = - 8   ? - 8 = a + 8d –(ii)

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Subtract (i) from (ii), - 8 – 4 = 8d – 2d ? - 12 = 6d ? d = -2
Put this value in (i),  - 8 = a + 8 x – 2 ? a = - 8 + 16 = 8
Let nth term of this AP is zero
t
n
=  0 ? a + (n -1)d ? 0 = 8 + (n - 1) x -2 ? 8 = 2 (n - 1) ? n - 1 = 4  ? n = 5
So, 5th term of this AP is zero
Q. 4. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution: Let first term = a and common difference = d
t
17
= a + 16d and t
10
= a + 9d
The 17th term of an AP exceeds its 10th term by 7
? a + 16d =  a + 9d – 7 ?a + 16d – a – 9d = 7 ?7d = 7  ?d = 1
Q.5. Which term of the AP: 3. 15, 27, 39, ………….. . will be 132 more than its 54th term?
Solution: a = 3, d = 12,
54th term of AP = a + 53d = 3 + 53 x 12 = 3 + 636 = 639
Let, an term will be 132 more than the 54th term
? a + (n - 1)d = 639 + 132
? 3 + ( n - 1) x 12 = 771
? ( n - 1) x 12 = 771 - 3 ? n - 1 =
768
12
= 64 ? n = 64 + 1 = 65
Q. 6. How many three digit numbers are divisible by 7?
Solution: Smallest three digit number divisible by 7 is 105
Greatest three digit number divisible by 7 is 994
a = 105 and an = 994 , d = 7
a
n
= a + (n – 1)d ? an = 994 = 105 + (n – 1)7 ?
889
7
= n - 1 ? n = 128
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Class 10 Arithmetic progression CBSE Test Paper- 2
Q.1. Prove that S
n
– S
n-1
= t
n

Solution:  S
n
=
?? 2
[2 a + ( n - 1) d ]
S
n-1
=
?? -1
2
[2 a + ( n – 1 - 1) d ] ?
?? -1
2
[2a + ( n – 2)d]
Now, S
n
– S
n-1
=
?? 2
[2 a + ( n - 1) d ] -
?? -1
2
[2a + ( n – 2)d]
=
?? [ 2?? + ?? -1 ?? -(?? -1)[2a + ( n – 2)d]
2

=
2 ?? ?? +?? ?? -?? ?? -2?? ?? -?? ?? -2 ?? +2?? +(?? -2)?? 2

?
2?? +2 ?? -1 ?? 2

= ?? + (?? - ?? )??
= ?? ??
Q.2. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29
th

term.
Solution: Given, n = 50; t
n
= 106
t
3
= a + (3-1)d ? 12 = a + 2d -------------(i)
t
n
= 106 ? 106 = a + 49d ------(ii)
Substract (i) from (ii) , 106 - 12 = 49d – 2d  ? 94 = 47d ? Or, d = 2
Put this value in (i), 12 = a + 2 x 2 ? a = 12 – 4 Hence, a = 8
And, T
29
= a + 28 d ? 8 + 28 x 2 = 64
Q. 3. If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is
zero?
Solution: the 3rd term of an AP = 4  ? 4 = a + 2d -----------(i)
The 9rd term of an AP = - 8   ? - 8 = a + 8d –(ii)

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Subtract (i) from (ii), - 8 – 4 = 8d – 2d ? - 12 = 6d ? d = -2
Put this value in (i),  - 8 = a + 8 x – 2 ? a = - 8 + 16 = 8
Let nth term of this AP is zero
t
n
=  0 ? a + (n -1)d ? 0 = 8 + (n - 1) x -2 ? 8 = 2 (n - 1) ? n - 1 = 4  ? n = 5
So, 5th term of this AP is zero
Q. 4. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution: Let first term = a and common difference = d
t
17
= a + 16d and t
10
= a + 9d
The 17th term of an AP exceeds its 10th term by 7
? a + 16d =  a + 9d – 7 ?a + 16d – a – 9d = 7 ?7d = 7  ?d = 1
Q.5. Which term of the AP: 3. 15, 27, 39, ………….. . will be 132 more than its 54th term?
Solution: a = 3, d = 12,
54th term of AP = a + 53d = 3 + 53 x 12 = 3 + 636 = 639
Let, an term will be 132 more than the 54th term
? a + (n - 1)d = 639 + 132
? 3 + ( n - 1) x 12 = 771
? ( n - 1) x 12 = 771 - 3 ? n - 1 =
768
12
= 64 ? n = 64 + 1 = 65
Q. 6. How many three digit numbers are divisible by 7?
Solution: Smallest three digit number divisible by 7 is 105
Greatest three digit number divisible by 7 is 994
a = 105 and an = 994 , d = 7
a
n
= a + (n – 1)d ? an = 994 = 105 + (n – 1)7 ?
889
7
= n - 1 ? n = 128

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Q.7. How many multiples of 4 lie between 10 and 250?
Solution: Smallest number divisible by 4 after 10 is 12,
The greatest number below 250 which is divisible by 4 is 248
a = 12, d = 4 and a
n
= 248
?  a + (n – 1) d = 248  ? 12 + (n – 1) x 4 = 248 ?
248-12
4
= ?? - 1 ? 59 +1 = n
n = 60 Thus,60 multiples of 4 lie between 10 and 250.
Q.8. For what value of n, are the nth terms of two APs: 63, 65, 67,… and 3, 10, 17,… equal?
Solution: in the first AP:  a = 63 and d = 2 and in the second AP:  a = 3 and d = 7
As per question,
63 + (n - 1) x 2 = 3 + (n – 1) x 7  ? 63 + 2n - 2 = 3 + 7n – 7 ?  61 + 2n  = - 4 + 7n
? 61 + 4 = 7n - 2n   ? 65 = 5n ? n = 13
Therefore, 13 terms of both these A.P.s are equal to each other.
Q.9. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution:  7th term exceeds the 5th term by 12,
? a + 6d = a + 4d +12 ? 2d = 12 ? d = 6
3
rd
term of Ap = 16 ? a + 2d = 16 ? a + 2 x 6 ? 16 – 12 = a ? a = 4
AP: 4, 10, 16, 22, 28, 34, 40, 46, ……..
Q. 10. Find the 20th term from the last term of the AP: 3, 8, 13, ……, 253.
Solution: a = 3, d = 5 and a
n
= 253 ? 253 = 3 + (n - 1) x 5 ?
253-3
5
+ 1 = ?? ? n = 51
From the last:  a = 253, d = -3 , n = 20
So, the 20th term from the last term =  a + 19d = 253 + 19 x -3 =  196
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Class 10 Arithmetic progression CBSE Test Paper- 2
Q.1. Prove that S
n
– S
n-1
= t
n

Solution:  S
n
=
?? 2
[2 a + ( n - 1) d ]
S
n-1
=
?? -1
2
[2 a + ( n – 1 - 1) d ] ?
?? -1
2
[2a + ( n – 2)d]
Now, S
n
– S
n-1
=
?? 2
[2 a + ( n - 1) d ] -
?? -1
2
[2a + ( n – 2)d]
=
?? [ 2?? + ?? -1 ?? -(?? -1)[2a + ( n – 2)d]
2

=
2 ?? ?? +?? ?? -?? ?? -2?? ?? -?? ?? -2 ?? +2?? +(?? -2)?? 2

?
2?? +2 ?? -1 ?? 2

= ?? + (?? - ?? )??
= ?? ??
Q.2. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29
th

term.
Solution: Given, n = 50; t
n
= 106
t
3
= a + (3-1)d ? 12 = a + 2d -------------(i)
t
n
= 106 ? 106 = a + 49d ------(ii)
Substract (i) from (ii) , 106 - 12 = 49d – 2d  ? 94 = 47d ? Or, d = 2
Put this value in (i), 12 = a + 2 x 2 ? a = 12 – 4 Hence, a = 8
And, T
29
= a + 28 d ? 8 + 28 x 2 = 64
Q. 3. If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is
zero?
Solution: the 3rd term of an AP = 4  ? 4 = a + 2d -----------(i)
The 9rd term of an AP = - 8   ? - 8 = a + 8d –(ii)

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Subtract (i) from (ii), - 8 – 4 = 8d – 2d ? - 12 = 6d ? d = -2
Put this value in (i),  - 8 = a + 8 x – 2 ? a = - 8 + 16 = 8
Let nth term of this AP is zero
t
n
=  0 ? a + (n -1)d ? 0 = 8 + (n - 1) x -2 ? 8 = 2 (n - 1) ? n - 1 = 4  ? n = 5
So, 5th term of this AP is zero
Q. 4. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution: Let first term = a and common difference = d
t
17
= a + 16d and t
10
= a + 9d
The 17th term of an AP exceeds its 10th term by 7
? a + 16d =  a + 9d – 7 ?a + 16d – a – 9d = 7 ?7d = 7  ?d = 1
Q.5. Which term of the AP: 3. 15, 27, 39, ………….. . will be 132 more than its 54th term?
Solution: a = 3, d = 12,
54th term of AP = a + 53d = 3 + 53 x 12 = 3 + 636 = 639
Let, an term will be 132 more than the 54th term
? a + (n - 1)d = 639 + 132
? 3 + ( n - 1) x 12 = 771
? ( n - 1) x 12 = 771 - 3 ? n - 1 =
768
12
= 64 ? n = 64 + 1 = 65
Q. 6. How many three digit numbers are divisible by 7?
Solution: Smallest three digit number divisible by 7 is 105
Greatest three digit number divisible by 7 is 994
a = 105 and an = 994 , d = 7
a
n
= a + (n – 1)d ? an = 994 = 105 + (n – 1)7 ?
889
7
= n - 1 ? n = 128

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Q.7. How many multiples of 4 lie between 10 and 250?
Solution: Smallest number divisible by 4 after 10 is 12,
The greatest number below 250 which is divisible by 4 is 248
a = 12, d = 4 and a
n
= 248
?  a + (n – 1) d = 248  ? 12 + (n – 1) x 4 = 248 ?
248-12
4
= ?? - 1 ? 59 +1 = n
n = 60 Thus,60 multiples of 4 lie between 10 and 250.
Q.8. For what value of n, are the nth terms of two APs: 63, 65, 67,… and 3, 10, 17,… equal?
Solution: in the first AP:  a = 63 and d = 2 and in the second AP:  a = 3 and d = 7
As per question,
63 + (n - 1) x 2 = 3 + (n – 1) x 7  ? 63 + 2n - 2 = 3 + 7n – 7 ?  61 + 2n  = - 4 + 7n
? 61 + 4 = 7n - 2n   ? 65 = 5n ? n = 13
Therefore, 13 terms of both these A.P.s are equal to each other.
Q.9. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution:  7th term exceeds the 5th term by 12,
? a + 6d = a + 4d +12 ? 2d = 12 ? d = 6
3
rd
term of Ap = 16 ? a + 2d = 16 ? a + 2 x 6 ? 16 – 12 = a ? a = 4
AP: 4, 10, 16, 22, 28, 34, 40, 46, ……..
Q. 10. Find the 20th term from the last term of the AP: 3, 8, 13, ……, 253.
Solution: a = 3, d = 5 and a
n
= 253 ? 253 = 3 + (n - 1) x 5 ?
253-3
5
+ 1 = ?? ? n = 51
From the last:  a = 253, d = -3 , n = 20
So, the 20th term from the last term =  a + 19d = 253 + 19 x -3 =  196

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Q. 11. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and the 10th
terms is 44. Find the first three terms of the AP.
Solution: The sum of the 4th and 8th terms of an AP is 24
? a + 3d + a + 7d = 24 ? 2a + 10d = 24 ? a + 5d = 12 ------(i)
Similarly, the sum of the 6th and the 10th terms is 44 ? a + 5d + a + 9d = 44
? 2a + 14d = 44 ? a + 7d = 22 -----------(ii)
From (ii) - (i) we get, a + 7d – a – 5d = 22 – 12 ? 2d = 10 ? d = 5
Put, d = 6 in (i) , a + 5 x 5 =  12 ? a = 12 – 25 = - 13
So, first three terms of AP : -13, - 8, - 3
Q. 12. Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an
increment of Rs. 200 each year. In which year did his income reached Rs. 7000.?
Solution: a
n
= a + (n - 1)d ? 7000 = 5000 + 200 (n - 1) ? 200 (n - 1) = 2000
? n - 1 = 10 ? n = 11
Q. 13. Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by
Rs. 1.75. If in the nth week, her savings become Rs. 20.75, find n.
Solution: a
n
= a + ( n - 1)d  ? 20.75 = 5 + 1.75 ( n – 1 )  ? 1.75 ( n – 1 ) = 15.75
? n - 1 = 9 ? n = 10
Q.14. Given a
12
= 37, d = 3, find a and S
12
.
Solution: a
12
= 37 ? a + 11d = 37 ?  a + 11 x 3 = 37 ? a = 37 – 33 = 4
Or, S
n
=
?? 2
[?? + ?? ] ? S
12
=
12
2
[ 4 + 37 ] = 6 x 41= 246
Q.15. Given a = 5, d = 3, an = 50, find n and Sn.
Solution: a
n
= 50 ? 5 + (n -1) x 3 = 50 ? n = 16
S
16
=
?? 2
[?? + ?? ] = 8 [5 + 50] = 440
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