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1. Chapter : Arithmetic Progressions Website: www.letstute.com
Arithmetic Progressions
Problems based on
Arithmetic Progressions

2. Q) Show that -3, 0, 3, 6, 9, …. is an AP. Find its 25th term and
the general term.
Solution: We have 0 - (-3) = 3 - 0 = 6 - 3 = 9 - 6 = 3, which is a
constant. Therefore, the given sequence is an AP with a common
difference = 3.
a = first term = - 3 and d = common difference = 3
an = a + (n - 1)d
a25 = - 3 + (25 - 1) (3)
Problems based on
Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com

3. a25 = -3 + (24) (3)
a25 = -3 + (72) = 69
an = a + (n - 1) dGeneral term,
an = -3 + (n - 1) 3
an = -3 + 3n - 3
an = 3n - 6
Hence, a25 = 69 and an = 3n - 6
Chapter : Arithmetic Progressions Website: www.letstute.com
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Arithmetic Progressions

4. Q) Show that the sequence defined by an = 2n - 1, is an AP.
Find its 11th term.
Solution: an = 2n - 1
Replacing n by n - 1 we get,
an – 1 = 2(n - 1 ) - 1
Now, an - an – 1 = 2n - 1 - [2(n - 1) - 1]
= 2n - 1 - 2n + 2 + 1 = 2
Thus, the given sequence is an AP with a constant difference 2.
Chapter : Arithmetic Progressions Website: www.letstute.com
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Arithmetic Progressions

5. an = 2n - 1
a11 = 2 x 11 - 1
a11 = 22 - 1 = 21
Chapter : Arithmetic Progressions Website: www.letstute.com
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Arithmetic Progressions
Hence, a11 = 21

6. Q) How many terms are there in the AP 2, 9, 16, … 261 ?
Solution:
an = a + (n – 1) d
Thus, the given AP has 38 terms.
a = first term = 2 and d = common difference = 9 – 2 = 7
Suppose there are n terms in the given AP, then nth term = 261
261 = 2 + (n – 1) 7
261 = 2 + 7n – 7
7n = 266
n = 38
Chapter : Arithmetic Progressions Website: www.letstute.com
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Arithmetic Progressions

7. Q) Find the value of the middle term (s) of the AP -11, -7, -3,
….49
Solution:
an = a + (n - 1) d
a = first term = -11, an = 49
49 = -11 + (n - 1) x 4
60 = 4n - 4
64 = 4n
n = 16
d = common difference = -7 (-11) = 4
Now
Chapter : Arithmetic Progressions Website: www.letstute.com
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Arithmetic Progressions

8. As n is an even number, there will be two middle terms
𝟏𝟔
𝟐
𝐭𝐡
and
th
i.e 8th term and the 9th term
a8 = a + 7d = -11 + (7 x 4) = 17
a9 = a + 8d = -11 + (8 x 4) = 21
Hence, the values of two middle terms are 17 and 21,
respectively.
Chapter : Arithmetic Progressions Website: www.letstute.com
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Arithmetic Progressions
𝟏𝟔
𝟐
+ 𝟏

9. Q) Is 63 a term of the AP -1, 4, 9, 14,……?
Solution: a = first term = -1 and
d = common difference = 4 - (-1) = 4 + 1 = 5
Let the nth term of the given AP be 63
Then an = 63
 a + (n - 1) d = 63
 -1+ 5n - 5 = 63
 5n = 69
Chapter : Arithmetic Progressions Website: www.letstute.com
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Arithmetic Progressions
-1 + (n - 1)5 = 63

10.  n = = 13
69
5
4
5
Number of terms cannot be a fraction.
Thus 63 is not a term of the AP -1, 4, 9, 14, …..
Chapter : Arithmetic Progressions Website: www.letstute.com
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Arithmetic Progressions