Sequence and series questions are given here, along with their solutions to help students understand the concept by problem-solving. The questions presented here are based on Class X and XI syllabi of NCERT (CBSE) with practice questions for the exercise of students. These questions will benefit them to master the skills of solving questions regarding sequence and series, which not only help in their examination but also for future competitive examinations. Learn more about sequence and series.
A sequence is a succession of numbers arranged according to a given rule in an orderly manner. For example, a sequence of square numbers like 1, 4, 9, 16,… in each sequence, there is a general formula to find the next terms in the sequence. A progression is a sequence that follows a particular pattern. An arithmetic progression is a progression in which each term differs from its preceding term by a constant number. A series can be generalised as the sum of terms of the given sequence.
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Sequence and Series Questions with Solutions
Here are given some questions with their solutions for practice.
1. If 7 times the 7th term of an AP is equal to 11 times its 11th term, show that the 18th term of the AP is zero.
Solution: Let a be the first term and d be the common difference of the AP.
7th term of the AP = a + (7 – 1) d = a + 6d
11th term of the AP = a + (11 – 1) d = a + 10d
According to the question,
7 × [a + 6d] = 11 ×[a + 10d]
⇒ 7a + 42d = 11a + 110d
⇒ 4a + 68d = 0
⇒ a + 17d = 0
⇒ a + (18 – 1)d = 0
⇒ 18th term of AP = 0
2. Find the 28th term from the end of the AP 6, 9, 12, 15, 18, …, 102.
Solution: Let the nth term of the AP be 102 where
First term, a = 6
Common Difference, d = 3
Tn = a + (n – 1)d = 102
⇒ 6 + (n – 1)3 = 102
⇒ 6 + 3n – 3 = 102
⇒ 3n = 102 – 3
⇒ 3n = 99 ⇒ n = 33
Therefore, there are 33 terms in the given AP. Then the 28th term from the end is the 6th term from the beginning,
T6 = 6 + (6 – 1)3 = 6 + 15 = 21
Hence, the 28th term from the end is 21.
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If there are n terms in an A.P., then the rth term from the end is [(n – r) + 1]th term from the beginning of the given A.P. |
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3. How many 3-digits numbers are divisible by 7?
Solution: The first 3-digit number divisible by 7 is 105 and the last 3-digit number is 994
Let the nth term be 994, the first term is 105 and common difference is 7
Tn = a + (n – 1)d = 994
⇒ 105 + 7n – 7 = 994
⇒ 7n = 994 -98 ⇒ n = 128
Therefore, there are 128 3-digit numbers which are divisible by 7.
4. The first, second and the last term of an AP are a, b and c, respectively. Prove that the sum of the AP is [(a + c)(b + c – 2a)/2(b – a)].
Solution: Let d be the common difference of the given AP and there are n terms. Then d = b – a
Last term = Tn = c = a + (n – 1)d
⇒ c = a + (n – 1)(b – a) ⇒ n = (b + c – 2a)/(b – a)
Now, sum of n terms of the AP = n/2 ( first term + last term) = [(b + c – 2a)/2(b – a)] × (a + c)
⇒ Sn = [(a + c)(b + c – 2a)/2(b – a)]
Tn = a + (n – 1)d
Sn = n/2[a + l] |
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5. Find the sum of all odd integers from 1 to 1001.
Solution: Clearly, this will form an AP with first term a = 1, common difference d = 2 and last term l = 1001
Let there be n odd numbers from 1 to 1001. Then,
Tn = a + (n -1)d = 1 + (n – 1) 2 = 1001
⇒ -1 + 2n = 1001
⇒ 2n = 1002
⇒ n = 501
Therefore, sum of 501 terms is
S501 = 501/2[1 + 1001] = 251001
A1 = [a + {(b – a)/(n + 1)}], A2 = [a + {2(b – a)/(n + 1)}], …, An = [a + {n(b – a)/(n + 1)}] |
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6. Insert six arithmetic means between 15 and -13.
Solution: Let A1, A2, A3, A4, A5, A6 are six AM between 15 and -13, such that 15, A1, A2, A3, A4, A5, A6, -13 are in AP
Common difference d = (b – a)/(n + 1) = (-13 – 15)/(6 + 1) = -4
Then A1 = 15 + d = 15 – 4 = 11
A2 = 15 + 2d = 15 – 8 = 7
A3 = 15 + 3d = 15 – 12 = 3
A4 = 15 + 4d = 15 – 16 = -1
A5 = 15 + 5d = 15 – 20 = -5
A6 = 15 + 6d = 15 – 24 = -9
Hence, the required six AMs between 15 and -13 are
11, 7, 3, -1, -5 and -9
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7. The sum of some terms of a GP is 315. Its first term is 5 and the common ratio is 2. Find the number of its terms and the last term.
Solution: Let the sum of n terms of the GP be 315. Then,
Sn = {a(rn – 1)/(r – 1)} (since r = 2 > 1)
⇒ 315 = {5(2n – 1)/(2 -1)}
⇒ (2n – 1) = 63 ⇒ 2n = 64
⇒ n = 6
Last term = arn – 1 = 5.26 – 1 = 5 × 32 = 160
Hence, the given GP contains 6 terms and its last term is 160
Also Read:
8. Find the common ratio of a GP whose sum of infinite terms is 8 and its second term is 2.
Solution: Let a be the first term of the GP. Then,
ar = 2 ⇒ r = a/2, then,
a/(1 – r) = 8 ⇒ a2/(a – 2)
⇒ a2 – 8a + 16 = 0
⇒ ( a- 4)2 = 0
⇒ a = 4
Therefore, r = 2/a = ½
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9. If A is the AM and G1, G2 be the two GM between any two numbers then prove that
2A = (G12/G2) + (G22/G1).
Solution: Let the given numbers be a and b. Then,
A = (a + b)/2 ⇒ a + b = 2A.
And a, G1, G2, b are in GP.
Therefore, G1/a = G2/G1 = b/G2
⇒ a = G12/G2 and b = G22/G1
⇒ a + b = (G12/G2 ) + (G22/G1)
⇒ 2A = (G12/G2 ) + (G22/G1).
10. Find the sum to n terms of the series whose nth term is n(n + 3).
Solution: We have Tk = k(k + 3) = k2 + 3k
∴ sum of n terms is given by
Practice Questions
1. Insert two geometric means between 9 and 243.
2. If p, q, r are in AP then prove that pth, qth and rth term of any GP are in GP.
3. The first term of a GP is 27 and its 8th term is 1/81. Find the sum of the first 10 terms of the GP.
4. Find the sum of the infinite geometric series (√2 + 1) + 1 + ( √2 – 1) + …
5. Find the sum of integers from 1 to 100 divisible by 2 or 5.
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