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Mathematics, (www.tiwariacademy.com : A step towards free education), , (Chapter – 9) (Sequences and Series), (Class – XI), , Exercise 9.2, Question 1:, Find the sum of odd integers from 1 to 2001., Answer 1:, The odd integers from 1 to 2001 are 1, 3, 5 …1999, 2001., This sequence forms an A.P., Here, first term, a = 1, Common difference, d = 2, , Thus, the sum of odd numbers from 1 to 2001 is 1002001., Question 2:, Find the sum of all natural numbers lying between 100 and 1000, which are, multiples of 5., , 1, www.tiwariacademy.com

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Mathematics, (www.tiwariacademy.com : A step towards free education), , (Chapter – 9) (Sequences and Series), (Class – XI), , Answer 2:, The natural numbers lying between 100 and 1000, which are multiples of 5,, are 105, 110, … 995., , Thus, the sum of all natural numbers lying between 100 and 1000, which, are multiples of 5, is 98450., Question 3:, In an A.P, the first term is 2 and the sum of the first five terms is one-fourth, of the next five terms. Show that 20th term is –112., , Answer 3:, First term = 2, Let d be the common difference of the A.P., , 2, www.tiwariacademy.com

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Mathematics, (www.tiwariacademy.com : A step towards free education), , (Chapter – 9) (Sequences and Series), (Class – XI), , Therefore, the A.P. is 2, 2 + d, 2 + 2d, 2 + 3d …, Sum of first five terms = 10 + 10d, Sum of next five terms = 10 + 35d, According to the given condition,, , Thus, the 20th term of the A.P. is –112., Question 4:, How many terms of the A.P., , are needed to give the sum –25?, , Answer 4:, Let the sum of n terms of the given A.P. be –25., It is known that,, , ,, Where n = number of terms, a = first term, and d = common difference, Here, a = –6, , Therefore, we obtain, , 3, www.tiwariacademy.com

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Mathematics, (www.tiwariacademy.com : A step towards free education), , (Chapter – 9) (Sequences and Series), (Class – XI), , Question 5:, In an A.P., if pth term is 1/q and qth term is 1/p, prove that the sum of first, pq terms is ½ (pq +1), where p ≠ q., Answer 5:, It is known that the general term of an A.P. is an = a + (n –, 1)d, , According to the given information,, , Subtracting (2) from (1), we obtain, , 4, www.tiwariacademy.com

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Mathematics, (www.tiwariacademy.com : A step towards free education), , (Chapter – 9) (Sequences and Series), (Class – XI), , Putting the value of d in (1), we obtain, , Thus, the sum of first pq terms of the A.P. is, , ., , Question 6:, If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116., Find the last term, , 5, www.tiwariacademy.com

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Mathematics, (www.tiwariacademy.com : A step towards free education), , (Chapter – 9) (Sequences and Series), (Class – XI), , Answer 6:, Let the sum of n terms of the given A.P. be 116., , Here, a = 25 and d = 22 – 25 = – 3, , However,, n cannot be equal to, , . Therefore, n = 8, , a8 = Last term = a + (n – 1)d = 25 + (8 – 1) (– 3), = 25 + (7) (– 3) = 25 – 21, =4, Thus, the last term of the A.P. is 4., , Question 7:, Find the sum to n terms of the A.P., whose kth term is 5k + 1., , 6, www.tiwariacademy.com

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Mathematics, (www.tiwariacademy.com : A step towards free education), , (Chapter – 9) (Sequences and Series), (Class – XI), , Answer 7:, It is given that the kth term of the A.P. is 5k + 1., kth term = ak = a + (k – 1)d, a + (k – 1)d = 5k + 1 a + kd – d = 5k + 1, Comparing the coefficient of k, we obtain d = 5 a – d = 1, ⇒a–5=1, ⇒a=6, , Question 8:, If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants,, find the common difference., Answer 8:, It is known that:, According to the given condition,, , 7, www.tiwariacademy.com

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Mathematics, (www.tiwariacademy.com : A step towards free education), , (Chapter – 9) (Sequences and Series), (Class – XI), , Comparing the coefficients of n2 on both sides, we obtain, , ∴d=2q, Thus, the common difference of the A.P. is 2q., , Question 9:, The sums of n terms of two arithmetic progressions are in the ratio 5n + 4:, 9n + 6. Find the ratio of their 18th terms., , Answer 9:, Let a1, a2, and d1, d2 be the first terms and the common difference of the first, and second arithmetic progression respectively., According to the given condition,, , 8, www.tiwariacademy.com

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Mathematics, (www.tiwariacademy.com : A step towards free education), , (Chapter – 9) (Sequences and Series), (Class – XI), , Substituting n = 35 in (1), we obtain, , From (2) and (3), we obtain, , Thus, the ratio of 18th term of both the A.P.s is 179: 321., Question 10:, If the sum of first p terms of an A.P. is equal to the sum of the first q terms,, then find the sum of the first (p + q) terms., Answer 10:, Let a and d be the first term and the common difference of the A.P., respectively., , 9, www.tiwariacademy.com

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Mathematics, (www.tiwariacademy.com : A step towards free education), , (Chapter – 9) (Sequences and Series), (Class – XI), , Here,, , According to the given condition,, , Thus, the sum of the first (p + q) terms of the A.P. is 0., , 10, www.tiwariacademy.com

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Mathematics, (www.tiwariacademy.com : A step towards free education), , (Chapter – 9) (Sequences and Series), (Class – XI), , Question 11:, Sum of the first p, q and r terms of an A.P. are a, b and c, respectively., Prove that, Answer 11:, Let a1 and d be the first term and the common difference of the A.P., respectively., According to the given information,, , Subtracting (2) from (1), we obtain, , 11, www.tiwariacademy.com

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Mathematics, (www.tiwariacademy.com : A step towards free education), , (Chapter – 9) (Sequences and Series), (Class – XI), , Subtracting (3) from (2), we obtain, , Equating both the values of d obtained in (4) and (5), we obtain, , 12, www.tiwariacademy.com

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Mathematics, (www.tiwariacademy.com : A step towards free education), , (Chapter – 9) (Sequences and Series), (Class – XI), , Dividing both sides by pqr, we obtain, , Thus, the given result is proved., Question 12:, The ratio of the sums of m and n terms of an A.P. is m2: n2. Show that the, ratio of mth and nth term is (2m – 1): (2n – 1)., , Answer 12:, Let a and b be the first term and the common difference of the A.P., respectively. According to the given condition,, , 13, www.tiwariacademy.com

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Mathematics, (www.tiwariacademy.com : A step towards free education), , (Chapter – 9) (Sequences and Series), (Class – XI), , Putting m = 2m – 1 and n = 2n – 1 in (1), we obtain, , From (2) and (3), we obtain, , Thus, the given result is proved., Question 13:, If the sum of n terms of an A.P. is 2n2 + 5n and its mth term is 164, find the, value of m., Answer 13:, Let a and b be the first term and the common difference of the A.P., respectively., , 14, www.tiwariacademy.com

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Mathematics, (www.tiwariacademy.com : A step towards free education), , (Chapter – 9) (Sequences and Series), (Class – XI), , am = a + (m – 1)d = 164 … (1), Sum of n terms:, Here,, , Comparing the coefficient of n2 on both sides, we obtain, , Comparing the coefficient of n on both sides, we obtain, , Therefore, from (1), we obtain, 8 + (m – 1) 6 = 164, ⇒ (m – 1) 6 = 164 – 8 = 156, ⇒ m – 1 = 26, ⇒ m = 27, Thus, the value of m is 27., Question 14:, Insert five numbers between 8 and 26 such that the resulting sequence is an, A.P., Answer 14:, Let A1, A2, A3, A4, and A5 be five numbers between 8 and 26, such that 8, A1, A2, A3, A4, A5, 26 is an A.P., , 15, www.tiwariacademy.com

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Mathematics, (www.tiwariacademy.com : A step towards free education), , (Chapter – 9) (Sequences and Series), (Class – XI), , Here, a = 8, b = 26, n = 7, Therefore, 26 = 8 + (7 – 1) d, ⇒ 6d = 26 – 8 = 18, ⇒d=3, A1 = a + d = 8 + 3 = 11, A2 = a + 2d = 8 + 2 × 3 = 8 + 6 = 14, A3 = a + 3d = 8 + 3 × 3 = 8 + 9 = 17, A4 = a + 4d = 8 + 4 × 3 = 8 + 12 = 20, A5 = a + 5d = 8 + 5 × 3 = 8 + 15 = 23, Thus, the required five numbers between 8 and 26 are 11, 14, 17, 20,, and 23., Question 15:, If, , is the A.M. between a and b, then find the value of n., , Answer 15:, A.M. of a and b, According to the given condition,, , 16, www.tiwariacademy.com

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Mathematics, (www.tiwariacademy.com : A step towards free education), , (Chapter – 9) (Sequences and Series), (Class – XI), , Question 16:, Between 1 and 31, m numbers have been inserted in such a way that the, resulting sequence is an A.P. and the ratio of 7th and (m – 1)th numbers is 5:9., Find the value of m., Answer 16:, Let A1, A2, … Am be m numbers such that 1, A1, A2, … Am, 31 is an A.P., Here, a = 1, b = 31, n = m + 2, ∴ 31 = 1 + (m + 2 – 1) (d), ⇒ 30 = (m + 1) d, , A1 = a + d, A2 = a + 2d, A3 = a + 3d …, ∴ A7 = a + 7d, , 17, www.tiwariacademy.com

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Mathematics, (www.tiwariacademy.com : A step towards free education), , (Chapter – 9) (Sequences and Series), (Class – XI), , Am–1 = a + (m – 1) d, According to the given condition,, , Thus, the value of m is 14., Question 17:, A man starts repaying a loan as first installment of Rs. 100. If he increases, the installment by Rs 5 every month, what amount he will pay in the 30 th, installment?, , Answer 17:, The first installment of the loan is Rs 100., The second installment of the loan is Rs 105 and so on., The amount that the man repays every month forms an A.P., The A.P. is 100, 105, 110 …, , 18, www.tiwariacademy.com

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Mathematics, (www.tiwariacademy.com : A step towards free education), , (Chapter – 9) (Sequences and Series), (Class – XI), , First term, a = 100, Common difference, d = 5, A30 = a + (30 – 1)d, = 100 + (29) (5), = 100 + 145, = 245, Thus, the amount to be paid in the 30th installment is Rs 245., Question 18:, The difference between any two consecutive interior angles of a polygon is, 5°. If the smallest angle is 120°, find the number of the sides of the polygon., Answer 18:, The angles of the polygon will form an A.P. with common difference d as 5°, and first term a as 120°., It is known that the sum of all angles of a polygon with n sides is 180°(n – 2)., , 19, www.tiwariacademy.com