# Class 11 Mathematics Chapter 9: Sequence and Series - NCERT Solutions

This page focuses on the complete NCERT solutions for Class 11 Mathematics Chapter 9: Sequence and Series. This section provides detailed, easy-to-understand solutions for all the questions from this chapter. These Sequence and Series question answers will offer you valuable insights and explanations.

Sequence is a collection of objects taken one by one in which repetitions may present but order matters. It can have any number of terms. When these terms are added it is called a series. We get some general expressions to solve sequences and series. It solves very tedious calculations which are very difficult to solve manually. This chapter consists of arithmetic progression and mean, geometric progression and mean, general terms, sum of n terms, arithmetic and geometric series, infinite G.P, relation between A.M and G.M, some special series.

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### Exercise 1

•  Q1 Write the first five terms of the sequences whose nth term is an=n(n+2) Ans: an = n(n+2) Substituting n = 1, 2, 3, 4, and 5, we obtain  a1 = 1(1+2) = 3  a2 = 2(2+2) = 8  a3 = 3(3+2) = 15  a4 = 4(4+2) = 24  a5 = 5(5+2) = 35 Therefore, the required terms are 3, 8, 15, 24, and 35. Q2 Write the first five terms of the sequences whose nth term is \begin{align} a_n = \frac {n}{n+1}\end{align} Ans: \begin{align} a_n = \frac {n}{n+1}\end{align} Substituting n = 1, 2, 3, 4, 5, we obtain \begin{align} a_1 = \frac {1}{1+1}=\frac{1}{2},a_2 = \frac {2}{2+1}=\frac{2}{3},a_3 = \frac {3}{3+1}=\frac{3}{4},a_4 = \frac {4}{4+1}=\frac{4}{5},a_5 = \frac {5}{5+1}=\frac{5}{6}\end{align} Therefore, the required terms are \begin{align} \frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},and \frac{5}{6}\end{align}

### Exercise 2

•  Q1 Find the sum of odd integers from 1 to 2001. Ans: 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   Here,   \begin{align} a + (n - 1)d = 2001 \end{align}   \begin{align} => 1 + (n - 1)(2) = 2001 \end{align}   \begin{align} => 2n -2 = 2000 \end{align}   \begin{align} => n = 1001 \end{align}   \begin{align} S_n = \frac {n}{2}\left[2a + (n -1)d\right]\end{align}   \begin{align} \therefore S_n = \frac {1001}{2}\left[2 × 1 + (1001 -1)×2\right]\end{align}   \begin{align} = \frac {1001}{2}\left[2 + 1000×2\right]\end{align}   \begin{align} = \frac {1001}{2} × 2002\end{align}   \begin{align} =1001 × 1001 \end{align}   \begin{align} = 1002001 \end{align}   Thus, the sum of odd numbers from 1 to 2001 is 1002001. Q2 Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5. Ans: The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995. This sequence forms an A.P.   Here, first term, a = 105   Common difference, d = 5   Here,   \begin{align} a + (n - 1)d = 995 \end{align}   \begin{align} => 105 + (n - 1)5 = 995 \end{align}   \begin{align} =>  (n - 1)5 = 995 - 105 = 890 \end{align}   \begin{align} => n -1 = 178 \end{align}   \begin{align} => n = 179 \end{align}   \begin{align} S_n = \frac {n}{2}\left[2a + (n -1)d\right]\end{align}   \begin{align} \therefore S_n = \frac {179}{2}\left[2 × (105) + (179 -1)×(5)\right]\end{align}   \begin{align} = \frac {179}{2}\left[2(105) + (178)(5)\right]\end{align}   \begin{align} = 179\left[105 + (89)5\right]\end{align}   \begin{align} = (179)\left[105 + 445\right]\end{align}   \begin{align} =179 × 550 \end{align}   \begin{align} = 98450 \end{align}   Thus, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450. Q3 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. Ans: First term = 2 Let d be the common difference of the A.P. 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, \begin{align} => 10 + 10d = \frac{1}{4}(10 + 35d) \end{align} \begin{align} => 40 + 40d = 10 + 35d \end{align} \begin{align} => 30 = - 5d \end{align} \begin{align} => d = - 6 \end{align} \begin{align} \therefore a_{20} = a + (20 -1)d = 2 + (19)(-6) = 2 - 114 = -112\end{align} Thus, the 20th term of the A.P. is –112. Q4 How many terms of the A.P. \begin{align} -6, -\frac{11}{2}, -5, ... \end{align} are needed to give the sum –25? Ans: Let the sum of n terms of the given A.P. be –25. It is known that, \begin{align} S_n = \frac {n}{2}\left[2a + (n -1)d\right]\end{align}, where n = number of terms, a = first term, and d = common difference Here, a = –6 \begin{align} d = -\frac{11}{2} + 6 = \frac{-11 + 12}{2} = \frac{1}{2}\end{align} Therefore, we obtain \begin{align} -25 = \frac {n}{2}\left[2 × (-6) + (n -1)×\frac{1}{2}\right]\end{align} \begin{align} => -50 = n\left[-12  + \frac{n}{2} - \frac{1}{2}\right]\end{align}   \begin{align} => -50 = n\left[-\frac{25}{2} + \frac{n}{2}\right]\end{align}   \begin{align} => -100 = n\left(-25 + n\right)\end{align}   \begin{align} => n^2 - 25n + 100 = 0\end{align}   \begin{align} => n^2 - 5n -20n + 100 = 0\end{align}   \begin{align} => n(n - 5)- 20(n - 5) = 0\end{align}   \begin{align} => n = 20 \; or\; 5\end{align} Q5 In an A.P., if pth term is \begin{align} \frac{1}{q} \; and \;qth\; term \; is\; \frac{1}{p}\end{align} , prove that the sum of first pq terms is \begin{align} \frac{1}{2}(pq+1) \; where \; p ≠ q \end{align} Ans: It is known that the general term of an A.P. is an = a + (n – 1)d ∴ According to the given information, \begin{align}p^{th} \; term= a_p=a+(p-1)d=\frac{1}{q}  \;\; ...(1)\end{align} \begin{align}q^{th} \; term= a_q=a+(q-1)d=\frac{1}{p}  \;\; ...(2)\end{align} Subtracting (2) from (1), we obtain \begin{align} (p-1)d - (q-1)d=\frac{1}{q}-\frac{1}{p} \end{align} \begin{align} ⇒(p-1-q+1)d = \frac{p-q}{pq} \end{align} \begin{align} ⇒(p-q)d = \frac{p-q}{pq} \end{align} \begin{align} ⇒d = \frac{1}{pq} \end{align} Putting the value of d in (1), we obtain \begin{align} a + (p-1)\frac{1}{pq}= \frac{1}{q}\end{align} \begin{align} ⇒a = \frac{1}{q}-\frac{1}{q}+\frac{1}{pq}=\frac{1}{pq} \end{align} \begin{align} \therefore S_{pq} = \frac{pq}{2}\left[2a + (pq-1)d\right]\end{align} \begin{align}  = \frac{pq}{2}\left[\frac{2}{pq} + (pq-1)\frac{1}{pq}\right]\end{align} \begin{align}  = 1 + \frac{1}{2}(pq-1)\end{align} \begin{align}  =\frac{1}{2}pq + 1 - \frac{1}{2}=\frac{1}{2}pq + \frac{1}{2}\end{align} \begin{align}  =\frac{1}{2}(pq+1) \end{align} Thus, the sum of first pq terms of the A.P. is \begin{align}  =\frac{1}{2}(pq+1). \end{align} Q6 If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term Ans: Let the sum of n terms of the given A.P. be 116. \begin{align}  S_n=\frac{n}{2}\left[2a + (n-1)d\right] \end{align} Here, a = 25 and d = 22 – 25 = – 3 \begin{align}  \therefore S_n=\frac{n}{2}\left[2 × 25 + (n-1)(-3)\right] \end{align} \begin{align}  ⇒ 116=\frac{n}{2}\left[50 -3n +3\right] \end{align} \begin{align}  ⇒ 232=n(53-3n)=53n -3n^2 \end{align} \begin{align}  ⇒ 3n^2 -53n + 232 =0\end{align} \begin{align}  ⇒ 3n^2 -24n -29n + 232 =0\end{align} \begin{align}  ⇒ 3n(n-8) -29(n-8) =0\end{align} \begin{align}  ⇒ (n-8)(3n-29) =0\end{align} \begin{align}  ⇒ n=8 \;or\;n=\frac{29}{3} \end{align} However, n cannot be equal to \begin{align}  \frac{29}{3}. \end{align} Therefore, n = 8 \begin{align}  \therefore a_8 = Last \; Term = a + (n-1)d= 25 + (8-1)(-3) \end{align} \begin{align}  = 25 + (7)(-3)=25-21 \end{align} \begin{align}  = 4 \end{align} Thus, the last term of the A.P. is 4. Q7 Find the sum to n terms of the A.P., whose kth term is 5k + 1. Ans: 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 \begin{align}   S_n = \frac{n}{2}\left[2a + (n-1)d\right] \end{align} \begin{align}   = \frac{n}{2}\left[2(6) + (n-1)(5)\right] \end{align} \begin{align}   = \frac{n}{2}\left[12 + 5n -5\right] \end{align} \begin{align}   = \frac{n}{2}\left(5n + 7\right) \end{align} Q8 If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants, find the common difference Ans: It is known that, \begin{align}   S_n= \frac{n}{2}\left[2a + (n-1)d\right] \end{align} According to the given condition, \begin{align}    \frac{n}{2}\left[2a + (n-1)d\right] = pn + qn^2 \end{align} \begin{align}   ⇒ \frac{n}{2}\left[2a + nd-d\right] = pn + qn^2 \end{align} \begin{align}   ⇒ na + n^2\frac{d}{2} - n.\frac{d}{2}= pn + qn^2 \end{align} Comparing the coefficients of n2 on both sides, we obtain \begin{align}   \frac{d}{2} = q \end{align} \begin{align}  \therefore d = 2q \end{align} Thus, the common difference of the A.P. is 2q. Q9 The sums of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n + 6. Find the ratio of their 18th terms. Ans: 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,   \begin{align}  \frac{Sum \;of \;n \;terms \;of \;first\; A.P.}{Sum\; of \;n\; terms \;of \;second \;A.P.} = \frac{5n+4}{9n+6} \end{align} \begin{align}  ⇒\frac{\frac{n}{2}\left[2a_1 + (n-1)d_1\right]}{\frac{n}{2}\left[2a_2 + (n-1)d_2\right]} = \frac{5n+4}{9n+6} \end{align} \begin{align}  ⇒\frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{5n+4}{9n+6} \;\;\;\;...(1)\end{align} Substituting n = 35 in (1), we obtain \begin{align}  ⇒\frac{2a_1 + 34d_1}{2a_2 + 34d_2} = \frac{5(35)+4}{9(35)+6} \end{align} \begin{align}  ⇒\frac{a_1 + 17d_1}{a_2 + 17d_2} = \frac{179}{321} \;\;\;\;...(2)\end{align} \begin{align}  \frac{18^{th} \;term \;of\; first\; A.P.}{18^{th} \;term \;of\; second\; A.P.}=\frac{a_1 + 17d_1}{a_2 + 17d_2}  \;\;\;\;...(3)\end{align} From (2) and (3), we obtain \begin{align}  \frac{18^{th} \;term \;of\; first\; A.P.}{18^{th} \;term \;of\; second\; A.P.}=\frac{179}{321}\end{align} Thus, the ratio of 18th term of both the A.P.s is 179: 321. Q10 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. Ans: Let a and d be the first term and the common difference of the A.P. respectively. Here, \begin{align} S_p = \frac{p}{2}\left[2a+(p-1)d\right]\end{align} \begin{align} S_q = \frac{q}{2}\left[2a+(q-1)d\right]\end{align} According to the given condition, \begin{align} \frac{p}{2}\left[2a+(p-1)d\right]=\frac{q}{2}\left[2a+(q-1)d\right]\end{align} \begin{align} ⇒p\left[2a+(p-1)d\right]=q\left[2a+(q-1)d\right]\end{align} \begin{align} ⇒2ap + pd(p-1)=2aq+qd(q-1)\end{align} \begin{align} ⇒2a(p-q) +d[p(p-1)-q(q-1)]=0\end{align} \begin{align} ⇒2a(p-q) +d[p^2 -p -q^2 +q]=0\end{align} \begin{align} ⇒2a(p-q) +d[(p-q)(p+q)-(p-q)]=0\end{align} \begin{align} ⇒2a(p-q) +d[(p-q)(p+q-1)]=0\end{align} \begin{align} ⇒2a +d(p+q-1)=0\end{align} \begin{align} ⇒d=\frac{-2a}{p+q-1} \;\;\;\;...(1)\end{align} \begin{align} \therefore S_{p+q}=\frac{p+q}{2}\left[2a+(p+q-1).d\right] \end{align} \begin{align} ⇒ S_{p+q}=\frac{p+q}{2}\left[2a+(p+q-1).\left(\frac{-2a}{p+q-1}\right)\right] \;\;\;\; [From (1)] \end{align} \begin{align}=\frac{p+q}{2}\left[2a-2a\right]\end{align} \begin{align}=0\end{align} Thus, the sum of the first (p + q) terms of the A.P. is 0. Q11 Sum of the first p, q and r terms of an A.P. are a, b and c, respectively. Prove that Ans: Let a1 and d be the first term and the common difference of the A.P. respectively. According to the given information, Q12 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). Ans: Let a and b be the first term and the common difference of the A.P. respectively. According to the given condition, Q13 If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m. Ans: Let a and b be the first term and the common difference of the A.P. respectively. 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. Q14 Insert five numbers between 8 and 26 such that the resulting sequence is an A.P. Ans: 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. 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. Q15 If is the A.M. between a and b, then find the value of n. Ans: A.M. of a and b =  According to the given condition, Q16 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)thnumbers is 5:9. Find the value of m. Ans: 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 Am–1 = a + (m – 1) d According to the given condition, Thus, the value of m is 14. Q17 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 30th installment? Ans: 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, … 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. Q18 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. Ans: 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). So

### Exercise 3

•  Q1 Find the 20th and nthterms of the G.P. Ans: Here, a = First Term =  r = common ratio = Q2 Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2. Ans: Common ratio, r = 2 Let a be the first term of the G.P. ∴ a8 = ar 8–1 = ar7 ⇒ ar7 = 192 a(2)7 = 192 a(2)7 = (2)6 (3) Q3 The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps. Ans: Let a be the first term and r be the common ratio of the G.P. According to the given condition, a5 = a r5–1 = a r4 = p … (1) a8 = a r8–1 = a r7 = q … (2) a11 = a r11–1 = a r10 = s … (3) Dividing equation (2) by (1), we obtain Dividing equation (3) by (2), we obtain Equating the values of r3 obtained in (4) and (5), we obtain Thus, the given result is proved. Q4 The 4th term of a G.P. is square of its second term, and the first term is –3. Determine its 7th term. Ans: Let a be the first term and r be the common ratio of the G.P. ∴ a = –3 It is known that, an = arn–1 ∴a4 = ar3 = (–3) r3 a2 = a r1 = (–3) r According to the given condition, (–3) r3 = [­(–3) r]2 ⇒ –3r3 = 9 r2 ⇒ r = –3 a7 = a r 7–1 = a r6 = (–3) (–3)6 = – (3)7 = –2187 Thus, the seventh term of the G.P. is –2187. Q5 Which term of the following sequences: Ans: (a) The given sequence is  Here, a = 2 and r =  Let the nth term of the given sequence be 128. Thus, the 13th term of the given sequence is 128. (b) The given sequence is  Here, a=  Let the nth term of the given sequence be 729. Thus, the 12th term of the given sequence is 729. (c) The given sequence is  Here,  a= Let the nth term of the given sequence be . Thus, the 9th term of the given sequence is . Q6 For what values of x, the numbers are in G.P? Ans: The given numbers are . Common ratio =  Also, common ratio =  so Thus, for x = ± 1, the given numbers will be in G.P. Q7 Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015 … Ans: The given G.P. is 0.15, 0.015, 0.00015, … Here, a = 0.15 and r = Q8 Find the sum to n terms in the geometric progression Ans: The given G.P. is  Here, a = Q9 Find the sum to n terms in the geometric progression 1,-a, a2,-a3, ... (if a ≠ -1) Ans: The given G.P. is 1,-a, a2,-a3, ......... Here, first term = a1 = 1 Common ratio = r = – a Q10 Find the sum to n terms in the geometric progression x3, x5, x7 ... (if x ≠ ±1) Ans: The given G.P. is  x3, x5, x7 ........ Here, a = x3 and r = x2 Q11 Evaluate Ans: Q12 The sum of first three terms of a G.P. is  and their product is 1. Find the common ratio and the terms. Ans: Let  be the first three terms of the G.P. From (2), we obtain a3 = 1 ⇒ a = 1 (Considering real roots only) Substituting a = 1 in equation (1), we obtain Q13 How many terms of G.P. 3, 32, 33, … are needed to give the sum 120? Ans: The given G.P. is 3, 32, 33, … Let n terms of this G.P. be required to obtain the sum as 120. Here, a = 3 and r = 3 ∴  ∴  n = 4 Thus, four terms of the given G.P. are required to obtain the sum as 120. Q14 The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P. Ans: Let the G.P. be a, ar, ar2, ar3, … According to the given condition, a + ar + ar2 = 16 and ar3 + ar4 + ar5 = 128 ⇒ a (1 + r + r2) = 16 … (1) ar3(1 + r + r2) = 128 … (2) Dividing equation (2) by (1), we obtain Substituting r = 2 in (1), we obtain a (1 + 2 + 4) = 16 ⇒ a (7) = 16 Q15 Given a G.P. with a = 729 and 7th term 64, determine S7. Ans: a = 729 a7 = 64 Let r be the common ratio of the G.P. It is known that, an = a rn–1 a7 = ar7–1 = (729)r6 ⇒ 64 = 729 r6 Also, it is known that, Q16 Find a G.P. for which sum of the first two terms is –4 and the fifth term is 4 times the third term. Ans: Let a be the first term and r be the common ratio of the G.P. According to the given conditions, From (1), we obtain ..... Q17 If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P. Ans: Let a be the first term and r be the common ratio of the G.P. According to the given condition, a4 = a r3 = x … (1) a10 = a r9 = y … (2) a16 = a r15 = z … (3) Dividing (2) by (1), we obtain Dividing (3) by (2), we obtain ∴   Thus, x, y, z are in G. P. Q18 Find the sum to n terms of the sequence, 8, 88, 888, 8888… Ans: The given sequence is 8, 88, 888, 8888… This sequence is not a G.P. However, it can be changed to G.P. by writing the terms as Sn = 8 + 88 + 888 + 8888 + …………….. to n terms Q19 Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, . Ans: Required sum = 2 × 128 + 4 × 32 + 8 × 8 + 16 × 2 + 32 × Here, 4, 2, 1,  is a G.P. First term, a = 4 Common ratio, r = It is known that, Q20 Show that the products of the corresponding terms of the sequences a,ar,ar2, ...arn-1 and A, AR, AR2, ,,,ARn-1 form a G.P, and find the common ratio. Ans: It has to be proved that the sequence, aA, arAR, ar2AR2, …arn–1ARn–1, forms a G.P. Thus, the above sequence forms a G.P. and the common ratio is rR. Q21 Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4th by 18. Ans: Let a be the first term and r be the common ratio of the G.P. a1 = a, a2 = ar, a3 = ar2, a4 = ar3 By the given condition, a3 = a1 + 9 ⇒ ar2 = a + 9 … (1) a2 = a4 + 18 ⇒ ar = ar3 + 18 … (2) From (1) and (2), we obtain a(r2 ­­– 1) = 9 … (3) ar (1– r2) = 18 … (4) Dividing (4) by (3), we obtain Substituting the value of r in (1), we obtain 4a = a + 9 ⇒ 3a = 9 ∴ a = 3 Thus, the first four numbers of the G.P. are 3, 3(– 2), 3(–2)2, and 3­(–2)3 i.e., 3¸–6, 12, and –24. Q22 If the pth, qth and rth terms of a G.P. are a, b and c, respectively. Prove that aq-rbr-pcp-q=1 Ans: Let A be the first term and R be the common ratio of the G.P. According to the given information, ARp–1 = a ARq–1 = b ARr–1 = c aq–r br–p cp–q = Aq–r × R(p–1) (q–r) × Ar–p × R(q–1) (r-p) × Ap–q × R(r –1)(p–q) = Aq­ – r + r – p + p – q × R (pr – pr – q + r) + (rq – r + p – pq) + (pr – p – qr + q) = A0 × R0 = 1 Thus, the given result is proved. Q23 If the first and the nth term of a G.P. are a ad b, respectively, and if P is the product of n terms, prove that P2 = (ab)n. Ans: The first term of the G.P is a and the last term is b. Therefore, the G.P. is a, ar, ar2, ar3, … arn–1, where r is the common ratio. b = arn–1 … (1) P = Product of n terms = (a) (ar) (ar2) … (arn–1) = (a × a ×…a) (r × r2 × …rn–1) = an r 1 + 2 +…(n–1) … (2) Here, 1, 2, …(n – 1) is an A.P. ∴1 + 2 + ……….+ (n – 1) Thus, the given result is proved. Q24 Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from Ans: Let a be the first term and r be the common ratio of the G.P. Since there are n terms from (n +1)th to (2n)th term, Sum of terms from(n + 1)th to (2n)th term = a n +1 = ar n + 1 – 1 = arn Thus, required ratio =  Thus, the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th term is . Q25 If a, b, c and d are in G.P. show that Ans: a, b, c, d are in G.P. Therefore, bc = ad … (1) b2 = ac … (2) c2 = bd … (3) It has to be proved that, (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc – cd)2 R.H.S. = (ab + bc + cd)2 = (ab + ad + cd)2 [Using (1)] = [ab + d (a + c)]2 = a2b2 + 2abd (a + c) + d2 (a + c)2 = a2b2 +2a2bd + 2acbd + d2(a2 + 2ac + c2) = a2b2 + 2a2c2 + 2b2c2 + d2a2 + 2d2b2 + d2c2 [Using (1) and (2)] = a2b2 + a2c2 + a2c2 + b2c2 + b2c2 + d2a2 + d2b2 + d2b2 + d2c2 = a2b2 + a2c2 + a2d2 + b2 × b2 + b2c2 + b2d2 + c2b2 + c2 × c2 + c2d2 [Using (2) and (3) and rearranging terms] = a2(b2 + c2 + d2) + b2 (b2 + c2 + d2) + c2 (b2+ c2 + d2) = (a2 + b2 + c2) (b2 + c2 + d2) = L.H.S. ∴ L.H.S. = R.H.S. ∴ Q26 Insert two numbers between 3 and 81 so that the resulting sequence is G.P. Ans: Let G1 and G2 be two numbers between 3 and 81 such that the series, 3, G1, G2, 81, forms a G.P. Let a be the first term and r be the common ratio of the G.P. ∴81 = (3) (r)3 ⇒ r3 = 27 ∴ r = 3 (Taking real roots only) For r = 3, G1 = ar = (3) (3) = 9 G2 = ar2 = (3) (3)2 = 27 Thus, the required two numbers are 9 and 27. Q27 Find the value of n so that  may be the geometric mean between a and b. Ans: G. M. of a and b is . By the given condition,  Squaring both sides, we obtain Q28 The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio Ans: Let the two numbers be a and b. G.M. =  According to the given condition, Also, Adding (1) and (2), we obtain Substituting the value of a in (1), we obtain Thus, the required ratio is . Q29 If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are . Ans: It is given that A and G are A.M. and G.M. between two positive numbers. Let these two positive numbers be a and b. ∴  From (1) and (2), we obtain a + b = 2A … (3) ab = G2 … (4) Substituting the value of a and b from (3) and (4) in the identity (a – b)2 = (a + b)2 – 4ab, we obtain (a – b)2 = 4A2 – 4G2 = 4 (A2–G2) (a – b)2 = 4 (A + G) (A – G) From (3) and (5), we obtain Substituting the value of a in (3), we obtain Thus, the two numbers are . Q31 What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually? Ans: The amount deposited in the bank is Rs 500. At the end of first year, amount =  = Rs 500 (1.1) At the end of 2nd year, amount = Rs 500 (1.1) (1.1) At the end of 3rd year, amount = Rs 500 (1.1) (1.1) (1.1) and so on ∴Amount at the end of 10 years = Rs 500 (1.1) (1.1) … (10 times) = Rs 500(1.1)10 Q32 If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation. Ans: Let the root of the quadratic equation be a and b. According to the given condition, The quadratic equation is given by, x2– x (Sum of roots) + (Product of roots) = 0 x2 – x (a + b) + (ab) = 0 x2 – 16x + 25 = 0 [Using (1) and (2)] Thus, the required quadratic equation is x2 – 16x + 25 = 0

### Exercise 4

•  Q1 Find the sum to n terms of the series 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + … Ans: The given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + … nth term, an = n ( n + 1) ∴ Q2 Find the sum to n terms of the series 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + … Ans: The given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + … nth term, an = n ( n + 1) ( n + 2) = (n2 + n) (n + 2) = n3 + 3n2 + 2n ∴ Q3 Find the sum to n terms of the series 3 × 12 + 5 × 22 + 7 × 32 + … Ans: The given series is 3 ×12 + 5 × 22 + 7 × 32 + … nth term, an = ( 2n + 1) n2 = 2n3 + n2 Q4 Find the sum to n terms of the series Ans: The given series is  nth term, an =  Adding the above terms column wise, we obtain Q5 Find the sum to n terms of the series 52 + 62 + 72 + ... + 202 Ans: The given series is 52 + 62 + 72 + … + 202 nth term, an = ( n + 4)2 = n2 + 8n + 16 16th term is (16 + 4)2 = (20)2 Q6 Find the sum to n terms of the series 3 × 8 + 6 × 11 + 9 × 14 +… Ans: The given series is 3 × 8 + 6 × 11 + 9 × 14 + … an = (nth term of 3, 6, 9 …) × (nth term of 8, 11, 14, …) = (3n) (3n + 5) = 9n2 + 15n Q7 Find the sum to n terms of the series 12 + (12 + 22) + (12 + 22 + 32) + … Ans: The given series is 12 + (12 + 22) + (12 + 22 + 33 ) + … an = (12 + 22 + 33 +…….+ n2) Q8 Find the sum to n terms of the series whose nth term is given by n (n + 1) (n + 4). Ans: an = n (n + 1) (n + 4) = n(n2 + 5n + 4) = n3 + 5n2 + 4n Q9 Find the sum to n terms of the series whose nth terms is given by n2 + 2n Ans: an = n2 + 2n The above series 2, 22, 23, … is a G.P. with both the first term and common ratio equal to 2. Therefore, from (1) and (2), we obtain Q10 Find the sum to n terms of the series whose nth terms is given by (2n – 1)2 Ans: an = (2n – 1)2 = 4n2 – 4n + 1

### Exercise 5

•  Q1 Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term. Ans: Let a and d be the first term and the common difference of the A.P. respectively. It is known that the kth term of an A. P. is given by ak = a + (k –1) d ∴ am + n = a + (m + n –1) d am – n = a + (m – n –1) d am = a + (m –1) d ∴ am + n + am – n = a + (m + n –1) d + a + (m – n –1) d = 2a + (m + n –1 + m – n –1) d = 2a + (2m – 2) d = 2a + 2 (m – 1) d =2 [a + (m – 1) d] = 2am Thus, the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term. Q2 If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers. Ans: Let the three numbers in A.P. be a – d, a, and a + d. According to the given information, (a – d) + (a) + (a + d) = 24 … (1) ⇒ 3a = 24 ∴ a = 8 (a – d) a (a + d) = 440 … (2) ⇒ (8 – d) (8) (8 + d) = 440 ⇒ (8 – d) (8 + d) = 55 ⇒ 64 – d2 = 55 ⇒ d2 = 64 – 55 = 9 ⇒ d = ± 3 Therefore, when d = 3, the numbers are 5, 8, and 11 and when d = –3, the numbers are 11, 8, and 5. Thus, the three numbers are 5, 8, and 11. Q3 Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that S3 = 3 (S2– S1) Ans: Let a and b be the first term and the common difference of the A.P. respectively. Therefore, From (1) and (2), we obtain Hence, the given result is proved. Q4 Find the sum of all numbers between 200 and 400 which are divisible by 7. Ans: The numbers lying between 200 and 400, which are divisible by 7, are 203, 210, 217, ­­­­­­­­… 399 ∴First term, a = 203 Last term, l = 399 Common difference, d = 7 Let the number of terms of the A.P. be n. ∴ an = 399 = a + (n –1) d ⇒ 399 = 203 + (n –1) 7 ⇒ 7 (n –1) = 196 ⇒ n –1 = 28 ⇒ n = 29 Thus, the required sum is 8729. Q5 Find the sum of integers from 1 to 100 that are divisible by 2 or 5. Ans: The integers from 1 to 100, which are divisible by 2, are 2, 4, 6… 100. This forms an A.P. with both the first term and common difference equal to 2. ⇒100 = 2 + (n –1) 2 ⇒ n = 50 The integers from 1 to 100, which are divisible by 5, are 5, 10… 100. This forms an A.P. with both the first term and common difference equal to 5. ∴100 = 5 + (n –1) 5 ⇒ 5n = 100 ⇒ n = 20 The integers, which are divisible by both 2 and 5, are 10, 20, … 100. This also forms an A.P. with both the first term and common difference equal to 10. ∴100 = 10 + (n –1) (10) ⇒ 100 = 10n ⇒ n = 10 ∴Required sum = 2550 + 1050 – 550 = 3050 Thus, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050. Q6 Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder. Ans: The two-digit numbers, which when divided by 4, yield 1 as remainder, are 13, 17, … 97. This series forms an A.P. with first term 13 and common difference 4. Let n be the number of terms of the A.P. It is known that the nth term of an A.P. is given by, an = a + (n –1) d ∴97 = 13 + (n –1) (4) ⇒ 4 (n –1) = 84 ⇒ n – 1 = 21 ⇒ n = 22 Sum of n terms of an A.P. is given by, Thus, the required sum is 1210. Q7 If f is a function satisfying f(x +y) = f(x) f(y) for all x,y N  such that f(1) = 3  and  , find the value of n. Ans: It is given that, f (x + y) = f (x) × f (y) for all x, y ∈ N … (1) f (1) = 3 Taking x = y = 1 in (1), we obtain f (1 + 1) = f (2) = f (1) f (1) = 3 × 3 = 9 Similarly, f (1 + 1 + 1) = f (3) = f (1 + 2) = f (1) f (2) = 3 × 9 = 27 f (4) = f (1 + 3) = f (1) f (3) = 3 × 27 = 81 ∴ f (1), f (2), f (3), …, that is 3, 9, 27, …, forms a G.P. with both the first term and common ratio equal to 3. It is known that,  It is given that,  Thus, the value of n is 4. Q8 The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms. Ans: Let the sum of n terms of the G.P. be 315. It is known that,  It is given that the first term a is 5 and common ratio r is 2. ∴Last term of the G.P = 6th term = ar6 – 1 = (5)(2)5 = (5)(32) = 160 Thus, the last term of the G.P. is 160. Q9 The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P. Ans: Let a and r be the first term and the common ratio of the G.P. respectively. ∴ a = 1 a3 = ar2 = r2 a5 = ar4 = r4 ∴ r2 + r4 = 90 ⇒ r4 + r2 – 90 = 0 ∴ r = ± 3                                  ( Taking real roots) Thus, the common ratio of the G.P. is ±3. Q10 The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers. Ans: Let the three numbers in G.P. be a, ar, and ar2. From the given condition, a + ar + ar2 = 56 ⇒ a (1 + r + r2) = 56 a – 1, ar – 7, ar2 – 21 forms an A.P. ∴(ar – 7) – (a – 1) = (ar2 – 21) – (ar – 7) ⇒ ar – a – 6 = ar2 – ar – 14 ⇒ar2 – 2ar + a = 8 ⇒ar2 – ar – ar + a = 8 ⇒a(r2 + 1 – 2r) = 8 ⇒ a (r – 1)2 = 8 … (2) ⇒7(r2 – 2r + 1) = 1 + r + r2 ⇒7r2 – 14 r + 7 – 1 – r – r2 = 0 ⇒ 6r2 – 15r + 6 = 0 ⇒ 6r2 – 12r – 3r + 6 = 0 ⇒ 6r (r – 2) – 3 (r – 2) = 0 ⇒ (6r – 3) (r – 2) = 0 When r = 2, a = 8 When  Therefore, when r = 2, the three numbers in G.P. are 8, 16, and 32. When , the three numbers in G.P. are 32, 16, and 8. Thus, in either case, the three required numbers are 8, 16, and 32. Q11 A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio. Ans: Let the G.P. be T1, T2, T3, T4, … T2n. Number of terms = 2n According to the given condition, T1 + T2 + T3 + …+ T2n = 5 [T1 + T3 + … +T2n–1] ⇒ T1 + T2 + T3 + … + T2n – 5 [T1 + T3 + … + T2n–1] = 0 ⇒ T2 + T4 + … + T2n = 4 [T1 + T3 + … + T2n–1] Let the G.P. be a, ar, ar2, ar3, … Thus, the common ratio of the G.P. is 4. Q12 The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms. Ans: Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d. Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d Sum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d] = 4a + (4n – 10) d According to the given condition, 4a + 6d = 56 ⇒ 4(11) + 6d = 56 [Since a = 11 (given)] ⇒ 6d = 12 ⇒ d = 2 ∴ 4a + (4n –10) d = 112 ⇒ 4(11) + (4n – 10)2 = 112 ⇒ (4n – 10)2 = 68 ⇒ 4n – 10 = 34 ⇒ 4n = 44 ⇒ n = 11 Thus, the number of terms of the A.P. is 11. Q13 If Ans: It is given that, From (1) and (2), we obtain Thus, a, b, c, and d are in G.P. Q14 Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P2Rn = Sn Ans: Let the G.P. be a, ar, ar2, ar3, … arn – 1… According to the given information, Hence P2 Rn= Sn

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