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NCERT Exemplar · Class 11 Mathematics Sequence and Series

36 questions · 36 still being checked

EXERCISE 9.3 11–20 (part 2 of 4)

  1. Exercise 11

    Find the sum of the series (33−23)+(53−43)+(73−63)+…\left(3^3-2^3\right)+\left(5^3-4^3\right)+\left(7^3-6^3\right)+\ldots to
    (i)
    n\displaystyle n terms
    (ii)
    10\displaystyle 10 terms

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    NCERT’s answer
    (i)
    \(\displaystyle 4 n^3+9 n^2+6 n\)
    (ii)
    $\displaystyle 4960$
    The \(\displaystyle k\)th term is \(\displaystyle (2k+1)^3-(2k)^3\).\[(2k+1)^3-(2k)^3=12k^2+6k+1 \] \[S_n=\sum_{k=1}^{n}\left(12k^2+6k+1\right)=12\cdot\frac{n(n+1)(2n+1)}{6}+6\cdot\frac{n(n+1)}{2}+n \] \[S_n=2n(n+1)(2n+1)+3n(n+1)+n=4n^3+9n^2+6n \]For \(\displaystyle n=10\): \[S_{10}=4000+900+60=4960 \]Answer: (i) \(\displaystyle n\left(4n^2+9n+6\right)\); (ii) \(\displaystyle 4960\)

  2. Exercise 12

    Find the rth \displaystyle r^{\text {th }} term of an A.P. sum of whose first n\displaystyle n terms is 2n+3n2\displaystyle 2 n+3 n^2. [Hint: an=Sn−Sn−1\displaystyle a_n=\mathrm{S}_n-\mathrm{S}_{n-1}]

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    NCERT’s answer
    \(\displaystyle \mathrm{T}_r=6 r-1\)
    For \(\displaystyle r\ge 2\), \(\displaystyle a_r=S_r-S_{r-1}\):\[a_r=\left(2r+3r^2\right)-\left[2(r-1)+3(r-1)^2\right] \] \[a_r=2+3(2r-1)=6r-1 \]The first term: \[a_1=S_1=5=6(1)-1 \]Answer: \(\displaystyle a_r=6r-1\)

  3. Exercise 13

    If A is the arithmetic mean and G1,G2\displaystyle \mathrm{G}_1, \mathrm{G}_2 be two geometric means between any two numbers, then prove that 2A=G12G2+G22G12 \mathrm{A}=\frac{\mathrm{G}_1^2}{\mathrm{G}_2}+\frac{\mathrm{G}_2^2}{\mathrm{G}_1}

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    Let the numbers be \(\displaystyle a\) and \(\displaystyle b\), so \(\displaystyle A=\dfrac{a+b}{2}\). Then \(\displaystyle a, G_1, G_2, b\) are in G.P. with ratio \(\displaystyle \rho\):\[G_1=a\rho,\quad G_2=a\rho^2,\quad b=a\rho^3 \] \[\frac{G_1^2}{G_2}=\frac{a^2\rho^2}{a\rho^2}=a \] \[\frac{G_2^2}{G_1}=\frac{a^2\rho^4}{a\rho}=a\rho^3=b \] \[\frac{G_1^2}{G_2}+\frac{G_2^2}{G_1}=a+b=2A \]Answer: Proved.

  4. Exercise 14

    If θ1,θ2,θ3,…,θn\displaystyle \theta_1, \theta_2, \theta_3, \ldots, \theta_n are in A.P., whose common difference is d\displaystyle d, show that sec⁡θ1sec⁡θ2+sec⁡θ2sec⁡θ3+…+sec⁡θn−1sec⁡θn=tan⁡θn−tan⁡θ1sin⁡d.\sec \theta_1 \sec \theta_2+\sec \theta_2 \sec \theta_3+\ldots+\sec \theta_{n-1} \sec \theta_n=\frac{\tan \theta_n-\tan \theta_1}{\sin d} .

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    Since \(\displaystyle \theta_{k+1}-\theta_k=d\):\[\tan\theta_{k+1}-\tan\theta_k=\frac{\sin(\theta_{k+1}-\theta_k)}{\cos\theta_k\cos\theta_{k+1}}=\frac{\sin d}{\cos\theta_k\cos\theta_{k+1}} \] \[\sec\theta_k\sec\theta_{k+1}=\frac{\tan\theta_{k+1}-\tan\theta_k}{\sin d} \]Adding for \(\displaystyle k=1,\ldots,n-1\), the sum telescopes: \[\text{LHS}=\frac{(\tan\theta_2-\tan\theta_1)+(\tan\theta_3-\tan\theta_2)+\ldots+(\tan\theta_n-\tan\theta_{n-1})}{\sin d} \] \[\text{LHS}=\frac{\tan\theta_n-\tan\theta_1}{\sin d} \]Answer: Proved.

  5. Exercise 15

    If the sum of p\displaystyle p terms of an A.P. is q\displaystyle q and the sum of q\displaystyle q terms is p\displaystyle p, show that the sum of p+q\displaystyle p+q terms is −(p+q)\displaystyle -(p+q). Also, find the sum of first p−q\displaystyle p-q terms (p>q)\displaystyle (p>q).

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    Let the first term be \(\displaystyle a\) and the common difference \(\displaystyle d\). Take \(\displaystyle p>q\); the conditions are symmetric in \(\displaystyle p\) and \(\displaystyle q\).\[S_p-S_q=a_{q+1}+\ldots+a_p=\frac{p-q}{2}\left[a_{q+1}+a_p\right] \] \[a_{q+1}+a_p=2a+(p+q-1)d \] \[q-p=\frac{p-q}{2}\left[2a+(p+q-1)d\right]\Rightarrow 2a+(p+q-1)d=-2 \] \[S_{p+q}=\frac{p+q}{2}\left[2a+(p+q-1)d\right]=-(p+q) \]For \(\displaystyle S_{p-q}\), first find \(\displaystyle d\) from \(\displaystyle S_p=q\): \[2a+(p-1)d=\frac{2q}{p} \] \[qd=-2-\frac{2q}{p}\Rightarrow d=-\frac{2(p+q)}{pq} \]Then \[2a+(p-q-1)d=\left[2a+(p+q-1)d\right]-2qd=-2+\frac{4(p+q)}{p}=\frac{2(p+2q)}{p} \] \[S_{p-q}=\frac{p-q}{2}\cdot\frac{2(p+2q)}{p}=\frac{(p-q)(p+2q)}{p} \]Answer: \(\displaystyle S_{p+q}=-(p+q)\); \(\displaystyle S_{p-q}=\dfrac{(p-q)(p+2q)}{p}\)

  6. Exercise 16

    If pth ,qth \displaystyle p^{\text {th }}, q^{\text {th }}, and rth \displaystyle r^{\text {th }} terms of an A.P. and G.P. are both a,b\displaystyle a, b and c\displaystyle c respectively, show that ab−c.bc−a.ca−b=1a^{b-c} . b^{c-a} . c^{a-b}=1

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    Let the A.P. have first term \(\displaystyle A\), common difference \(\displaystyle D\), and the G.P. first term \(\displaystyle \alpha\), ratio \(\displaystyle \rho\).A.P.: \[a=A+(p-1)D,\quad b=A+(q-1)D,\quad c=A+(r-1)D \] \[b-c=(q-r)D,\quad c-a=(r-p)D,\quad a-b=(p-q)D \]G.P.: \[a=\alpha\rho^{p-1},\quad b=\alpha\rho^{q-1},\quad c=\alpha\rho^{r-1} \] \[a^{b-c}b^{c-a}c^{a-b}=\alpha^{(b-c)+(c-a)+(a-b)}\,\rho^{(p-1)(b-c)+(q-1)(c-a)+(r-1)(a-b)} \]The exponent of \(\displaystyle \alpha\) is \(\displaystyle 0\). For \(\displaystyle \rho\), since \(\displaystyle (b-c)+(c-a)+(a-b)=0\): \[(p-1)(b-c)+(q-1)(c-a)+(r-1)(a-b)=p(b-c)+q(c-a)+r(a-b) \] \[=D\left[p(q-r)+q(r-p)+r(p-q)\right]=0 \] \[a^{b-c}b^{c-a}c^{a-b}=\alpha^0\rho^0=1 \]Answer: Proved.
  7. Choose the correct answer out of the four given options in each of the Exercises $\displaystyle 17$ to $\displaystyle 26$ (M.C.Q.).

    Exercise 17

    If the sum of n\displaystyle n terms of an A.P. is given by Sn=3n+2n2\displaystyle \mathrm{S}_n=3 n+2 n^2, then the common difference of the A.P. is
    (A)
    3\displaystyle 3 (B) 2\displaystyle 2 (C) 6\displaystyle 6 (D) 4\displaystyle 4

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    NCERT’s answer
    D
    (D) \(\displaystyle 4\). Use \(\displaystyle a_n = S_n - S_{n-1}\).\[a_n = (3n + 2n^2) - \left[3(n-1) + 2(n-1)^2\right] \]\[a_n = 3 + 2(2n - 1) = 4n + 1 \]\[d = a_{n+1} - a_n = 4 \]

  8. Exercise 18

    The third term of G.P. is 4. The product of its first 5\displaystyle 5 terms is
    (A)
    43\displaystyle 4^3
    (B)
    44\displaystyle 4^4
    (C)
    45\displaystyle 4^5
    (D)
    None of these

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    NCERT’s answer
    C
    (C) \(\displaystyle 4^5\). The middle term of the five is the third.\[ar^2 = 4 \]\[a \cdot ar \cdot ar^2 \cdot ar^3 \cdot ar^4 = a^5 r^{10} \]\[a^5 r^{10} = (ar^2)^5 = 4^5 \]

  9. Exercise 19

    If 9\displaystyle 9 times the 9th \displaystyle 9^{\text {th }} term of an A.P. is equal to 13\displaystyle 13 times the 13th \displaystyle 13^{\text {th }} term, then the 22nd \displaystyle 22^{\text {nd }} term of the A.P. is
    (A)
    0\displaystyle 0 (B) 22\displaystyle 22
    (C)
    220\displaystyle 220
    (D)
    198\displaystyle 198

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    NCERT’s answer
    A
    (A) \(\displaystyle 0\).\[9\,a_9 = 13\,a_{13} \]\[9(a + 8d) = 13(a + 12d) \]\[9a + 72d = 13a + 156d \]\[a = -21d \]\[a_{22} = a + 21d = 0 \]

  10. Exercise 20

    If x,2y,3z\displaystyle x, 2 y, 3 z are in A.P., where the distinct numbers x,y,z\displaystyle x, y, z are in G.P. then the common ratio of the G.P. is
    (A)
    3\displaystyle 3 (B) 13\displaystyle \frac{1}{3}
    (C)
    2\displaystyle 2 (D) 12\displaystyle \frac{1}{2}

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    NCERT’s answer
    B
    (B) \(\displaystyle \tfrac{1}{3}\). Take the G.P. as \(\displaystyle x,\, xr,\, xr^2\), so \(\displaystyle y = xr,\ z = xr^2\).\[x,\ 2y,\ 3z \text{ in A.P.} \Rightarrow 4y = x + 3z \]\[4xr = x + 3xr^2 \]\[3r^2 - 4r + 1 = 0 \]\[(3r - 1)(r - 1) = 0 \]\[r \neq 1 \ \text{(distinct)} \Rightarrow r = \tfrac{1}{3} \]