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NCERT Exemplar · Class 11 Mathematics Sequence and Series

36 questions · 36 still being checked

EXERCISE 9.3 21–30 (part 3 of 4)

  1. Choose the correct answer out of the four given options in each of the Exercises $\displaystyle 17$ to $\displaystyle 26$ (M.C.Q.).

    Exercise 21

    If in an A.P., Sn=qn2\displaystyle \mathrm{S}_n=q n^2 and Sm=qm2\displaystyle \mathrm{S}_m=q m^2, where Sr\displaystyle \mathrm{S}_r denotes the sum of r\displaystyle r terms of the A.P., then Sq\displaystyle \mathrm{S}_q equals
    (A)
    q32\displaystyle \frac{q^3}{2}
    (B)
    mnq\displaystyle m n q
    (C)
    q3\displaystyle q^3
    (D)
    (m+n)q2\displaystyle (m+n) q^2

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    NCERT’s answer
    C
    (C) \(\displaystyle q^3\). For an A.P., \(\displaystyle \dfrac{S_r}{r} = a + \dfrac{(r-1)d}{2}\); take \(\displaystyle m \neq n\).\[a + \frac{(n-1)d}{2} = qn, \qquad a + \frac{(m-1)d}{2} = qm \]\[\frac{(n-m)d}{2} = q(n - m) \Rightarrow d = 2q \]\[a = qn - (n-1)q = q \]\[S_q = \frac{q}{2}\left[2q + (q-1)\,2q\right] = q^3 \]

  2. Exercise 22

    Let Sn\displaystyle \mathrm{S}_n denote the sum of the first n\displaystyle n terms of an A.P. If S2n=3 Sn\displaystyle \mathrm{S}_{2 n}=3 \mathrm{~S}_n then S3n:Sn\displaystyle \mathrm{S}_{3 n}: \mathrm{S}_n is equal to
    (A)
    4\displaystyle 4 (B) 6\displaystyle 6 (C) 8\displaystyle 8 (D) 10\displaystyle 10

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    B
    (B) \(\displaystyle 6\).\[S_{2n} = 3S_n \Rightarrow n\left[2a + (2n-1)d\right] = \frac{3n}{2}\left[2a + (n-1)d\right] \]\[4a + (4n-2)d = 6a + (3n-3)d \]\[2a = (n+1)d \]\[S_n = \frac{n}{2}\left[(n+1)d + (n-1)d\right] = n^2 d \]\[S_{3n} = \frac{3n}{2}\left[(n+1)d + (3n-1)d\right] = 6n^2 d \]\[S_{3n} : S_n = 6 : 1 \]

  3. Exercise 23

    The minimum value of 4x+41−x,x∈R\displaystyle 4^x+4^{1-x}, x \in \mathrm{R}, is
    (A)
    2\displaystyle 2 (B) 4\displaystyle 4 (C) 1\displaystyle 1 (D) 0\displaystyle 0

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    NCERT’s answer
    B
    (B) \(\displaystyle 4\). Apply A.M. \(\displaystyle \geq\) G.M. to the two positive terms.\[4^x + 4^{1-x} \geq 2\sqrt{4^x \cdot 4^{1-x}} \]\[2\sqrt{4^{x + 1 - x}} = 2\sqrt{4} = 4 \]\[\text{Equality when } 4^x = 4^{1-x} \Rightarrow x = \tfrac{1}{2} \]

  4. Exercise 24

    Let Sn\displaystyle \mathrm{S}_n denote the sum of the cubes of the first n\displaystyle n natural numbers and sn\displaystyle s_n denote the sum of the first n\displaystyle n natural numbers. Then ∑r=1n Srsr\displaystyle \sum_{r=1}^n \frac{\mathrm{~S}_r}{s_r} equals
    (A)
    n(n+1)(n+2)6\displaystyle \frac{n(n+1)(n+2)}{6}
    (B)
    n(n+1)2\displaystyle \frac{n(n+1)}{2}
    (C)
    n2+3n+22\displaystyle \frac{n^2+3 n+2}{2}
    (D)
    None of these

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    NCERT’s answer
    A
    (A) \(\displaystyle \dfrac{n(n+1)(n+2)}{6}\).\[S_r = \left[\frac{r(r+1)}{2}\right]^2, \qquad s_r = \frac{r(r+1)}{2} \]\[\frac{S_r}{s_r} = \frac{r(r+1)}{2} = \frac{r^2 + r}{2} \]\[\sum_{r=1}^n \frac{S_r}{s_r} = \frac12\left[\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\right] \]\[= \frac{n(n+1)}{2}\cdot\frac{(2n+1) + 3}{6} = \frac{n(n+1)(n+2)}{6} \]

  5. Exercise 25

    If tn\displaystyle t_n denotes the nth term of the series 2+3+6+11+18+…\displaystyle 2+3+6+11+18+\ldots then t50\displaystyle t_{50} is
    (A)
    492−1\displaystyle 49^2-1
    (B)
    492\displaystyle 49^2
    (C)
    502+1\displaystyle 50^2+1
    (D)
    492+2\displaystyle 49^2+2

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    NCERT’s answer
    D
    (D) \(\displaystyle 49^2+2\)The differences of successive terms are the odd numbers \(\displaystyle 1, 3, 5, 7, \ldots\)\[t_n = 2 + \sum_{k=1}^{n-1}(2k-1) \] \[t_n = 2 + (n-1)^2 \] \[t_5 = 2 + 4^2 = 18 \quad \text{(matches the series)} \] \[t_{50} = 2 + 49^2 \]Answer: \(\displaystyle 49^2+2\), option (D).

  6. Exercise 26

    The lengths of three unequal edges of a rectangular solid block are in G.P. The volume of the block is 216 cm3\displaystyle 216 \mathrm{~cm}^3 and the total surface area is 252 cm2\displaystyle 252 \mathrm{~cm}^2. The length of the longest edge is
    (A)
    12\displaystyle 12 cm
    (B)
    6\displaystyle 6 cm
    (C)
    18\displaystyle 18 cm
    (D)
    3\displaystyle 3 cm

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    NCERT’s answer
    A
    (A) \(\displaystyle 12\) cmTake the edges as \(\displaystyle \dfrac{a}{r},\; a,\; ar\).\[\frac{a}{r}\cdot a \cdot ar = a^3 = 216 \Rightarrow a = 6 \] \[2\left(\frac{a^2}{r} + a^2 + a^2 r\right) = 252 \] \[36\left(r + \frac1r + 1\right) = 126 \Rightarrow r + \frac1r = \frac52 \] \[2r^2 - 5r + 2 = 0 \Rightarrow (2r-1)(r-2) = 0 \Rightarrow r = 2 \ \text{or} \ \tfrac12 \]Either value gives the edges \(\displaystyle 3, 6, 12\) (unequal, as required).\[2(3\cdot 6 + 6\cdot 12 + 12\cdot 3) = 2(126) = 252 \]Answer: longest edge \(\displaystyle =12\) cm, option (A).
  7. Fill in the blanks in the Exercises $\displaystyle 27$ to 29.

    Exercise 27

    For a,b,c\displaystyle a, b, c to be in G.P. the value of a−bb−c\displaystyle \frac{a-b}{b-c} is equal to ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    \(\displaystyle \frac{a}{b}\) or \(\displaystyle \frac{b}{c}\)
    \(\displaystyle \dfrac{a}{b}\) (equivalently \(\displaystyle \dfrac{b}{c}\))Let the common ratio be \(\displaystyle r\), so \(\displaystyle b = ar\), \(\displaystyle c = ar^2\).\[\frac{a-b}{b-c} = \frac{a(1-r)}{ar(1-r)} \] \[= \frac1r = \frac{a}{b} \]Answer: \(\displaystyle \dfrac{a}{b}\)

  8. Exercise 28

    The sum of terms equidistant from the beginning and end in an A.P. is equal to ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    First term + last term
    the sum of the first and the last termsIn an A.P. \(\displaystyle a, a+d, \ldots\) of \(\displaystyle n\) terms, the \(\displaystyle k\)th term from the beginning and the \(\displaystyle k\)th term from the end are \(\displaystyle a_k\) and \(\displaystyle a_{n-k+1}\).\[a_k + a_{n-k+1} = [a + (k-1)d] + [a + (n-k)d] \] \[= 2a + (n-1)d \] \[= a_1 + a_n \]Answer: the sum of the first and last terms.

  9. Exercise 29

    The third term of a G.P. is 4\displaystyle 4, the product of the first five terms is ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    \(\displaystyle 4^5\)
    \(\displaystyle 1024\)Let the first term be \(\displaystyle a\) and the common ratio \(\displaystyle r\).\[ar^2 = 4 \] \[a \cdot ar \cdot ar^2 \cdot ar^3 \cdot ar^4 = a^5 r^{10} \] \[= (ar^2)^5 = 4^5 = 1024 \]Answer: \(\displaystyle 1024\)
  10. State whether statement in Exercises $\displaystyle 30$ to $\displaystyle 34$ are True or False.

    Exercise 30

    Two sequences cannot be in both A.P. and G.P. together.

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    NCERT’s answer
    F
    FalseA non-zero constant sequence is both an A.P. and a G.P.\[5, 5, 5, \ldots : \quad d = 5 - 5 = 0 \] \[r = \frac{5}{5} = 1 \]Common difference constant and common ratio constant, so one sequence can be both.