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SolveItNCERT · CBSE · NEET

NCERT Exemplar · Class 11 Mathematics Sequence and Series

36 questions · 36 still being checked

EXERCISE 9.3 31–36 (part 4 of 4)

  1. State whether statement in Exercises $\displaystyle 30$ to $\displaystyle 34$ are True or False.

    Exercise 31

    Every progression is a sequence but the converse, i.e., every sequence is also a progression need not necessarily be true.

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    NCERT’s answer
    T
    TrueEvery A.P. or G.P. is a sequence, but a sequence need not have a constant difference or ratio. The Fibonacci sequence is one such:\[1, 1, 2, 3, 5, 8, \ldots \] \[1 - 1 = 0 \neq 2 - 1 = 1 \quad \text{(not an A.P.)} \] \[\frac{1}{1} \neq \frac{2}{1} \quad \text{(not a G.P.)} \]

  2. Exercise 32

    Any term of an A.P. (except first) is equal to half the sum of terms which are equidistant from it.

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    NCERT’s answer
    T
    TrueTake the term \(\displaystyle a_n = a + (n-1)d\) and two terms \(\displaystyle k\) places before and after it.\[a_{n-k} + a_{n+k} = [a + (n-k-1)d] + [a + (n+k-1)d] \] \[= 2[a + (n-1)d] = 2a_n \] \[a_n = \tfrac12\left(a_{n-k} + a_{n+k}\right) \]

  3. Exercise 33

    The sum or difference of two G.P.s, is again a G.P.

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    NCERT’s answer
    F
    False. Take the G.P.s \(\displaystyle 1,2,4,\dots\) and \(\displaystyle 1,3,9,\dots\). Their sum is \[2,\ 5,\ 13,\ \dots \] \[\frac{5}{2} \ne \frac{13}{5} \] There is no common ratio, so the sum is not a G.P. Answer: False

  4. Exercise 34

    If the sum of n\displaystyle n terms of a sequence is quadratic expression then it always represents an A.P.

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    NCERT’s answer
    F
    False. Let \(\displaystyle S_n = an^2+bn+c\). Then \[a_1 = S_1 = a+b+c \] \[a_n = S_n - S_{n-1} = 2an - a + b, \quad n \ge 2 \] \[a_2 - a_1 = 2a - c, \qquad a_3 - a_2 = 2a \] The differences agree only if \(\displaystyle c = 0\). For \(\displaystyle S_n = n^2+1\): \[a_1 = 2,\quad a_2 = 3,\quad a_3 = 5 \] \[3 - 2 = 1 \ne 2 = 5 - 3 \] Answer: False
  5. Match the questions given under Column I with their appropriate answers given under the Column II.

    Exercise 35

    Column IColumn II
    (a) 4,1,14,116\displaystyle 4,1, \frac{1}{4}, \frac{1}{16}(i) A.P.
    (b) 2\displaystyle 2, 3\displaystyle 3, 5\displaystyle 5, 7\displaystyle 7(ii) sequence
    (c) 13\displaystyle 13, 8\displaystyle 8, 3\displaystyle 3, -2\displaystyle 2, -7\displaystyle 7(iii) G.P.

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    (a)-(iii), (b)-(ii), (c)-(i)
    (a)
    Consecutive ratios are equal, so it is a G.P.
    \[\frac{1}{4} = \frac{1/4}{1} = \frac{1/16}{1/4} \]
    (b)
    Neither differences nor ratios are constant, so it is only a sequence.
    \[3-2 = 1 \ne 2 = 5-3, \qquad \frac{3}{2} \ne \frac{5}{3} \]
    (c)
    Consecutive differences are equal, so it is an A.P.
    \[8-13 = 3-8 = -2-3 = -7-(-2) = -5 \]
    Answer: (a)-(iii), (b)-(ii), (c)-(i)

  6. Exercise 36

    Column IColumn II
    (a) 12+22+32+…+n2\displaystyle 1^2+2^2+3^2+\ldots+n^2(i) (n(n+1)2)2\displaystyle \left(\frac{n(n+1)}{2}\right)^2
    (b) 13+23+33+…+n3\displaystyle 1^3+2^3+3^3+\ldots+n^3(ii) n(n+1)\displaystyle n(n+1)
    (c) 2+4+6+…+2n\displaystyle 2+4+6+\ldots+2 n(iii) n(n+1)(2n+1)6\displaystyle \frac{n(n+1)(2 n+1)}{6}
    (d) 1+2+3+…+n\displaystyle 1+2+3+\ldots+n(iv) n(n+1)2\displaystyle \frac{n(n+1)}{2}

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    NCERT’s answer
    (a)
    \(\displaystyle \leftrightarrow\) (iii)
    (b)
    \(\displaystyle \leftrightarrow\) (i)
    (c)
    \(\displaystyle \leftrightarrow\) (ii)
    (d)
    \(\displaystyle \leftrightarrow\) (iv)
    (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
    (a)
    Sum of squares of the first \(\displaystyle n\) natural numbers:
    \[1^2+2^2+\dots+n^2 = \frac{n(n+1)(2n+1)}{6} \]
    (b)
    Sum of cubes equals the square of the sum of the numbers:
    \[1^3+2^3+\dots+n^3 = \left(\frac{n(n+1)}{2}\right)^2 \]
    (d)
    Sum of the first \(\displaystyle n\) natural numbers:
    \[1+2+\dots+n = \frac{n(n+1)}{2} \]
    (c)
    Twice the sum in (d):
    \[2+4+\dots+2n = 2\cdot\frac{n(n+1)}{2} = n(n+1) \]
    Answer: (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)