Ex. no. 2.4 class 10 Samacheer kalvi

Exercise 2.4 - Sequences and Series Solutions

Exercise 2.4: Sequences

A complete guide with questions and step-by-step solutions.

Questions

  1. Find the next three terms of the following sequence.
    1. $8, 24, 72, \dots$
    2. $5, 1, -3, \dots$
    3. $\frac{1}{4}, \frac{2}{9}, \frac{3}{16}, \dots$
  2. Find the first four terms of the sequences whose $n^{th}$ terms are given by
    1. $a_n = n^3 - 2$
    2. $a_n = (-1)^{n+1}n(n+1)$
    3. $a_n = 2n^2 - 6$
  3. Find the $n^{th}$ term of the following sequences
    1. $2, 5, 10, 17, \dots$
    2. $0, \frac{1}{2}, \frac{2}{3}, \dots$
    3. $3, 8, 13, 18, \dots$
  4. Find the indicated terms of the sequences whose $n^{th}$ terms are given by
    1. $a_n = \frac{5n}{n+2}$; find $a_6$ and $a_{13}$
    2. $a_n = -(n^2 - 4)$; find $a_4$ and $a_{11}$
  5. Find $a_8$ and $a_{15}$ whose $n^{th}$ term is $a_n = \begin{cases} \frac{n^2 - 1}{n+3} & \text{if } n \text{ is even, } n \in \mathbb{N} \\ \frac{n^2}{2n+1} & \text{if } n \text{ is odd, } n \in \mathbb{N} \end{cases}$
  6. If $a_1 = 1, a_2 = 1$ and $a_n = 2a_{n-1} + a_{n-2}$ for $n \ge 3, n \in \mathbb{N}$, then find the first six terms of the sequence.

Questions with Detailed Solutions

  1. Find the next three terms of the following sequence.
    1. $8, 24, 72, \dots$
    2. $5, 1, -3, \dots$
    3. $\frac{1}{4}, \frac{2}{9}, \frac{3}{16}, \dots$

    (i) Sequence: $8, 24, 72, \dots$

    This is a geometric sequence. To find the common ratio, we divide a term by its preceding term: $\frac{24}{8} = 3$ and $\frac{72}{24} = 3$. The common ratio is 3.

    $a_1 = 8$

    $a_2 = 8 \times 3 = 24$

    $a_3 = 24 \times 3 = 72$

    The next three terms are:

    $a_4 = a_3 \times 3 = 72 \times 3 = 216$

    $a_5 = a_4 \times 3 = 216 \times 3 = 648$

    $a_6 = a_5 \times 3 = 648 \times 3 = 1944$

    The next three terms are 216, 648, and 1944.

    (ii) Sequence: $5, 1, -3, \dots$

    This is an arithmetic sequence. To find the common difference, we subtract a term from its succeeding term: $1 - 5 = -4$ and $-3 - 1 = -4$. The common difference is -4.

    $a_1 = 5$

    $a_2 = 5 - 4 = 1$

    $a_3 = 1 - 4 = -3$

    The next three terms are:

    $a_4 = a_3 - 4 = -3 - 4 = -7$

    $a_5 = a_4 - 4 = -7 - 4 = -11$

    $a_6 = a_5 - 4 = -11 - 4 = -15$

    The next three terms are -7, -11, and -15.

    (iii) Sequence: $\frac{1}{4}, \frac{2}{9}, \frac{3}{16}, \dots$

    Let's analyze the numerator and denominator separately. The numerators are $1, 2, 3, \dots$, which corresponds to $n$. The denominators are $4, 9, 16, \dots$, which are squares: $2^2, 3^2, 4^2, \dots$. This corresponds to $(n+1)^2$.

    So, the formula for the $n^{th}$ term is $a_n = \frac{n}{(n+1)^2}$.

    $a_1 = \frac{1}{(1+1)^2} = \frac{1}{4}$

    $a_2 = \frac{2}{(2+1)^2} = \frac{2}{9}$

    $a_3 = \frac{3}{(3+1)^2} = \frac{3}{16}$

    The next three terms are:

    $a_4 = \frac{4}{(4+1)^2} = \frac{4}{25}$

    $a_5 = \frac{5}{(5+1)^2} = \frac{5}{36}$

    $a_6 = \frac{6}{(6+1)^2} = \frac{6}{49}$

    The next three terms are $\frac{4}{25}, \frac{5}{36}, \frac{6}{49}$.
  2. Find the first four terms of the sequences whose $n^{th}$ terms are given by...

    (i) $a_n = n^3 - 2$

    $a_1 = 1^3 - 2 = 1 - 2 = -1$

    $a_2 = 2^3 - 2 = 8 - 2 = 6$

    $a_3 = 3^3 - 2 = 27 - 2 = 25$

    $a_4 = 4^3 - 2 = 64 - 2 = 62$

    The first four terms are -1, 6, 25, 62.

    (ii) $a_n = (-1)^{n+1}n(n+1)$

    $a_1 = (-1)^{1+1}(1)(1+1) = (-1)^2(1)(2) = 1 \times 2 = 2$

    $a_2 = (-1)^{2+1}(2)(2+1) = (-1)^3(2)(3) = -1 \times 6 = -6$

    $a_3 = (-1)^{3+1}(3)(3+1) = (-1)^4(3)(4) = 1 \times 12 = 12$

    $a_4 = (-1)^{4+1}(4)(4+1) = (-1)^5(4)(5) = -1 \times 20 = -20$

    The first four terms are 2, -6, 12, -20.

    (iii) $a_n = 2n^2 - 6$

    $a_1 = 2(1)^2 - 6 = 2(1) - 6 = -4$

    $a_2 = 2(2)^2 - 6 = 2(4) - 6 = 8 - 6 = 2$

    $a_3 = 2(3)^2 - 6 = 2(9) - 6 = 18 - 6 = 12$

    $a_4 = 2(4)^2 - 6 = 2(16) - 6 = 32 - 6 = 26$

    The first four terms are -4, 2, 12, 26.
  3. Find the $n^{th}$ term of the following sequences...

    (i) $2, 5, 10, 17, \dots$

    Let's examine the terms:

    $a_1 = 2 = 1^2 + 1$

    $a_2 = 5 = 2^2 + 1$

    $a_3 = 10 = 3^2 + 1$

    $a_4 = 17 = 4^2 + 1$

    The pattern is $n^2 + 1$.

    The $n^{th}$ term is $a_n = n^2 + 1$.

    (ii) $0, \frac{1}{2}, \frac{2}{3}, \dots$

    Let's rewrite the terms:

    $a_1 = 0 = \frac{1-1}{1}$

    $a_2 = \frac{1}{2} = \frac{2-1}{2}$

    $a_3 = \frac{2}{3} = \frac{3-1}{3}$

    The pattern is $\frac{n-1}{n}$.

    The $n^{th}$ term is $a_n = \frac{n-1}{n}$.

    (iii) $3, 8, 13, 18, \dots$

    This is an Arithmetic Progression (A.P.) because the difference between consecutive terms is constant.

    First term, $a = 3$.

    Common difference, $d = 8 - 3 = 5$.

    The formula for the $n^{th}$ term of an A.P. is $t_n = a + (n-1)d$.

    $t_n = 3 + (n-1)5$

    $t_n = 3 + 5n - 5$

    $t_n = 5n - 2$

    The $n^{th}$ term is $a_n = 5n - 2$.
  4. Find the indicated terms of the sequences whose $n^{th}$ terms are given by...

    (i) $a_n = \frac{5n}{n+2}$; find $a_6$ and $a_{13}$

    For $a_6$, we substitute $n=6$:

    $a_6 = \frac{5(6)}{6+2} = \frac{30}{8} = \frac{15}{4}$

    For $a_{13}$, we substitute $n=13$:

    $a_{13} = \frac{5(13)}{13+2} = \frac{65}{15} = \frac{13}{3}$

    $a_6 = \frac{15}{4}$ and $a_{13} = \frac{13}{3}$.

    (ii) $a_n = -(n^2 - 4)$; find $a_4$ and $a_{11}$

    For $a_4$, we substitute $n=4$:

    $a_4 = -(4^2 - 4) = -(16 - 4) = -12$

    For $a_{11}$, we substitute $n=11$:

    $a_{11} = -(11^2 - 4) = -(121 - 4) = -117$

    $a_4 = -12$ and $a_{11} = -117$.
  5. Find $a_8$ and $a_{15}$ whose $n^{th}$ term is $a_n = \begin{cases} \frac{n^2 - 1}{n+3} & \text{if } n \text{ is even, } n \in \mathbb{N} \\ \frac{n^2}{2n+1} & \text{if } n \text{ is odd, } n \in \mathbb{N} \end{cases}$

    To find $a_8$, we note that $n=8$ is an even number. So we use the first formula:

    $a_n = \frac{n^2 - 1}{n+3}$

    $a_8 = \frac{8^2 - 1}{8+3} = \frac{64 - 1}{11} = \frac{63}{11}$

    To find $a_{15}$, we note that $n=15$ is an odd number. So we use the second formula:

    $a_n = \frac{n^2}{2n+1}$

    $a_{15} = \frac{15^2}{2(15)+1} = \frac{225}{30+1} = \frac{225}{31}$

    $a_8 = \frac{63}{11}$ and $a_{15} = \frac{225}{31}$.
  6. If $a_1 = 1, a_2 = 1$ and $a_n = 2a_{n-1} + a_{n-2}$ for $n \ge 3, n \in \mathbb{N}$, then find the first six terms of the sequence.

    We are given the first two terms and a recurrence relation to find the subsequent terms.

    Given:

    $a_1 = 1$

    $a_2 = 1$

    Finding the next terms for $n \ge 3$:

    For $n=3$:

    $a_3 = 2a_{3-1} + a_{3-2} = 2a_2 + a_1 = 2(1) + 1 = 3$

    For $n=4$:

    $a_4 = 2a_{4-1} + a_{4-2} = 2a_3 + a_2 = 2(3) + 1 = 6 + 1 = 7$

    For $n=5$:

    $a_5 = 2a_{5-1} + a_{5-2} = 2a_4 + a_3 = 2(7) + 3 = 14 + 3 = 17$

    For $n=6$:

    $a_6 = 2a_{6-1} + a_{6-2} = 2a_5 + a_4 = 2(17) + 7 = 34 + 7 = 41$

    The first six terms are 1, 1, 3, 7, 17, 41.