If a sequence a n is defined as -an- cases lendcases -where n - N- then the sum of first four terms is 47A- 7-48B- 1C- 2 49D- 4 -3-48

The correct option is D 4948a_n=1n,when n is odd 1n^2,when n is even So, a_1 =1, a_2= 14, a_3=13, nbsp;a_4= 116 Hence, the required sum = 1 14+13 116 ...

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Standard XII

Mathematics

Sequence

If a sequence...

Question

If a sequence ${a_{n}}$ is defined as
$a_{n} = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{1}{n}, when n is odd - \frac{1}{n^{2}}, when n is even \end{matrix}$
where $n \in N$ , then the sum of first four terms is

\frac{49}{48}

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\frac{47}{48}

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Solution

The correct option is A $\frac{49}{48}$
$a_{n} = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{1}{n}, when n is odd - \frac{1}{n^{2}}, when n is even \end{matrix}$
So,
$a_{1} = 1, a_{2} = - \frac{1}{4}, a_{3} = \frac{1}{3}, a_{4} = - \frac{1}{16}$

Hence, the required sum
$= 1 - \frac{1}{4} + \frac{1}{3} - \frac{1}{16} = \frac{49}{48}$

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If a sequence {an} is defined as an=⎧⎪ ⎪⎨⎪ ⎪⎩1n,when n is odd−1n2,when n is even where n∈N, then the sum of first four terms is

The correct option is A 4948an=⎧⎪ ⎪⎨⎪ ⎪⎩1n,when n is odd−1n2,when n is even So, a1=1,a2=−14,a3=13,a4=−116 Hence, the required sum =1−14+13−116=4948

If a sequence ${a_{n}}$ is defined as
$a_{n} = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{1}{n}, when n is odd - \frac{1}{n^{2}}, when n is even \end{matrix}$
where $n \in N$ , then the sum of first four terms is