The sum of the first 25 terms of the series 1-3-1-4-1-2-1-3-1-2-1-4-1-3-1-4-1-2-1-3-is

The correct option is D 1/3 Sum of the first 6 terms ⇒1/3+1/4 1/2 1/3+1/2 1/4=0 ∴ Sum nbsp; of nbsp;first nbsp;24 nbsp; nbsp;terms ...

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Byju's Answer

Standard XII

Mathematics

Summation by Sigma Method

The sum of th...

Question

The sum of the first 25 terms of the series
$\frac{1}{3} + \frac{1}{4} - \frac{1}{2} - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} + \frac{1}{4} - \frac{1}{2} - \frac{1}{3} + . . . . .$ is

\frac{1}{2}

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- \frac{1}{4}

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\frac{1}{3}

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\frac{1}{4}

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Solution

The correct option is D $\frac{1}{3}$
$\frac{1}{3} + \frac{1}{4} - \frac{1}{2} - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} + \frac{1}{4} - \frac{1}{2} - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + . . . .$
$\Rightarrow (\frac{1}{3} + \frac{1}{4} - \frac{1}{2} - \frac{1}{3} + \frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} + \frac{1}{4} - \frac{1}{2} - \frac{1}{3} + \frac{1}{2} - \frac{1}{4}) + . . . .$
As we can observe, the series repeat itself after every $6$ th term.
Therefore, sum of the first 6 terms
$\Rightarrow \frac{1}{3} + \frac{1}{4} - \frac{1}{2} - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} = 0$
Hence, sum of first $24$ terms = $0 \times 4 = 4$
Now, $25$ th term of the series will be $\frac{1}{3}$
Hence, sum of first 25 terms = $0 + \frac{1}{3} = \frac{1}{3}$

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Similar questions

Q. Assertion :

Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})}{1 + a_{1} b_{1}} & \frac{(1 + a_{1} b_{2}) (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2})}{1 + a_{1} b_{2}} & \frac{(1 + a_{1} b_{3}) (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3})}{1 + a_{1} b_{3}} \frac{(1 + a_{2} b_{1}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1})}{1 + a_{2} b_{1}} & \frac{(1 + a_{2} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2})}{1 + a_{2} b_{2}} & \frac{(1 + a_{2} b_{3}) (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3})}{1 + a_{2} b_{3}} \frac{(1 + a_{3} b_{1}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1})}{1 + a_{3} b_{1}} & \frac{(1 + a_{3} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2})}{1 + a_{3} b_{2}} & \frac{(1 + a_{3} b_{3}) (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3})}{1 + a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Δ = 0

Reason:

Δ

can be written as product of two determinants.

The product of the following series $(1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + . . .)$ $\times$ $(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + . . .)$ is

Q. Consider the sequence

a_{n}

given by

a_{1} = \frac{1}{2}, a_{n} + 1 = a_{n}^{2} + a_{n}

Let

S_{n} = \frac{1}{a_{1} + 1} + \frac{1}{a_{2} + 1} + . . . . . . . . . . . + \frac{1}{a_{n} + 1}

then find the value of

[S_{2012}]

, where [.] denotes greatest integer function.

Q. STATEMENT 1: In a

Δ A B C

, if

r_{1}, r_{2}, r_{3}

are in

H . P

. then

\frac{r_{2}}{r} = \frac{1}{3}

STATEMENT 2: In a

Δ A B C, \frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} = \frac{2}{r}

$l o g_{e} (n + 1) - l o g_{e} (n - 1) = 4 a [(\frac{1}{n}) + (\frac{1}{3 n^{3}}) + (\frac{1}{5 n^{5}}) + . . . \infty]$ Find $8 a$ .

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The sum of the first 25 terms of the series13+14−12−13+12−14+13+14−12−13+..... is

The sum of the first 25 terms of the series
$\frac{1}{3} + \frac{1}{4} - \frac{1}{2} - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} + \frac{1}{4} - \frac{1}{2} - \frac{1}{3} + . . . . .$ is