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नीचे दिए गए प्रश्न के लिए अनुच्छेद

Sum of an infinite GP is defined when common ratio of the GP lies between –1 and 1.

एक अनन्त गु.श्रे. का योगफल परिभाषित है जब गु.श्रे. का सार्व अनुपात –1 तथा 1 के मध्य स्थित है।

Q. Sum of the sequence $1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, . . . \infty$ is

प्रश्न - अनुक्रम $1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, . . . \infty$ का योगफल है

\frac{3}{2}

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\frac{4}{3}

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\frac{2}{3}

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\frac{5}{2}

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Solution

The correct option is A $\frac{3}{2}$
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Q. Paragraph for below question

नीचे दिए गए प्रश्न के लिए अनुच्छेद

Sum of an infinite GP is defined when common ratio of the GP lies between –1 and 1.

एक अनन्त गु.श्रे. का योगफल परिभाषित है जब गु.श्रे. का सार्व अनुपात –1 तथा 1 के मध्य स्थित है।

Q. If S = 1 + 2x + 4x² + 8x³ + … ∞ is finite, then

प्रश्न - यदि S = 1 + 2x + 4x² + 8x³ + … ∞ परिमित है, तब

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

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.

If $f = \frac{x}{1 + x^{2}} + \frac{1}{3} {(\frac{x}{1 + x^{2}})}^{3} + \frac{1}{5} {(\frac{x}{1 + x^{2}})}^{5} + . . .$ to $\infty$ and $g = x - \frac{2}{3} x^{3} + \frac{1}{5} x^{5} + \frac{1}{7} x^{7} - \frac{2}{9} x^{9} + . . .$ , then $f = d \times g$ . Find 4d.

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.

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The correct option is A 32Solution

The correct option is A $\frac{3}{2}$
Solution