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The numbers x − 6, 2x and x2 are in G.P. Find nth term

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प्रश्न

The numbers x − 6, 2x and x2 are in G.P. Find nth term

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उत्तर

nth term = arn−1, where a = 4, r = `x^2/(2x)` = `100/20` = 5

= 4(5)n−1

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पाठ 2: Sequences and Series - Exercise 2.1 [पृष्ठ २८]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
पाठ 2 Sequences and Series
Exercise 2.1 | Q 15. (iii) | पृष्ठ २८

संबंधित प्रश्‍न

Which term of the following sequence:

`1/3, 1/9, 1/27`, ...., is `1/19683`?


Find the sum to indicated number of terms in the geometric progressions 1, – a, a2, – a3, ... n terms (if a ≠ – 1).


Find a G.P. for which sum of the first two terms is –4 and the fifth term is 4 times the third term.


Find :

the 12th term of the G.P.

\[\frac{1}{a^3 x^3}, ax, a^5 x^5 , . . .\]


Find : 

nth term of the G.P.

\[\sqrt{3}, \frac{1}{\sqrt{3}}, \frac{1}{3\sqrt{3}}, . . .\]


Find :

the 10th term of the G.P.

\[\sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}, . . .\]


Which term of the G.P.: `sqrt3, 3, 3sqrt3`, ... is 729?


The fourth term of a G.P. is 27 and the 7th term is 729, find the G.P.


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Evaluate the following:

\[\sum^{11}_{n = 1} (2 + 3^n )\]


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If S1, S2, S3 be respectively the sums of n, 2n, 3n terms of a G.P., then prove that \[S_1^2 + S_2^2\] = S1 (S2 + S3).


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\[0 . 6\overline8\]


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\[\frac{1}{a^2 + b^2}, \frac{1}{b^2 - c^2}, \frac{1}{c^2 + d^2} \text { are in G . P } .\]


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Answer the following:

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Answer the following:

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Answer the following:

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Answer the following:

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The third term of G.P. is 4. The product of its first 5 terms is ______.


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Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.


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