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Select the correct answer from the given alternative. Which term of the geometric progression 1, 2, 4, 8, ... is 2048 - Mathematics and Statistics

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प्रश्न

Select the correct answer from the given alternative.

Which term of the geometric progression 1, 2, 4, 8, ... is 2048

विकल्प

  • 10th 

  • 11th 

  • 12th 

  • 13th 

MCQ
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उत्तर

12th 

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2: Sequences and Series - Miscellaneous Exercise 2.1 [पृष्ठ ४१]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
अध्याय 2 Sequences and Series
Miscellaneous Exercise 2.1 | Q I. (4) | पृष्ठ ४१

संबंधित प्रश्न

Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.


If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.


If the first and the nth term of a G.P. are a ad b, respectively, and if P is the product of n terms, prove that P2 = (ab)n.


Which term of the G.P. :

\[\sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}, \frac{1}{4\sqrt{2}}, . . . \text { is }\frac{1}{512\sqrt{2}}?\]


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


Which term of the progression 18, −12, 8, ... is \[\frac{512}{729}\] ?

 

The 4th term of a G.P. is square of its second term, and the first term is − 3. Find its 7th term.


If \[\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx}\] (x ≠ 0), then show that abc and d are in G.P.


Find the sum of the following geometric series:

\[\sqrt{2} + \frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}} + . . .\text { to 8  terms };\]


Find the sum of the following geometric series:

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The common ratio of a G.P. is 3 and the last term is 486. If the sum of these terms be 728, find the first term.


How many terms of the G.P. `3, 3/2, 3/4` ..... are needed to give the sum `3069/512`?


Find the sum of the following series to infinity:

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Find the sum of the terms of an infinite decreasing G.P. in which all the terms are positive, the first term is 4, and the difference between the third and fifth term is equal to 32/81.


The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an A.P. Find the numbers.


If a, b, c are in G.P., prove that:

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If a, b, c, d are in G.P., prove that:

(b + c) (b + d) = (c + a) (c + d)


If a, b, c, d are in G.P., prove that:

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


If (a − b), (b − c), (c − a) are in G.P., then prove that (a + b + c)2 = 3 (ab + bc + ca)


If pth, qth, rth and sth terms of an A.P. be in G.P., then prove that p − q, q − r, r − s are in G.P.


If \[\frac{1}{a + b}, \frac{1}{2b}, \frac{1}{b + c}\] are three consecutive terms of an A.P., prove that a, b, c are the three consecutive terms of a G.P.


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Find : `sum_("r" = 1)^oo 4(0.5)^"r"`


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