Q82.If three successive terms of a G.P. with common ratio ๐๐> 1 are the length of the sides of a triangle and ๐ denotes the greatest integer less than or equal to r, then 3๐+ โ๐ is equal to: 2๐
What This Question Tests
This question combines the concept of a geometric progression with the triangle inequality theorem to determine the possible range for the common ratio, and then evaluates an expression involving the greatest integer function.
Concepts Tested
Formulas Used
Triangle inequality: a+b>c, b+c>a, a+c>b
Properties of GP: a, ar, ar^2
๐ NCERT Sections This Tests
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2.2 A regular hexagon of side 10 cm has a charge 5 mC at each of its vertices. Calculate the potential at the centre of the hexagon.
8.2 โ Name The Following Compounds According To Iupac System Of Nomenclature:
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8.2 Name the following compounds according to IUPAC system of nomenclature: (i) CH3CH(CH3)CH2CH2CHO (ii) CH3CH2COCH(C2H5)CH2CH2Cl (iii) CH3CH=CHCHO (iv) CH3COCH2COCH3 (v) CH3CH(CH3)CH2C(CH3)2COCH3 (vi) (CH3)3CCH2COOH (vii) OHCC6H4CHO-p
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5.12 Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?
๐ Question Details
- Chapter
- Sequences & Series
- Topic
- Geometric Progression and Triangle Inequality
- Year
- 2024
- Shift
- 01 Feb Shift 2
- Q Number
- Q82
- Type
- Numerical
- NCERT Ref
- Class 11 Mathematics Ch 9: Sequences & Series, Class 11 Mathematics Ch 3: Trigonometric Functions (for integer function context)
More from this Chapter
Q86.In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of this progression equals (1) 1 2 (1 โโ5) (2) 21 โ5 (3) โ5 (4) 12 (โ5 โ1)
Q88.The sum of the series 2! 1 โ13! + 4!1 โโฆ upto infinity is (1) eโ2 (2) eโ1 (3) eโ1/2 (4) e1/2
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Q76.The first two terms of a geometric progression add up to 12. The sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is (1) โ4 (2) โ12 (3) 12 (4) 4