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Ask QuestionPosted by Pranjal Tak 7 years, 9 months ago
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Posted by Sakshi Gothwal 7 years, 9 months ago
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Tanisha Sharma 7 years, 9 months ago
Posted by Aruhi Sharma 7 years, 9 months ago
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Sia ? 6 years, 6 months ago
Let a be the first term and d be the common difference of the given AP. Then, in general the mth and nth terms can be written as
Tm = a + (m - 1) d and Tn = a + (n - 1)d respectively.
According to the question,we are given that,
(m.Tm) = (n.Tn)
{tex}\Rightarrow{/tex} m.{a + (m - 1)d} = n.{a + (n - 1)d}
{tex}\Rightarrow{/tex} a.(m - n) + {(m2 - n2) - (m - n)} . d = 0
{tex}\Rightarrow{/tex} (m - n).{a + (m + n - 1)}d.
{tex}\Rightarrow{/tex} (m - n).Tm+n = 0
{tex}\Rightarrow{/tex}Tm+n = 0 [{tex}\because{/tex} (m-n){tex}\neq{/tex}0].
Hence, the (m + n)th term is zero.
Posted by Jatin Rao 7 years, 9 months ago
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Posted by Astha Kumari 5 years, 8 months ago
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Posted by Aditya Makwana 7 years, 9 months ago
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Daljit Singh 7 years, 9 months ago
Ayush Agrawal 7 years, 9 months ago
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Amanpreet Singh 7 years, 9 months ago
Posted by Sahil Goria 7 years, 9 months ago
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Posted by Rohit Dixit 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
In the given figure we have {tex}P A \perp A C{/tex} and {tex}Q B \perp A C{/tex}.
{tex}\Rightarrow Q B \| P A{/tex}
In {tex}\triangle PAC{/tex} and {tex}\triangle Q B C{/tex}, we have
{tex}\angle QCB= \angle PCA{/tex} ( Common )
{tex}\angle QBC= \angle PAC{/tex} ( both are 90o ).
So by AA similarity rule , {tex}\triangle Q B C \sim \triangle P A C{/tex}.
{tex}\therefore \frac { Q B } { P A } = \frac { B C } { A C }{/tex}
{tex}\Rightarrow \frac { z } { x } = \frac { b } { a + b }{/tex}. .....................................(i) [by the property of similar triangles]
In {tex}\triangle RAC{/tex} , {tex}Q B \| R C{/tex}.
So, {tex}\triangle Q B A \sim \triangle R C A{/tex}.
{tex}\therefore \frac { Q B } { R C } = \frac { A B } { A C }{/tex}
{tex}\Rightarrow \frac { z } { y } = \frac { a } { a + b }{/tex}. .....................................(ii) [by the property of similar triangles]
Form (i) and (ii), we obtain
{tex}\frac { z } { x } + \frac { z } { y }{/tex}{tex}= \left( \frac { b } { a + b } + \frac { a } { a + b } \right) = 1{/tex}
{tex}\Rightarrow \quad \frac { z } { x } + \frac { z } { y } = 1{/tex}
{tex}\Rightarrow \frac { 1 } { x } + \frac { 1 } { y } = \frac { 1 } { z }{/tex}
or {tex}\frac { 1 } { x } + \frac { 1 } { y } = \frac { 1 } { z }{/tex}.
Hence proved.
Posted by Thomas Sangala 7 years, 9 months ago
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Keshav Bindal 7 years, 9 months ago
Posted by Charu Vatsal 7 years, 9 months ago
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Posted by Gautam Agrawal 7 years, 9 months ago
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Posted by Komal Khichar 7 years, 9 months ago
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Komal Khichar 7 years, 9 months ago
Posted by Parwati Sahu 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
In a right triangle BAC,
BC2 = AB2 + AC2 …….. By Pythagoras theorem
= (14)2 + (14)2 = 2(14)2
{tex}\Rightarrow{/tex} BC = {tex}14\sqrt2{/tex} cm
{tex}\Rightarrow{/tex} Radius of the semicircle
= {tex}\;\frac{14\sqrt2}2{/tex} cm = {tex}7\sqrt2{/tex} cm
Required area = Area BPCQB
= Area BCQB – Area BCPB
= Area BCQB – (Area BACPB - Area of {tex}\Rightarrow{/tex}BAC)
{tex}= \;\frac{180}{360}\mathrm\pi\left(7\sqrt2\right)^2\;-\left[\frac{90}{360}\mathrm\pi(14)^2\;-\frac{14\times\;14}2\right]{/tex}
{tex}\;\frac12\times\frac{22}7\times98\;-\left[\frac14\times\frac{22}7\times196\;-98\right]{/tex}
= 154 – (154 – 98) = 98 cm2
Posted by Abbas Panched Wala 7 years, 9 months ago
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Komal Khichar 7 years, 9 months ago
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Keshav Bindal 7 years, 9 months ago
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