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Ask QuestionPosted by Rupesh Kumar 7 years, 9 months ago
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Posted by Anurag Jagetiya 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Let a be the first term and d be the common difference of the given A.P.
Clearly, in an A.P. consisting of 11 terms, {tex} \left( \frac { 11 + 1 } { 2 } \right) ^ { t h }{/tex} i.e. 6th term is the middle term.
{tex}\text{ it is given that the middle term =30}{/tex}
So a+5d=30 .....(1)
.{tex}S_{11}=\frac{11}{2}(2a+10d){/tex}
= 11(a+5d)
But a+5d=30 from (1)
Hence S11 = 11 × 30= 330
Posted by Rahul Kumar Tiwari 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
L.C.M of 8, 15 and 21.
8 = 2 {tex}\times{/tex} 2 {tex}\times{/tex} 2
15 = 3 {tex}\times{/tex} 5
21 = 3 {tex}\times{/tex} 7
L.C.M of 8, 15 and 21 = 23 {tex}\times{/tex} 3 {tex}\times{/tex} 5 {tex}\times{/tex} 71= 840
If we divide 110000 by 840, we will find out that it is not exactly divisible and we get 800 as remainder.
Thus the number nearest to 110000 but greater than 100000 which is exactly divisible by 840 and also divisible by 8, 15 and 21 is = 110000 - 840 = 109200
Hence 109200 is exactly divisible by 8, 15 and 21.
Posted by Abdul Hafiz 7 years, 9 months ago
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Sankalp Kumar 7 years, 9 months ago
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Darshanaa Yadav 7 years, 9 months ago
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Gurman Singh 7 years, 9 months ago
Posted by Ishaan Gupta 5 years, 8 months ago
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Sia ? 6 years, 5 months ago
Let AB be a chord of a circle with centre O, and let AP and BP be the tangents at A and B respectively. Let the two tangents AP and BP meets at P.
Now Join OP. Suppose OP meets AB at C and the chord AB=AC+BC.
We have to prove that {tex}\angle P A C = \angle P B C{/tex}
In two triangles PCA and PCB, we have
PA = PB [{tex} \because{/tex}Tangents from an external point are equal]
{tex}\angle A P C = \angle B P C{/tex} [{tex} \because {/tex} PA and PB are equally inclined to OP]
and, PC = PC [Common]
So, by SAS-criterion of congruence, we obtain
{tex}\Delta P A C \cong \Delta P B C{/tex}
{tex}\Rightarrow \quad \angle P A C = \angle P B C{/tex}
Posted by Aflah Ek 7 years, 9 months ago
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Mental Roy 7 years, 9 months ago
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Darshanaa Yadav 7 years, 9 months ago
Posted by Mahima Verma 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Let us assume that point P divides the line segment AB in the ratio (k : 1)
Using section formula,
P(4, m) = {tex}\frac{6k+2}{k+1} , {/tex}{tex}\frac{-3k+3}{k+1}{/tex}
{tex}\therefore{/tex} {tex}\frac{6 \mathrm{k}+2}{\mathrm{k}+1}{/tex} = 4 {tex}\Rightarrow 6k+2=4k+4 \Rightarrow 2k=2{/tex}
{tex}\Rightarrow{/tex} k = 1
{tex}\therefore{/tex} The ratio is 1 : 1
Now, m = {tex}\frac{-3k+3}{k+1}{/tex} = {tex}\frac{-3+3}{2}{/tex} = 0
Posted by Sanjay Vishwakarma 7 years, 9 months ago
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