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Sia ? 6 years, 6 months ago
{tex}\style{font-family:Tahoma}{\style{font-size:8px}{\begin{array}{l}Let\;P(x,y,z)\;be\;any\;point\;which\;is\;equidis\tan t\;from\;A(0,2,3)\;and\;B\;(2,\;-2,1),\\PA\;=\;PB\\\Rightarrow PA^2\;=\;PB^2\\\Rightarrow\sqrt{\left(x-0\right)^2\;+\left(y-2\right)^2+\left(z-3\right)^2}\;=\sqrt{\left(x-2\right)^2\;+\left(y+2\right)^2+\left(z-1\right)^2}\\\Rightarrow\;4x\;-\;8y\;-\;4z\;+4\;=\;0\;or\;x\;-\;2y\;-\;z\;+1\;=\;0\\Hence\;the\;required\;locus\;is \;x\;-\;2y\;-\;z+1\;=\;0\end{array}}}{/tex}
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Sia ? 6 years, 6 months ago
Let three consecutive numbers be x, (x + 1) and (x + 2)
Let x = 6q + r 0 {tex}\leq r < 6{/tex}
{tex}\therefore x = 6 q , 6 q + 1,6 q + 2,6 q + 3,6 q + 4,6 q + 5{/tex}
{tex}\text { product of } x ( x + 1 ) ( x + 2 ) = 6 q ( 6 q + 1 ) ( 6 q + 2 ){/tex}
if x = 6q then which is divisible by 6
{tex}\text { if } x = 6 q + 1{/tex}
{tex}= ( 6 q + 1 ) ( 6 q + 2 ) ( 6 q + 3 ){/tex}
{tex}= 2 ( 3 q + 1 ) .3 ( 2 q + 1 ) ( 6 q + 1 ){/tex}
{tex}= 6 ( 3 q + 1 ) \cdot ( 2 q + 1 ) ( 6 q + 1 ){/tex}
which is divisible by 6
if x = 6q + 2
{tex}= ( 6 q + 2 ) ( 6 q + 3 ) ( 6 q + 4 ){/tex}
{tex}= 3 ( 2 q + 1 ) .2 ( 3 q + 1 ) ( 6 q + 4 ){/tex}
{tex}= 6 ( 2 q + 1 ) \cdot ( 3 q + 1 ) ( 6 q + 1 ){/tex}
Which is divisible by 6
{tex}\text { if } x = 6 q + 3{/tex}
{tex}= ( 6 q + 3 ) ( 6 q + 4 ) ( 6 q + 5 ){/tex}
{tex}= 6 ( 2 q + 1 ) ( 3 q + 2 ) ( 6 q + 5 ){/tex}
which is divisible by 6
{tex}\text { if } x = 6 q + 4{/tex}
{tex}= ( 6 q + 4 ) ( 6 q + 5 ) ( 6 q + 6 ){/tex}
{tex}= 6 ( 6 q + 4 ) ( 6 q + 5 ) ( q + 1 ){/tex}
which is divisible by 6
if x = 6q + 5
{tex}= ( 6 q + 5 ) ( 6 q + 6 ) ( 6 q + 7 ){/tex}
{tex}= 6 ( 6 q + 5 ) ( q + 1 ) ( 6 q + 7 ){/tex}
which is divisible by 6
{tex}\therefore {/tex} the product of any three natural numbers is divisible by 6.
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Sia ? 6 years, 6 months ago
Let a, and A be the first terms and d and D be the common difference of two A.Ps
Then, according to the question,
{tex}\frac { S _ { n } } { S _ { n } ^ { \prime } } = \frac { \frac { n } { 2 } [ 2 a + ( n - 1 ) d ] } { \frac { n } { 2 } [ 2 A + ( n - 1 ) D ] } = \frac { 7 n + 1 } { 4 n + 27 }{/tex}
or, {tex}\frac { 2 a + ( n - 1 ) d } { 2 A + ( n - 1 ) D } = \frac { 7 n + 1 } { 4 n + 27 }{/tex}
or,{tex}\frac { a + \left( \frac { n - 1 } { 2 } \right) d } { A + \left( \frac { n - 1 } { 2 } \right) D } = \frac { 7 n + 1 } { 4 n + 27 }{/tex}
Putting, {tex}\frac { n - 1 } { 2 } = m - 1{/tex}
{tex}n-1 = 2m - 2{/tex}
{tex}n= 2m - 2 + 1{/tex}
or, {tex}n = 2m - 1{/tex}
{tex}\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \frac { 7 ( 2 m - 1 ) + 1 } { 4 ( 2 m - 1 ) + 27 }{/tex}
{tex}\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \frac { 14 m - 7 + 1 } { 8 m - 4 + 27 }{/tex}
{tex}\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \frac { 14 m - 6 } { 8 m + 23 }{/tex}
Hence, {tex}\frac { a _ { m } } { A _ { m } } = \frac { 14 m - 6 } { 8 m + 23 }{/tex}
Posted by Priyanka Nanda 7 years, 9 months ago
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Anmol Yadav 7 years, 9 months ago
1Thank You