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Sia ? 6 years, 1 month ago
{tex}\frac{4x}{7}{/tex} - {tex}\frac{2}{3}{/tex} = {tex}\frac{11}{2}{/tex}
{tex}\Rightarrow{/tex} {tex}\frac{4x}{7}{/tex}= {tex}\frac{11}{2}{/tex} + {tex}\frac{2}{3}{/tex}
{tex}\Rightarrow{/tex} {tex}\frac{4x}{7}{/tex} = {tex}\frac{37}{6}{/tex}
{tex}\Rightarrow{/tex} 4x = {tex}\frac{37}{6}\times7{/tex}
{tex}\Rightarrow{/tex} x = {tex}\frac{37}{6}\times\frac{7}{4}{/tex}
{tex}\Rightarrow{/tex} x ={tex}\frac{259}{24}{/tex}
Posted by Pawan Y 6 years, 1 month ago
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Sia ? 6 years, 1 month ago
Given: {tex}{S_1} = {n \over 2}\left[ {2a + (n - 1)d} \right]{/tex} …..(i)
{tex}{S_2} = {{2n} \over 2}\left[ {2a + (2n - 1)d} \right]{/tex}…..(ii)
And {tex}{S_3} = {{3n} \over 2}\left[ {2a + (3n - 1)d} \right]{/tex}
Now, {tex}{S_2} - {S_1} = {{2n} \over 2}\left[ {2a + (2n - 1)d} \right] - {n \over 2}\left[ {2a + (n - 1)d} \right]{/tex}
{tex}\Rightarrow {S_2} - {S_1} = (n - {n \over 2})2a + \left[ {n(2n - 1) - {n \over 2}(n - 1)} \right]d{/tex}
{tex}= na + {1 \over 2}\left[ {4{n^2} - 2n - {n^2} + n} \right]d{/tex}
{tex}= {n \over 2}\left[ {2a + (3n - 1)d} \right] = {1 \over 3}\left\{ {{{3n} \over 2}\left[ {2a + (3n - 1)d} \right]} \right\} = {1 \over 3}{S_3}{/tex}
{tex}\Rightarrow {/tex} 3(S2 - S1) = S3
Hence proved.
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Sia ? 6 years, 1 month ago
√3 cosx +sinx = √2
√3 cosx = √2 -sinx
Now square both sides,
3 cos²x = 2 + sin²x -2√2 sinx
3(1-sin²x) = 2 + sin²x -2√2 sinx
3-3sin²x = 2 + sin²x -2√2 sinx
0 = -1 + 4sin²x -2√2 sinx
4sin²x - 2√2 sinx - 1 = 0
let y = sinx, then,
4y² -2√2y -1 =0
y = {2√2±√(8+16)}/8
y = {2√2 ±2√6}/8
y = {√2 ±√6}/4
sinx = {√2 ±√6}/4
x = sin inverse {√2 ±√6}/4
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Nikhil Nishad 6 years, 1 month ago
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