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Sia ? 6 years, 6 months ago
According to the question,we have,
a=9. Therefore, common difference d =17-9=8
let the required number of terms be n.
Therefore, Sn=636
{tex}\Rightarrow{/tex}{tex}\frac{n}{2}{/tex}[2a+(n-1)d]=636
{tex}\Rightarrow{/tex}{tex}\frac{n}{2}{/tex}[2(9)+(n-1)8]=636
{tex}\Rightarrow{/tex}n[18+8n-8]=1272
{tex}\Rightarrow{/tex}8n2+10n-1272=0
{tex}\Rightarrow{/tex}4n2+5n-636=0
{tex}\Rightarrow{/tex}4n2+53n-48n-636=0
{tex}\Rightarrow{/tex}n(4n+53)-12(4n+53)=0
{tex}\Rightarrow{/tex}(4n+53)(n-12)=0
{tex}\Rightarrow{/tex}4n+53=0 or n-12=0
{tex}\Rightarrow{/tex}n={tex}\frac{{ - 53}}{4}{/tex} or n=12
Since number of terms cannot neither be negative nor fraction, n=12
hence, the required number of terms is 12.
Posted by Sankar Kumar 7 years, 4 months ago
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Posted by Ashwin Edakkara 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let speed of passenger train be x km/h
{tex}\therefore {/tex} speed of superfast train = (x + 16) km/h
By question, {tex}T _ { \text { passenger } } = \frac { 192 } { x }{/tex} and {tex}\mathrm { T } _ { \text { superfast } } = \frac { 192 } { ( x + 16 ) }{/tex}
or, {tex}\frac { 192 } { x } - \frac { 192 } { x + 16 } = 2{/tex}
or, {tex}192 ( x + 16 ) - 192 x = 2 \left( x ^ { 2 } + 16 x \right){/tex}
or, {tex}192 x + 192 \times 16 - 192 x = 2 \left( x ^ { 2 } + 16 x \right){/tex}
192x + 3072 - 192x {tex}{/tex}{tex} = 2 \left( x ^ { 2 } + 16 x \right){/tex}( divide throughout by 2, we get,
96x + 1536 - 96x {tex} = \left( x ^ { 2 } + 16 x \right){/tex}
or x(x + 48) - 32(x + 48) = 0
or, (x - 32) (x + 48) = 0
or, x = 32 or - 48
Since speed can't be negative, therefore - 48 is not possible.
{tex}\therefore {/tex} Speed of passenger train = 32 km/h and Speed of fast train = 48 km/h
Posted by Ravikant Sharma 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
We have {tex}\frac{{16}}{x} - 1 = \frac{{15}}{{x + 1}}{/tex}
{tex} \Rightarrow \frac { 16 - x } { x } = \frac { 15 } { x + 1 }{/tex}
Cross multiply,
{tex}\Rightarrow{/tex} {tex}(16-x)(x+1)=15x{/tex}
{tex} \Rightarrow{/tex} {tex}16x+16-x^2-x=15x{/tex}
{tex} \Rightarrow{/tex} 15x - 16x - 16 + x2 + x = 0
{tex} \Rightarrow{/tex} x2 - 16 = 0
{tex} \Rightarrow{/tex} x2 = 16
{tex} \therefore{/tex} x = {tex} \pm{/tex}4
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Sia ? 6 years, 6 months ago
Since the point (3, a) lies on the line 2x - 3y = 5, we have
2 {tex}\times{/tex} 3 - 3 {tex}\times{/tex} a = 5
{tex}\Rightarrow{/tex} 6 - 3a = 5
{tex}\Rightarrow{/tex} 3a = 1
{tex}\Rightarrow{/tex} {tex}a = \frac { 1 } { 3 }{/tex}
Posted by Suhail Ghaffar 7 years, 4 months ago
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Posted by Ayush Kumar 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let the volume of the cylinder be 16 litres(a1).
Air removed by pump ={tex}{1 \over 4} \times 16 = 4litres{/tex}
Air present after first removal = 16-4=12 litres(a2)
Air again removed ={tex}{1 \over 4} \times 12 = 3litres{/tex}
Air present after second removal =12-3 =9 litres(a3)
The amount of air present in the cyinder is the series
16,12,9.....
a2-a1= 12-16 = -4
a3-a2=9-12= -3
Since the difference is not same .This is not A.P.
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