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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 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
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Sia ? 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
since 3 is the root of the equation, x = 3 must satisfy the equation.
Applying x = 3 in the equation {tex}{x^2} - 5x + 6 = 0{/tex}
gives, {tex}{\left( 3 \right)^2} - 5 ( 3) + 6 = 0{/tex}
{tex} \Rightarrow {/tex}{tex}9 - 15 + 6 = 0{/tex}
{tex} \Rightarrow {/tex}{tex}15 - 15 = 0{/tex}
{tex} \Rightarrow {/tex}{tex}0 = 0{/tex}
{tex} \Rightarrow {/tex}L.H.S. = R.H.S.
Hence, {tex}{x^2} - 5x + 6 = 0{/tex} is a required equation which has 3 as root.
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Shrish Kashyap 7 years, 5 months ago
0Thank You