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Ask QuestionPosted by Krishna Priya Vijayan 7 years, 3 months ago
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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 AP. Then, in general the mth and nth terms can be written as
Tm = a + (m - 1) d and Tn = a + (n - 1)d respectively.
According to the question,we are given that,
(m.Tm) = (n.Tn)
{tex}\Rightarrow{/tex} m.{a + (m - 1)d} = n.{a + (n - 1)d}
{tex}\Rightarrow{/tex} a.(m - n) + {(m2 - n2) - (m - n)} . d = 0
{tex}\Rightarrow{/tex} (m - n).{a + (m + n - 1)}d.
{tex}\Rightarrow{/tex} (m - n).Tm+n = 0
{tex}\Rightarrow{/tex}Tm+n = 0 [{tex}\because{/tex} (m-n){tex}\neq{/tex}0].
Hence, the (m + n)th term is zero.
Posted by Ranveer Singh 7 years, 3 months ago
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Posted by Roshini Rajesh 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Given polynomial is p(x) = 2x3 - 6x2 + 5x - 7
Let {tex}\alpha{/tex} = (a - b), {tex} \beta{/tex} = a and {tex}\gamma{/tex} = ( a + b)
Now, {tex} \alpha + \beta + \gamma{/tex} = {tex}-\frac{(-6)}{2}{/tex} = 3
⇒ ( a - b) + a + (a + b ) = 3
⇒ a - b + a + a + b = 3
⇒ 3a = 3
⇒ a = 3/3
⇒ a = 1
So the value of a in given polynomial is 1.
Posted by Bharathi Siva 7 years, 3 months ago
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Sia ? 6 years, 6 months ago

We have, AB = 16 cm
{tex} \therefore{/tex} AL = BL = 8 cm [Perpendicular from the centre of the circle to a chord bisects it]
In {tex}\triangle{/tex}OLB, we have,
OB2 = OL2 + LB2
{tex} \Rightarrow{/tex} 102 = OL2 + 82
{tex} \Rightarrow{/tex} OL2 = 100 - 64 = 36
{tex} \Rightarrow{/tex} OL = 6 cm
Let PL = x and PB = y. Then, OP = (x + 6) cm.
Now, {tex} \angle PBO = 90^\circ{/tex} as tangent makes a right angle with the radius of the circle at the point of contact.
In {tex}\triangle{/tex}'s PLB and OBP, we have,
PB2 = PL2 + BL2 and OP2 = OB2 + PB2
{tex} \Rightarrow{/tex} y2 = x2 + 64 and (x + 6)2 = 100 + y2 [Substituting the value of y2 in second equation]
{tex} \Rightarrow{/tex} (x + 6)2 = 100 + x2 + 64
{tex} \Rightarrow{/tex} 12x = 128
{tex}\Rightarrow x = \frac{{32}}{3}cm{/tex}
{tex} \therefore{/tex} y2 = x2 + 64
{tex} \Rightarrow {y^2} = {\left( {\frac{{32}}{3}} \right)^2} + 64 = \frac{{1600}}{9}{/tex}
{tex} \Rightarrow y = \frac{{40}}{3}cm{/tex}
Therefore, {tex} PA = PB = \frac{{40}}{3}cm{/tex}
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Dipesh Goyal 7 years, 3 months ago
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