NDA 2011
Exam NDA
Year 2011
Subject Mathematics
Chapter Sequences and Series NDA 2011 · Mathematics · Sequences and Series
If (1)/(b-a)+(1)/(b-c)=(1)/(a)+(1)/(c) then a, b, c are in
If
1 b − a + 1 b − c = 1 a + 1 c \frac{1}{b-a}+\frac{1}{b-c}=\frac{1}{a}+\frac{1}{c} b − a 1 + b − c 1 = a 1 + c 1 then
a , b , c a, b, c a , b , c are in
Options A AP B GP C HP D None of the above
Step-by-step solution
Given 1 b − a + 1 b − c = 1 a + 1 c ⇒ 1 b − a − 1 c = 1 a − 1 b − c ⇒ c − b + a ( b − a ) c = b − c − a a ( b − c ) ⇒ c − b + a ( b − a ) c = − ( c − b + a ) a ( b − c ) \begin{aligned} & \text{ Given } \quad \frac{1}{b-a}+\frac{1}{b-c}=\frac{1}{a}+\frac{1}{c} \\ & \Rightarrow \quad \frac{1}{b-a}-\frac{1}{c}=\frac{1}{a}-\frac{1}{b-c} \\ & \Rightarrow \quad \frac{c-b+a}{(b-a) c}=\frac{b-c-a}{a(b-c)} \\ & \Rightarrow \quad \frac{c-b+a}{(b-a) c}=\frac{-(c-b+a)}{a(b-c)}\end{aligned} Given b − a 1 + b − c 1 = a 1 + c 1 ⇒ b − a 1 − c 1 = a 1 − b − c 1 ⇒ ( b − a ) c c − b + a = a ( b − c ) b − c − a ⇒ ( b − a ) c c − b + a = a ( b − c ) − ( c − b + a ) ⇒ a ( b − c ) = − ( b − a ) c ⇒ a b − a c = − b c + a c ⇒ b ( a + c ) = 2 a c ⇒ b = 2 a c a + c . \begin{aligned}
\Rightarrow & a(b-c) & =-(b-a) c \\
\Rightarrow & a b-a c & =-b c+a c \\
\Rightarrow & b(a+c) & =2 a c \\
\Rightarrow & b & =\frac{2 a c}{a+c} .
\end{aligned} ⇒ ⇒ ⇒ ⇒ a ( b − c ) ab − a c b ( a + c ) b = − ( b − a ) c = − b c + a c = 2 a c = a + c 2 a c . C l e a r l y Clearly Cl e a r l y a, b, c$ are in H.P.
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