If (a+b)/(2) , b, (b+c)/(2) are in HP, then which one of the following is correct?

Correct answer: B. a, b, c are in GP (a+b)/(2), b, (b+c)/(2) are in H.P (2)/(b)=(2)/(a+b)+(2)/(b+c)=(2(a+2 b+c))/((a+b)(b+c)) aligned & (a+b)(b+c)=b(a+2 b+c) & a b+c a+b^2+b c=a b+2 b^2+b c & b^2=c a aligned a, b, c are in G.P.

NDA 2022

Exam: NDA
Year: 2022
Subject: Mathematics
Chapter: Sequences and Series

NDA 2022 · Mathematics · Sequences and Series

If (a+b)/(2) , b, (b+c)/(2) are in HP, then which one of the...

\frac{a+b}{2}

, b,

\frac{b+c}{2} are in HP, then which one of the following is correct?

Options

A $a, b, c$ are in AP
B $a, b, c$ are in GP
C $a+b, b+c, c+a$ are in GP
D $a+b, b+c, c+a$ are in AP

Correct answer

B. $a, b, c$ are in GP

Step-by-step solution

\because \frac{a+b}{2}, b, \frac{b+c}{2}

are in H.P

\therefore \quad \frac{2}{b}=\frac{2}{a+b}+\frac{2}{b+c}=\frac{2(a+2 b+c)}{(a+b)(b+c)}

\begin{aligned} & \Rightarrow \quad(a+b)(b+c)=b(a+2 b+c) \\ & \Rightarrow \quad a b+c a+b^2+b c=a b+2 b^2+b c \\ & \Rightarrow \quad b^2=c a\end{aligned}

\therefore \quad a, b, c

are in G.P.

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