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If a, b, c are in G.P. and a - b, c - a, b - c are in H.P., then the value of a + 4b + c  is equal to

 

Option: 1

1


Option: 2

0


Option: 3

2abc


Option: 4

b2+ac


Answers (1)

best_answer

Answer (2)

a,b,c are in G.P

Let r is common ratio

b=ar,c=ar^{2}

Now \left ( a-b \right ),\left ( c-a \right ),\left ( b-c \right ) are in H.P.

\frac{1}{a-b},\frac{1}{c-a},\frac{1}{b-c} are in A.P.

\Rightarrow \left ( b-c \right ), \left ( c-a \right ), \left ( a-b \right ), \left (b-c \right ), \left ( a-b \right ), \left ( c-a \right ) are in A.P.

Now  \left ( ar-ar^{2} \right )\left ( ar^{2}-a \right ),\left ( a-ar \right )\left ( ar-ar^{2} \right ), \left ( a-ar \right )\left ( ar^{2}-a \right ) are in A.P.

\left (r-r^{2} \right )\left (r^{2}-1 \right ), \left ( 1-r \right )\left (r-r^{2} \right ), \left ( 1-r \right )\left (r^{2}-1 \right ) are in A.P.

r\left ( 1-r \right )\left ( r-1 \right )\left ( r+r \right ),\left ( 1-r \right )r\left ( 1-r \right ),\left ( 1-r \right )\left ( r-1 \right ),\left ( r+1 \right ) are in A.P.

r\left ( 1+r\right ),-r\left ( r+1 \right ) are in A.P

r\left ( 1-r \right )\left ( r-1 \right )\left ( r+r \right ),\left ( 1-r \right )r\left ( 1-r \right )\left ( r-1 \right ),\left ( r+1 \right )are in A.P.

r\left ( 1+r \right ),-r,\left ( r+1 \right ) are in A.P.

Now,

-2r=r\left ( 1+r \right )+\left ( 1+r \right )

\Rightarrow -2r=\left ( 1+r \right )\left ( 1+r \right )

\Rightarrow r^{2}+4r+1=0\; \; \; \; \; \; \; \; \; \; \; ...(i)

Now value of  a+4b+c

a+4\left ( ar \right )+\left ( ar^{2} \right )=a\left ( 1+4r+r^{2} \right )

                                       =a\times 0=0

Posted by

Deependra Verma

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