If 2a-b-ca- a-2b-cb and a-b-2cc are in A-P- then

Given : 2a+b+ca, a+2b+cb,a+b+2cc are in A.P., so 2a+b+ca 1, a+2b+cb 1,a+b+2cc 1 are in A.P. ⇒a+b+ca, a+b+cb,a+b+cc are in A.P. ⇒1a, 1b,1c are in A.P. ...

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Byju's Answer

Standard XIII

Mathematics

Sum of n Terms

If 2a+b+ca, a...

Question

If $\frac{2 a + b + c}{a}, \frac{a + 2 b + c}{b}$ and $\frac{a + b + 2 c}{c}$ are in A.P., then

a, b, c

are in A.P.

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\frac{1}{a}, \frac{1}{b}, \frac{1}{c}

are in A.P.

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a, c, b

are in A.P.

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\frac{1}{a}, \frac{1}{c}, \frac{1}{b}

are in A.P.

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Solution

The correct option is B $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.
Given : $\frac{2 a + b + c}{a}, \frac{a + 2 b + c}{b}, \frac{a + b + 2 c}{c}$ are in A.P., so

$\frac{2 a + b + c}{a} - 1, \frac{a + 2 b + c}{b} - 1, \frac{a + b + 2 c}{c} - 1$ are in A.P.
$\Rightarrow \frac{a + b + c}{a}, \frac{a + b + c}{b}, \frac{a + b + c}{c}$ are in A.P.
$\Rightarrow \frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.

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If 2a+b+ca,a+2b+cb and a+b+2cc are in A.P., then

The correct option is B 1a,1b,1c are in A.P.Given : 2a+b+ca,a+2b+cb,a+b+2cc are in A.P., so 2a+b+ca−1,a+2b+cb−1,a+b+2cc−1 are in A.P. ⇒a+b+ca,a+b+cb,a+b+cc are in A.P. ⇒1a,1b,1c are in A.P.

If $\frac{2 a + b + c}{a}, \frac{a + 2 b + c}{b}$ and $\frac{a + b + 2 c}{c}$ are in A.P., then