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# Solve this

Question:

If $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in AP, prove that

(i) $\frac{(b+c)}{a}, \frac{(c+a)}{b}, \frac{(a+b)}{c}$ are in AP.

(ii) $\frac{(b+c-a)}{a}, \frac{(c+a-b)}{b}, \frac{(a+b-c)}{c}$ are in AP.

Solution:

(i) $\frac{(b+c)}{a}, \frac{(c+a)}{b}, \frac{(a+b)}{c}$ are in A.P.

To prove: $\frac{(b+c)}{a}, \frac{(c+a)}{b}, \frac{(a+b)}{c}$ are in A.P.

Given: $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.

Proof: $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.

If each term of an A.P. is multiplied by a constant, then the resulting sequence is also an A.P.

Multiplying the A.P. with (a + b + c )

$\Rightarrow \frac{(a+b+c)}{a}, \frac{(a+b+c)}{b}, \frac{(a+b+c)}{c}$ are also in A.P.

If a constant is subtracted from each term of an A.P., the resulting sequence is also an A.P.

Substracting the above A.P. with 1

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

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

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

Hence Proved

(ii) $\frac{(b+c-a)}{a}, \frac{(c+a-b)}{b}, \frac{(a+b-c)}{c}$ are in A.P.

To prove: $\frac{(b+c-a)}{a}, \frac{(c+a-b)}{b}, \frac{(a+b-c)}{c}$ are in A.P.

Given: $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.

Proof: $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.

If each term of an A.P. is multiplied by a constant, then the resulting sequence is also an A.P.

Multiplying the A.P. with ( a + b + c )

$\Rightarrow \frac{(a+b+c)}{a}, \frac{(a+b+c)}{b}, \frac{(a+b+c)}{c}$ are also in A.P.

If a constant is subtracted from each term of an A.P., the resulting sequence is also an A.P.

Substracting the above A.P. with 2

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

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

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

Hence Proved