Question

If a, b, c are in HP, then prove that log(a+c) + log(a-2b+c)=2log(a-c)​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: a,b,c \: are \: in \: HP$

$\rm\implies \:\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c} \: are \: in \: AP$

We know, Three numbers x, y, z are in AP iff 2y = x + z

So, using this result, we have

$\rm \: \dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}$

$\rm \: \dfrac{2}{b} = \dfrac{c + a}{ac}$

$\rm\implies \:b = \dfrac{2ac}{a + c} - - - (1)$

Now, Consider

$\rm \: log(a + c) + log(a - 2b + c)$

We know

$\boxed{\tt{ \: \: logx \: + \: logy \: = \: logxy \: }}$

So, using this result, we get

$\rm \: = \: log\bigg((a + c)(a - 2b + c) \bigg)$

can be rewritten as

$\rm \: = \: log\bigg((a + c)(a + c- 2b) \bigg)$

$\rm \: = \: log\bigg( {(a + c)}^{2} - 2b(a + c) \bigg)$

On substituting the value from equation (1), we get

$\rm \: = \: log\bigg( {(a + c)}^{2} - \dfrac{4ac}{a + c} (a + c) \bigg)$

$\rm \: = \: log\bigg( {(a + c)}^{2} - 4ac \bigg)$

$\rm \: = \: log\bigg( {a}^{2} + {c}^{2} + 2ac - 4ac \bigg)$

$\rm \: = \: log\bigg( {a}^{2} + {c}^{2} - 2ac \bigg)$

$\rm \: = \: log\bigg( {(a - c)}^{2} \bigg)$

We know,

$\boxed{\tt{ \: \: log {x}^{y} \: = \: y \: logx \: }}$

So, using this result ,we get

$\rm \: = \: 2log(a - c)$

Hence,

$\rm\implies \:\boxed{\sf{ log(a + c) + log(a - 2b + c) = 2log(a - c)}}$

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ADDITIONAL INFORMATION

If a and b are two positive real numbers then

1. Arithmetic mean (AM) between a and b is given by

$\boxed{\tt{ \: AM \: = \: \frac{a + b}{2} \: }}$

2. Geometric mean (GM) between a and b is given by

$\boxed{\tt{ \: GM \: = \: \sqrt{ab} \: }}$

3. Harmonic mean (HM) between a and b is given by

$\boxed{\tt{ \: HM \: = \: \frac{2ab}{a + b} \: }}$

4. Relationship between Arithmetic mean, Geometric mean and Harmonic mean

$\boxed{\tt{ \: AM \geqslant GM \geqslant HM \: }}$

$\boxed{\tt{ \: {GM}^{2} = AM \times HM \: }}$

$\boxed{\tt{ \: AM,GM,HM \: are \: in \: GP \: }}$

Answer 2

Question:-

If a, b, c are in H.P, then prove that log(a + c) +

log(a – 2b + c) = 2log (a – c).

Given:-

• a, b, c are in H.P.

To Find:-

• log(a + c) + log(a – 2b + c) = 2log (a – c).

Solution:-

We have: a, b, c are in H.P.

$\sf \large \therefore \frac{2}{b} = \frac{1}{a} + \frac{1}{c}$

$\sf \large⇒ \frac{2}{b} = \frac{a + c}{ac}$

$\sf \large⇒b = \frac{2ac}{a + c} \: \: \: ...(1)$

L.H.S. = log(a + c) + log(a – 2b + c).

$\sf \large = log \bigg \{(a + c).(a + c - 2b) \bigg \}$

$\sf \large = log \bigg \{(a + c)(a + c - \frac{4ac}{(a + c)} \bigg \} \: \: [From (1)]$

$\sf \large = log \bigg \{((a + c) {}^{2} - 4ac) \bigg \}$

$\sf \large = log(a - c) {}^{2}$

$\sf \large = 2log(a - c) = R.H.S.$

Answer:-

Hence, Proved that

log(a + c) + log(a – 2b + c) = 2log (a – c).

Hope you have satisfied. ⚘