Question

If 0<a<b<c and a, b c are in HP, then log(a+c), log(c-a), log(a+c-2b) are in

(a) GP

(b) HP

(c) AP

(d) none of these

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Answer 1

Answer:

d none of these pls follow and like kijiya

Answer 2

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

Given that

$\rm :\longmapsto\:0 < a< b< c$

and

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

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

Now, Consider,

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

On substituting the value of b, we get

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

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

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

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

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

We know,

$\red{ \boxed{ \sf{ \:logx + logy = log(xy)}}}$

So, using this, we get

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

$\rm \:  =  \: log {(c - a)}^{2}$

$\rm \:  =  \: 2 \: log(c - a)$

$\bf\implies \:log(a + c) + log(a + c - 2b) = 2log(c - a)$

$\bf\implies \:log(a+c), log(c-a), log(a+c-2b) \: are \: in \: AP$

• Hence, Option (c) is correct.