Question

If a b c are in ap and a b d are in gp then show that a, a-b d-c

Answer 1

Answer:

Step-by-step explanation:

2b=a+c

b^2=ad

From first multiplying a

2ab=a^2+ac

a^2-2ab=ac

Adding b^2

a^2+b^2-2ab=ac+b^2

(a+b)^2=ac+ad

Answer 2

Answer:

$\green { a, (a-b) \:and \: (d-c) \:are \: in \: G.P}$

Step-by-step explanation:

$Given \: a,b \:and \: are \: in \:A.P$

$b - a = c - b\: ( common \: difference )$

$\implies 2b = a + c$

/* Multiply each term by a , we get

$\implies 2ab = a^{2} + ac$

$\implies 2ab + b^{2} = a^{2} + b^{2} + ac$

$\implies b^{2} = a^{2} + b^{2} -2ab + ac$

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

$Given \: a,b \:and \: d \: are \: G.P$

$\frac{b}{a} = \frac{d}{b} \: ( common \:ratio )$

$\implies b^{2} = ad \: ---(2)$

$\implies ( a - b )^{2} + ac = ad \:[From \:(1)]$

$\implies ( a - b )^{2} + ac = ad$

$\implies ( a - b )^{2} = ad - ac$

$\implies (a-b)^{2} = a(d-c)$

$\implies \frac{(a-b)}{a} = \frac{(d-c)}{(a-b)}$

Therefore.,

$\green { a, (a-b) \:and \: (d-c) \:are \: in \: G.P}$

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