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If $a , b , c , d$ are in Arithmetic progession (AP,) and $a , c , d$ are in Geometric Progession (GP) then Prove that ${a}^{2} - {d}^{2}$ = $3 \left({b}^{2} - a d\right)$

Aug 12, 2018

For, $a , b , c , d$ in A.P we can write

$2 b = a + c \ldots .1$ (as, $c - b = b - a$)

$2 c = b + d \ldots .2$ (as, $d - c = c - b$)

Doing $1 + 2$,

$\left(b + c\right) = \left(a + d\right) \ldots .3$

Doing $1 - 2$

$3 \left(b - c\right) = \left(a - d\right) \ldots .4$

And, for $a , c , d$ in G.P we get,
${c}^{2} = a d \ldots 5$ (as, $\frac{c}{a} = \frac{d}{c}$)

L.H.S $=$

${a}^{2} - {d}^{2}$

$= \left(a + d\right) \left(a - d\right)$

=(b+c)×3(b-c) (putting values from $3$ & $4$)

$= 3 \left({b}^{2} - {c}^{2}\right)$

$= 3 \left({b}^{2} - a d\right)$ (from $5$)

$=$R.H.S

Aug 12, 2018

Kindly refer to The Explanation.

Explanation:

Given that, $a , b , c , d$ are in AP.

$\therefore b \text{ is the AM of "a and c}$.

$\therefore b = \frac{a + c}{2} , \mathmr{and} , a + c = 2 b \ldots \ldots \ldots . \left({\ast}_{1}\right)$.

Similarly, $b + d = 2 c \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left({\ast}_{2}\right)$.

Then, $\left({\ast}_{1}\right) + \left({\ast}_{2}\right) \Rightarrow \left(a + d\right) + \left(b + c\right) = 2 \left(b + c\right) ,$

$\mathmr{and} , \left(a + d\right) = \left(b + c\right) \ldots \ldots \ldots \ldots \ldots \left({\star}_{1}\right)$.

Also, $\left({\ast}_{1}\right) - \left({\ast}_{2}\right) \Rightarrow \left(a - d\right) - \left(b - c\right) = 2 \left(b - c\right) ,$

$\mathmr{and} , \left(a - d\right) = 3 \left(b - c\right) \ldots \ldots \ldots \ldots \left({\star}_{2}\right)$.

Utilising $\left({\star}_{1}\right) \mathmr{and} \left({\star}_{2}\right)$, we have,

${a}^{2} - {d}^{2} = \left(a + d\right) \left(a - d\right) = \left(b + c\right) \left\{3 \left(b - c\right)\right\} ,$

$i . e . , {a}^{2} - {d}^{2} = 3 \left({b}^{2} - {c}^{2}\right)$.

Here, we use the fact that $a , c , d$ are in GP, so that,

$\frac{c}{a} = \frac{d}{c} , \mathmr{and} , {c}^{2} = a d$.

Therefore, ${a}^{2} - {d}^{2} = 3 \left({b}^{2} - {c}^{2}\right) = 3 \left({b}^{2} - a d\right)$, as desired!