Let a , b, c , d and p be any non zero distinct real numbers such that
(a2 + b2 + c2)p2 – 2(ab + bc + cd)p + (b2 + c2 + d2) = 0. Then :
Solution
(a<sup>2</sup> + b<sup>2</sup> + c<sup>2</sup>)p<sup>2</sup> – 2(ab + bc + cd)p + (b<sup>2</sup> + c<sup>2</sup> + d<sup>2</sup>) = 0
<br><br>$\Rightarrow$ (a<sup>2</sup>p<sup>2</sup>
+ 2abp + b<sup>2</sup>
) + (b<sup>2</sup>p<sup>2</sup>
+ 2bcp + c<sup>2</sup>
) + (c<sup>2</sup>
p<sup>2</sup>
+ 2cdp + d<sup>2</sup>) = 0
<br><br>$\Rightarrow$ (ab + b)<sup>2</sup>
+ (bp + c)<sup>2</sup>
+ (cp + d)<sup>2</sup>
= 0
<br><br><b>Note :</b> If sum of two or more positive quantity is zero then they are all zero.
<br><br>$\therefore$ ap + b = 0 and bp + c = 0 and cp + d = 0
<br><br>p = $- {b \over a}$ = $- {c \over b}$ = $- {d \over c}$
<br><br>or ${b \over a}$ = ${c \over b}$ = ${d \over c}$
<br><br>$\therefore$ a, b, c, d are in G.P.
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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