JEE MAIN - Mathematics (2020 - 6th September Morning Slot - No. 4)
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 :
(a2 + b2 + c2)p2 – 2(ab + bc + cd)p + (b2 + c2 + d2) = 0. Then :
a, c, p are in G.P.
a, b, c, d are in G.P.
a, b, c, d are in A.P.
a, c, p are in A.P.
Explanation
(a2 + b2 + c2)p2 – 2(ab + bc + cd)p + (b2 + c2 + d2) = 0
$$ \Rightarrow $$ (a2p2 + 2abp + b2 ) + (b2p2 + 2bcp + c2 ) + (c2 p2 + 2cdp + d2) = 0
$$ \Rightarrow $$ (ab + b)2 + (bp + c)2 + (cp + d)2 = 0
Note : If sum of two or more positive quantity is zero then they are all zero.
$$ \therefore $$ ap + b = 0 and bp + c = 0 and cp + d = 0
p = $$ - {b \over a}$$ = $$ - {c \over b}$$ = $$ - {d \over c}$$
or $${b \over a}$$ = $${c \over b}$$ = $${d \over c}$$
$$ \therefore $$ a, b, c, d are in G.P.
$$ \Rightarrow $$ (a2p2 + 2abp + b2 ) + (b2p2 + 2bcp + c2 ) + (c2 p2 + 2cdp + d2) = 0
$$ \Rightarrow $$ (ab + b)2 + (bp + c)2 + (cp + d)2 = 0
Note : If sum of two or more positive quantity is zero then they are all zero.
$$ \therefore $$ ap + b = 0 and bp + c = 0 and cp + d = 0
p = $$ - {b \over a}$$ = $$ - {c \over b}$$ = $$ - {d \over c}$$
or $${b \over a}$$ = $${c \over b}$$ = $${d \over c}$$
$$ \therefore $$ a, b, c, d are in G.P.
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