Question -
Answer -
(i)┬а(ab тАУ cd) / (b2┬атАУc2) = (a + c) / b
Given that a, b, c arein GP.
By using the propertyof geometric mean,
b2┬а=ac
bc = ad
c2┬а=bd
Let us consider LHS:(ab тАУ cd) / (b2┬атАУ c2)
(ab тАУ cd) / (b2┬атАУc2) = (ab тАУ cd) / (ac тАУ bd)
= (ab тАУ cd)b / (ac тАУbd)b
= (ab2┬атАУbcd) / (ac тАУ bd)b
= [a(ac) тАУ c(c2)]/ (ac тАУ bd)b
= (a2c тАУ c3)/ (ac тАУ bd)b
= [c(a2┬атАУc2)] / (ac тАУ bd)b
= [(a+c) (ac тАУ c2)]/ (ac тАУ bd)b
= [(a+c) (ac тАУ bd)] /(ac тАУ bd)b
= (a+c) / b
= RHS
тИ┤┬аLHS = RHS
Hence proved.
(ii)┬а(a + b + c + d)2┬а=(a + b)2┬а+ 2(b + c)2┬а+ (c + d)2
Given that a, b, c arein GP.
By using the propertyof geometric mean,
b2┬а=ac
bc = ad
c2┬а=bd
Let us consider RHS:(a + b)2┬а+ 2(b + c)2┬а+ (c + d)2
Let us expand
(a + b)2┬а+2(b + c)2┬а+ (c + d)2┬а= (a + b)2┬а+2 (a+b) (c+d) + (c+d)2
= a2┬а+b2┬а+ 2ab + 2(c2┬а+ b2┬а+ 2cb) + c2┬а+d2┬а+ 2cd
= a2┬а+b2┬а+ c2┬а+ d2┬а+ 2ab + 2(c2┬а+b2┬а+ 2cb) + 2cd
= a2┬а+b2┬а+ c2┬а+ d2┬а+ 2(ab + bd + ac +cb +cd) [Since, c2┬а= bd, b2┬а= ac]
You can visualize theabove expression by making separate terms for (a + b + c)2┬а+ d2┬а+2d(a + b + c) = {(a + b + c) + d}2
тИ┤┬аRHS = LHS
Hence proved.
(iii)┬а(b + c) (b + d) = (c +a) (c + d)
Given that a, b, c arein GP.
By using the propertyof geometric mean,
b2┬а=ac
bc = ad
c2┬а=bd
Let us consider LHS:(b + c) (b + d)
Upon expansion we get,
(b + c) (b + d) = b2┬а+bd + cb + cd
= ac + c2┬а+ad + cd [by using property of geometric mean]
= c (a + c) + d (a +c)
= (a + c) (c + d)
= RHS
тИ┤┬аLHS = RHS
Hence proved.