If a, b, c are in G.P.,then | vedclass

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(B) If $a, b, c$ are in $G.P.$,then the common ratio is constant,so $b/a = c/b$,which implies $b^2 = ac$.Now,consider the expression $a(b^2 + c^2)$:$a

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$a(b^2 + c^2) = c(a^2 + b^2)$

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(B) If $a, b, c$ are in $G.P.$,then the common ratio is constant,so $b/a = c/b$,which implies $b^2 = ac$.Now,consider the expression $a(b^2 + c^2)$:$a(b^2 + c^2) = ab^2 + ac^2$Since $b^2 = ac$,substitute $b^2$ with $ac$:$a(ac) + ac^2 = a^2c + ac^2 = ac(a + c)$.Now,consider the expression $c(a^2 + b^2)$:$c(a^2 + b^2) = ca^2 + cb^2$Since $b^2 = ac$,substitute $b^2$ with $ac$:$ca^2 + c(ac) = ca^2 + ac^2 = ac(a + c)$.Since both expressions equal $ac(a + c)$,we have $a(b^2 + c^2) = c(a^2 + b^2)$.Alternatively,let $a=1, b=2, c=4$. Then $a(b^2 + c^2) = 1(4 + 16) = 20$ and $c(a^2 + b^2) = 4(1 + 4) = 20$. Thus,option $B$ is correct.

If $a, b, c$ are in $G.P.$,then

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