If a, b, c are in GP and 4a, 5b, 4c are in AP | vedclass

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(B) Given $a, b, c$ are in $GP$,we have $b^2 = ac$ .....$(1)$Given $4a, 5b, 4c$ are in $AP$,we have $2(5b) = 4a + 4c$,which simplifies to $10b = 4(a +

Vedclass · Accepted Answer

$73000$

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(B) Given $a, b, c$ are in $GP$,we have $b^2 = ac$ .....$(1)$Given $4a, 5b, 4c$ are in $AP$,we have $2(5b) = 4a + 4c$,which simplifies to $10b = 4(a + c)$,or $a + c = \frac{5b}{2}$ .....$(2)$We are given $a + b + c = 70$. Substituting $(2)$ into this,we get $\frac{5b}{2} + b = 70$,which implies $\frac{7b}{2} = 70$,so $b = 20$.Now,substitute $b = 20$ into $(2)$ to get $a + c = \frac{5(20)}{2} = 50$. Also,from $(1)$,$ac = b^2 = 20^2 = 400$.We have the quadratic equation $x^2 - (a+c)x + ac = 0$,which is $x^2 - 50x + 400 = 0$. Solving this,$(x - 40)(x - 10) = 0$,so ${a, c} = {10, 40}$.Thus,$a^3 + b^3 + c^3 = 10^3 + 20^3 + 40^3 = 1000 + 8000 + 64000 = 73000$.

If $a, b, c$ are in $GP$ and $4a, 5b, 4c$ are in $AP$ such that $a + b + c = 70$,then the value of $a^3 + b^3 + c^3$ is

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