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

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(B) Since $a, b, c$ are in $GP$,we have $b^2 = ac$ .....$(1)$Since $4a, 5b, 4c$ are in $AP$,the middle term is the average of the other two:$2(5b) = 4

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$73000$

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(B) Since $a, b, c$ are in $GP$,we have $b^2 = ac$ .....$(1)$Since $4a, 5b, 4c$ are in $AP$,the middle term is the average of the other two:$2(5b) = 4a + 4c$$10b = 4(a + c)$$a + c = \frac{5b}{2}$ .....$(2)$We are given $a + b + c = 70$. Substituting $(2)$ into this:$\frac{5b}{2} + b = 70$$\frac{7b}{2} = 70$$b = 20$Now,substitute $b = 20$ into $(2)$:$a + c = \frac{5(20)}{2} = 50$From $(1)$,$ac = b^2 = 20^2 = 400$.We have $a + c = 50$ and $ac = 400$. These are roots of the quadratic equation $x^2 - 50x + 400 = 0$:$(x - 40)(x - 10) = 0$So,${a, c} = {10, 40}$.Finally,calculate $a^3 + b^3 + c^3$:$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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