If a, b, and c are in both Arithmetic Progres | vedclass

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(D) Given that $a, b, c$ are in $AP$,we have $\frac{a + c}{2} = b$,which implies $a + c = 2b$. Given that $a, b, c$ are in $GP$,we have $b^2 = ac$. Sq

Vedclass · Accepted Answer

$a = b = c$

Vedclass · Answer

(D) Given that $a, b, c$ are in $AP$,we have $\frac{a + c}{2} = b$,which implies $a + c = 2b$. Given that $a, b, c$ are in $GP$,we have $b^2 = ac$. Squaring the first equation: $(a + c)^2 = (2b)^2 = 4b^2$. Substituting $b^2 = ac$ into the equation: $(a + c)^2 = 4ac$. This simplifies to $(a + c)^2 - 4ac = 0$,which is $(a - c)^2 = 0$. Therefore,$a = c$. Substituting $a = c$ into $a + c = 2b$,we get $c + c = 2b$,so $2c = 2b$,which means $b = c$. Thus,$a = b = c$.

If $a, b,$ and $c$ are in both Arithmetic Progression $(AP)$ and Geometric Progression $(GP)$,then........

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