If the 2^{nd}, 5^{th}, text{and } 9^{th} term | vedclass

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(D) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.The $2^{nd}$ term is $a + d$.The $5^{th}$ term is $a + 4d$.The $9^{th}$ t

Vedclass · Accepted Answer

$\frac{4}{3}$

Vedclass · Answer

(D) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.The $2^{nd}$ term is $a + d$.The $5^{th}$ term is $a + 4d$.The $9^{th}$ term is $a + 8d$.Since these terms are in $G.P.$, let the common ratio be $r$.Thus, $(a + 4d) = (a + d)r$ and $(a + 8d) = (a + d)r^2$.From the first equation, $a + 4d = ar + dr \implies a(1-r) = d(r-4) \implies a = \frac{d(r-4)}{1-r}$.Substituting this into the second equation: $\frac{d(r-4)}{1-r} + 8d = r^2 \left( \frac{d(r-4)}{1-r} + d ight)$.Dividing by $d$ (since $d 
eq 0$ for a non-constant $A.P.$):$\frac{r-4 + 8 - 8r}{1-r} = r^2 \left( \frac{r-4 + 1-r}{1-r} ight)$.$\frac{4-7r}{1-r} = r^2 \left( \frac{-3}{1-r} ight)$.$4 - 7r = -3r^2 \implies 3r^2 - 7r + 4 = 0$.$(3r - 4)(r - 1) = 0$.Since the $A.P.$ is non-constant, $d 
eq 0$, which implies $r 
eq 1$.Therefore, $r = \frac{4}{3}$.

If the $2^{nd}, 5^{th}, \text{and } 9^{th}$ terms of a non-constant $A.P.$ are in $G.P.$, then the common ratio of this $G.P.$ is:

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