Let $a_1, a_2, a_3, \ldots$ be a G.P. of increasing positive terms such that $a_2 \cdot a_3 \cdot a_4=64$ and $a_1+a_3+a_5=\frac{813}{7}$. Then $a_3+a_5+a_7$ is equal to :

Question

Accepted Answer

B. 1. Let the G.P. have first term $a$ and common ratio $r$. Use the given product condition to find $a$.

$$
a_2\cdot a_3\cdot a_4=\frac{a}{r}\cdot a\cdot ar=a^3=64 \Rightarrow a=4
$$

2. Use the sum condition $a_1+a_3+a_5=\frac{813}{7}$, writing these in terms of $a$ and $r$.

$$
\frac{a}{r^2}+a+ar^2=\frac{813}{7} \Rightarrow \frac{4}{r^2}+4+4r^2=\frac{813}{7}
$$

3. Let $t=r^2$ and clear denominators to form a quadratic in $t$.

$$
28t^2-785t+28=0
$$

4. Factor this quadratic.

$$
(28t-1)(t-28)=0 \Rightarrow t=28 	ext{ or } t=\frac{1}{28}
$$

Since the G.P. is increasing, $r^2=t=28$ is the valid choice.

5. Compute $a_3+a_5+a_7=a+ar^2+ar^4$, factoring out $r^2$ to reuse the earlier sum.

$$
a_3+a_5+a_7=r^2\left(a+\frac{a}{r^2}+ar^2ight) = r^2\left(a_1+a_3+a_5ight)
$$

6. Substitute the known values.

$$
a_3+a_5+a_7=28	imes\frac{813}{7}=3252
$$

Hence, the answer is **Option B: 3252**.

Let a1,a2,a3,…a_1, a_2, a_3, \ldotsa1​,a2​,a3​,… be a G.P. of increasing positive terms such that a2⋅a3⋅a4=64a_2 \cdot a_3 \cdot a_4=64a2​⋅a3​⋅a4​=64 and a1+a3+a5=8137a_1+a_3+a_5=\frac{813}{7}a1​+a3​+a5​=7813​. Then a3+a5+a7a_3+a_5+a_7a3​+a5​+a7​ is equal to :

Let $a_1, a_2, a_3, \ldots$ be a G.P. of increasing positive terms such that $a_2 \cdot a_3 \cdot a_4=64$ and $a_1+a_3+a_5=\frac{813}{7}$ . Then $a_3+a_5+a_7$ is equal to :