Sequence and Series — JEE Main practice

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.

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

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.

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

6. Substitute the known values.

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

Hence, the answer is Option B: 3252.

1. Since $a_1, \frac{a_2}{2}, \frac{a_3}{4}, \ldots$ is a G.P. with common ratio $\frac{1}{\sqrt{2}}$, the ratio between successive terms of this new sequence equals $\frac{1}{\sqrt{2}}$.

$$\frac{a_2}{2a_1}=\frac{a_3}{2a_2}=\ldots=\frac{a_{10}}{2a_9}=\frac{1}{\sqrt{2}}$$

2. This shows that $a_1, a_2, a_3, \ldots, a_{10}$ themselves form a G.P. with common ratio $\sqrt{2}$, since each ratio $\frac{a_{k+1}}{a_k}=2\times\frac{1}{\sqrt{2}}=\sqrt{2}$.

3. Apply the finite G.P. sum formula to the 10 terms $a_1,\ldots,a_{10}$.

$$\sum_{i=1}^{10}a_i=\frac{a_1\left((\sqrt{2})^{10}-1\right)}{\sqrt{2}-1}=62$$

4. Since $(\sqrt{2})^{10}=2^5=32$, substitute and solve for $a_1$.

$$\frac{a_1(32-1)}{\sqrt{2}-1}=62 \Rightarrow \frac{31a_1}{\sqrt{2}-1}=62$$

5. Solve for $a_1$.

$$a_1=\frac{62(\sqrt{2}-1)}{31}=2(\sqrt{2}-1)$$

Hence, the answer is Option B: $2(\sqrt{2}-1)$.

1. Expand the left side of the equation, since each term is a product of two linear expressions in $x$.

Grouping the $x^2$ terms, the linear terms in $x$, and the constant terms separately gives:

$$nx^2+2x(1+3+5+\ldots+(2n-1))+\sum_{r=1}^n(2r-2)(2r)=\frac{8n}{3}$$

2. Use the identity $1+3+5+\ldots+(2n-1)=n^2$ for the linear coefficient, and evaluate the constant sum using standard formulas.

$$\sum_{r=1}^n(2r-2)(2r)=4\cdot\frac{n(n+1)(2n-2)}{6}$$

3. Substitute both results back into the equation.

$$nx^2+2n^2x+4\cdot\frac{n(n+1)(2n-2)}{6}=\frac{8n}{3}$$

4. Divide through by $n$ and simplify to a cleaner quadratic in $x$.

$$3x^2+6nx+4(n^2-1)=8$$

5. Substitute $n=3$ (testing the given options) into this equation.

$$3x^2+18x+24=0$$

6. Solve this quadratic; dividing by 3 gives $x^2+6x+8=0$, which factors as $(x+2)(x+4)=0$.

$$x=-2 \text{ or } x=-4$$

Since $-2$ and $-4$ are two consecutive even integers, $n=3$ satisfies the condition.

Hence, the answer is Option A: 3.