If a₁ ( 0), a₂, a₃, a₄, a₅ are in a G.P., a₂ + a₄ = 2a₃ + 1 and 3a₂ + a₃ = 2a₄, then a₂ + a₄ + 2a₅

Q: Wait, why can we just divide by $ar$? What if $a$ or $r$ is zero?

That is a fantastic question! In a Geometric Progression, if the first term $a$ were zero, all terms would be zero, which makes the sequence trivial and usually excluded in such problems. Similarly, if $r$ were zero, the terms after $a_1$ would be zero, which would contradict the structure of the given equations. Since we are given $a_1 > 0$, we know $a \neq 0$, and for the equations to hold, $r$ cannot be zero. Thus, dividing by $ar$ is perfectly safe and mathematically sound.

Q: Why do we reject the negative root $r = -1$?

We don't reject it just because it's negative; we reject it because of the constraint $a_1 > 0$. When we substitute $r = -1$ into the first equation, we get an equation that forces $a$ to be $-1/4$. Since the problem explicitly states $a_1 (> 0)$, a negative value for $a$ is physically impossible within the context of this specific G.P. Always check your constraints!

Q: Can we use the first equation $a_2 + a_4 = 2a_3 + 1$ first?

You certainly can, but it is much more algebraically taxing. The first equation contains a constant term ($+1$), which prevents you from simply dividing by $ar$ to find $r$. You would end up with an equation involving both $a$ and $r$, forcing you to substitute $a$ in terms of $r$ or vice-versa. Always look for the 'cleanest' equation—the one that allows you to isolate a variable most easily—to start your journey.

Question

If $a_1 (> 0), a_2, a_3, a_4, a_5$ are in a G.P., $a_2 + a_4 = 2a_3 + 1$ and $3a_2 + a_3 = 2a_4$, then $a_2 + a_4 + 2a_5$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we are going to unravel a problem that is less about brute-force calculation and more about the beautiful, logical rhythm of a Geometric Progression (G.P.).

Imagine a sequence of numbers where each step is a multiplication by a constant factor, the common ratio $r$. We are given five terms: $a_1, a_2, a_3, a_4, a_5$.

Let us define them in the language of algebra: $a, ar, ar^2, ar^3, ar^4$. Our journey begins with a crucial constraint: $a_1 = a > 0$. Never ignore these constraints; they are the guardrails that keep us from falling into mathematical traps.

### The Quadratic Hunt

We have two equations to guide us. The second one, $3a_2 + a_3 = 2a_4$, is our golden key because it is homogeneous.

Let us substitute our G.P. terms into this:
$$3(ar) + (ar^2) = 2(ar^3)$$

Since $a > 0$ and $r 
eq 0$, we can divide the entire equation by $ar$. This leaves us with a clean, elegant quadratic equation:
$$3 + r = 2r^2 \quad \Rightarrow \quad 2r^2 - r - 3 = 0$$

Factoring this is like solving a puzzle: we look for two numbers that multiply to $-6$ and add to $-1$. Those are $-3$ and $2$.

The equation becomes:
$$2r^2 - 3r + 2r - 3 = 0 \quad \Rightarrow \quad r(2r - 3) + 1(2r - 3) = 0$$

This leads us to $(2r - 3)(r + 1) = 0$. We have two candidates for our ratio: $r = \frac{3}{2}$ and $r = -1$.

### The Filter of Reality

Now, we must test these candidates against our first equation: $a_2 + a_4 = 2a_3 + 1$.

If we test $r = -1$, we find:
$$a(-1) + a(-1)^3 = 2a(-1)^2 + 1 \quad \Rightarrow \quad -2a = 2a + 1 \quad \Rightarrow \quad -4a = 1$$

This gives us $a = -\frac{1}{4}$. Since our constraint was $a > 0$, we must reject this value as it is a ghost solution.

Now, let us test $r = \frac{3}{2}$. Substituting this into the first equation:
$$a\left(\frac{3}{2}ight) + a\left(\frac{3}{2}ight)^3 = 2a\left(\frac{3}{2}ight)^2 + 1$$

Expanding this, we get:
$$\frac{3a}{2} + \frac{27a}{8} = 2a\left(\frac{9}{4}ight) + 1$$

With a common denominator of $8$, this becomes:
$$\frac{12a + 27a}{8} = \frac{36a}{8} + 1 \quad \Rightarrow \quad \frac{39a - 36a}{8} = 1$$

This leads us to $\frac{3a}{8} = 1$, or $a = \frac{8}{3}$. This is positive, and thus, our valid starting point.

### Final Calculation

We have our parameters: $a = \frac{8}{3}$ and $r = \frac{3}{2}$. The question asks for the value of $a_2 + a_4 + 2a_5$.

Let us express this in terms of $a$ and $r$:
$$ar + ar^3 + 2ar^4$$

Substituting our values:
$$ar = \frac{8}{3} 	imes \frac{3}{2} = 4$$
$$ar^3 = \frac{8}{3} 	imes \frac{27}{8} = 9$$
$$2ar^4 = 2 	imes \frac{8}{3} 	imes \frac{81}{16} = 27$$

Adding these together:
$$4 + 9 + 27 = 40$$

We have arrived at our destination. The final answer is **40**.

Animated Solution for Mathematics - Sequence and Series: If $a_{1} (> 0), a_{2}, a_{3}, a_{4}, a_{5}$ are in a G.P., $a_{2} + a_{4} = 2 a_{3} + 1$ and $3 a_{2} + a_{3} = 2 a_{4}$ , then $a_{2} + a_{4} + 2 a_{5}$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Defining the G.P. Terms

Analyzing the Second Equation

Simplifying to a Quadratic Equation

Solving for the Common Ratio $r$

Testing Case 1: $r = - 1$

Rejecting Case 1

Testing Case 2: $r = 3/2$

Solving for the First Term $a$

Setting up the Target Expression

Calculating Individual Terms

Final Result and Conclusion

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Setup

The Quadratic Hunt

The Filter of Reality

Final Calculation

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