Let 3, a, b, c be in A.P. and 3, a - 1, b + 1, c + 9 be in G.P. Then, the arithmetic mean of a, b an

Q: Why can't I just use the property $T_2/T_1 = T_3/T_2$ for the G.P.?

You absolutely can! However, be very careful. Using $T_2/T_1 = T_3/T_2$ implies that $T_1$ and $T_2$ cannot be zero. If you use the cross-multiplied form $T_2^2 = T_1 \cdot T_3$, you avoid the risk of division by zero, which is exactly why we caught the invalid case where $d = -2$.

Q: How do I know which value of $d$ to pick if both satisfy the quadratic?

Great question! In JEE problems, always verify your roots. When $d = -2$, our G.P. terms become $3, 0, 0, 6$. A G.P. requires a common ratio $r$ such that $T_n = T_1 \cdot r^{n-1}$. If $T_2 = 0$ and $T_1 = 3$, then $r$ must be $0$. But if $r=0$, then $T_3$ must also be $0$, which it is, but $T_4$ would then have to be $0$ as well. Since $T_4 = 6$, the sequence $3, 0, 0, 6$ is not a valid G.P.

Q: Is there a faster way to find the arithmetic mean without finding $a, b,$ and $c$ individually?

Yes! Since $a, b, c$ are in A.P., the arithmetic mean is simply the middle term $b$. Since $b = 3 + 2d$, once you find the valid $d$, you just plug it into that expression. It saves a few seconds of calculation!

Q: What if the problem didn't specify the order of the A.P.?

The problem explicitly states '$3, a, b, c$ are in A.P.', which locks the sequence order. If the order were unknown, you would have to consider permutations, but in JEE, the sequence order is usually strictly defined by the comma placement.

Q: Why does the G.P. property $T_2^2 = T_1 \cdot T_3$ only use the first three terms?

It doesn't have to! You could also use $T_3^2 = T_2 \cdot T_4$. However, using the first three terms is usually algebraically simpler. If you are ever unsure, you can use both to verify your value of $d$!

Question

Let $3, a, b, c$ be in A.P. and $3, a - 1, b + 1, c + 9$ be in G.P. Then, the arithmetic mean of $a, b$ and $c$ is :

Options:

(A) -4

(B) -1

(C) 13

(D) 11

Sigma Percentile · Accepted Answer

### The Dance of Progressions: A Mathematical Journey

Welcome, fellow traveler of the JEE landscape. Today, we aren't just solving a problem; we are choreographing a dance between two fundamental structures of mathematics: the Arithmetic Progression (A.P.) and the Geometric Progression (G.P.).

Imagine these as two different languages. The A.P. is the language of constant addition, a steady climb, while the G.P. is the language of scaling, a rapid expansion or contraction. Our goal is to find the bridge that connects them.

### Phase 1: Setting the Stage

We start with the sequence $3, a, b, c$ in A.P. The beauty of an A.P. lies in its common difference, let's call it $d$.

If we start at $3$, every subsequent term is just a step of $d$ away. We can define our variables with elegant simplicity:

$$a = 3 + d$$
$$b = 3 + 2d$$
$$c = 3 + 3d$$

Think of $d$ as the 'DNA' of this sequence. Once we know $d$, we know everything about $a, b,$ and $c$. This is the power of parameterization—reducing a complex system to a single unknown.

### Phase 2: The Transformation

Now, the problem introduces a twist. We are told that $3, a-1, b+1, c+9$ form a G.P.

We must substitute our A.P. definitions into this new sequence. Let's look at the terms:

- $T_1 = 3$
- $T_2 = (3 + d) - 1 = 2 + d$
- $T_3 = (3 + 2d) + 1 = 4 + 2d$
- $T_4 = (3 + 3d) + 9 = 12 + 3d$

We have transformed the G.P. into a sequence defined entirely by $d$. The tension is palpable—will this lead to a solvable equation?

### Phase 3: The Geometric Bridge

For any three numbers $x, y, z$ to be in G.P., they must satisfy the property $y^2 = xz$. This is the 'Golden Rule' of geometric sequences.

Applying this to our first three terms, we get:

$$(2 + d)^2 = 3(4 + 2d)$$

Let's expand this carefully. On the left, we have the square of a binomial: $4 + d^2 + 4d$. On the right, we distribute the $3$ to get $12 + 6d$.

Equating them gives us a quadratic equation:

$$d^2 + 4d + 4 = 12 + 6d$$

Bringing everything to one side, we arrive at the heart of the problem:

$$d^2 - 2d - 8 = 0$$

### Phase 4: The Moment of Truth

Factorizing this quadratic is like solving a puzzle. We look for two numbers that multiply to $-8$ and add to $-2$. Those numbers are $-4$ and $2$.

Thus, we have:

$$(d - 4)(d + 2) = 0$$

This gives us two candidates for $d$: $4$ and $-2$. But wait! In mathematics, as in life, not every path is valid.

We must test these values. If $d = -2$, our G.P. terms become $3, 0, 0, 6$. A G.P. with a zero in the middle is a mathematical trap—it breaks the definition of a common ratio. Thus, we reject $d = -2$ and embrace $d = 4$.

### Phase 5: The Grand Finale

With $d = 4$ as our key, the values of $a, b,$ and $c$ fall into place like pieces of a clockwork mechanism:

$$a = 3 + 4 = 7$$
$$b = 3 + 2(4) = 11$$
$$c = 3 + 3(4) = 15$$

Finally, the arithmetic mean of $a, b,$ and $c$ is simply the average:

$$	ext{Mean} = \frac{7 + 11 + 15}{3} = \frac{33}{3} = 11$$

We have arrived at the destination. The beauty of this problem isn't just the answer **$11$**; it's the realization that by defining our variables correctly, we can navigate between the rigid steps of an A.P. and the exponential leaps of a G.P. with absolute confidence.

Animated Solution for Mathematics - Sequence and Series: Let $3, a, b, c$ be in A.P. and $3, a - 1, b + 1, c + 9$ be in G.P. Then, the arithmetic mean of $a, b$ and $c$ is :

Select Answer:

Visualized Solution

Defining $a, b, c$ in A.P.

Constructing the G.P. Sequence

Applying the G.P. Property

Expanding the Equation

Forming the Quadratic Equation

Solving the Quadratic Equation

Verifying the Common Difference

Calculating $a, b,$ and $c$

Final Arithmetic Mean

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

The Dance of Progressions

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let 3,a,b,c be in A.P. and 3,a−1,b+1,c+9 be in G.P. Then, the arithmetic mean of a,b and c is :