Let A₁, A₂, A₃ be the three A.P. with the same common difference d and having their first terms as A

Q: Why does the variable $a$ disappear during the determinant expansion?

It is a beautiful piece of mathematical symmetry! When you expand along the first column, the coefficient of $a$ involves the difference between the second and third rows of the second column. Since both are $17$, that term becomes $17 - 17 = 0$. It is the problem's way of telling you to focus on the relationship between $b$ and $c$.

Q: I tried to solve for $A$ and $d$ separately using the determinant equation, but I got stuck. Why?

That is the 'Aha!' moment of this problem. The determinant equation is designed to isolate the relationship between the terms of the APs, independent of the common difference $d$. It forces you to find the first term $A$ first, which then unlocks everything else.

Q: Is there a faster way to solve the determinant without full expansion?

Yes! You can use row operations. Subtracting the second row from the third row ($R_3 \to R_3 - R_2$) makes the determinant much easier to handle. Always look for ways to create zeros in a determinant before expanding!

Question

Let $A_1, A_2, A_3$ be the three A.P. with the same common difference $d$ and having their first terms as $A, A+1, A+2$, respectively. Let $a, b, c$ be the $7^{th}, 9^{th}, 17^{th}$ terms of $A_1, A_2, A_3$, respectively such that $\begin{vmatrix} a & 7 & 1 \ 2b & 17 & 1 \ c & 17 & 1 \end{vmatrix} + 70 = 0$. If $a = 29$, then the sum of first 20 terms of an AP whose first term is $c - a - b$ and common difference is $\frac{d}{12}$, is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path of JEE Advanced mathematics. Today, we are not just solving a problem; we are uncovering a hidden structure.

We are given three arithmetic progressions, $A_1, A_2, A_3$, all dancing to the same rhythm—the same common difference $d$. Their starting points are staggered: $A, A+1, A+2$.

At first glance, this looks like a mess of variables. But in the world of JEE, complexity is often just a mask for elegance.

### Decoding the APs

Let us translate the language of sequences into the language of algebra. For any arithmetic progression, the $n^{th}$ term is given by $T_n = a_1 + (n-1)d$.

Applying this to our three sequences:

- For $A_1$, the $7^{th}$ term is $a = A + 6d$.
- For $A_2$, the $9^{th}$ term is $b = (A+1) + 8d$.
- For $A_3$, the $17^{th}$ term is $c = (A+2) + 16d$.

We now have our three protagonists, $a, b,$ and $c$, expressed in terms of the foundational parameters $A$ and $d$. Keep these expressions close; they are the keys to our final destination.

### The Determinant Dance

Now, we face the beast: the determinant equation:

$$
\begin{vmatrix} a & 7 & 1 \ 2b & 17 & 1 \ c & 17 & 1 \end{vmatrix} + 70 = 0
$$

Do not let the matrix notation intimidate you. Determinants are just a way of encoding linear relationships. Let us expand this along the first row.

The first term is $a(17 \cdot 1 - 17 \cdot 1)$, which is $a(0) = 0$. Just like that, the variable $a$ vanishes! This is the problem rewarding your courage.

We are left with:

$$
-7(2b - c) + 1(2b \cdot 17 - c \cdot 17) + 70 = 0
$$

Simplifying this, we get $-14b + 7c + 34b - 17c + 70 = 0$, which reduces to $20b - 10c + 70 = 0$. Dividing by $10$, we arrive at the beautiful, compact relationship: $2b - c + 7 = 0$.

### The Algebraic Revelation

Now, we substitute our expressions for $b$ and $c$ back into this equation:

$$
2(A + 1 + 8d) - (A + 2 + 16d) + 7 = 0
$$

Watch closely as the magic happens. Expanding the brackets gives us $2A + 2 + 16d - A - 2 - 16d + 7 = 0$.

The $16d$ and $-16d$ cancel out perfectly! The constants $2$ and $-2$ also vanish. We are left with $A + 7 = 0$, which means $A = -7$.

We have found the first term! With $A = -7$ and the given $a = 29$, we find $29 = -7 + 6d$, leading to $6d = 36$, or $d = 6$.

### Final Calculation

We are in the home stretch. We know $A = -7$ and $d = 6$. We can now find $b$ and $c$:

- $b = -7 + 1 + 8(6) = 42$
- $c = -7 + 2 + 16(6) = 91$

The new AP has a first term $a' = c - a - b = 91 - 29 - 42 = 20$. Its common difference is $d' = \frac{d}{12} = \frac{6}{12} = 0.5$.

Finally, the sum of the first $20$ terms is:

$$
S_{20} = \frac{20}{2} [2(20) + (20-1)(0.5)] = 10 [40 + 9.5] = 10 [49.5] = 495
$$

And there it is. The complexity dissolved, the variables aligned, and the answer revealed. You have mastered the sequence. The final result is **495**.

Enter Numerical Value:

Visualized Solution

Defining the Terms $a, b, c$

Setting up the Determinant

Expanding the Determinant

Simplifying the Equation

Solving for Capital $A$

Finding the Common Difference $d$

Calculating $b$ and $c$

Parameters of the New A.P.

Final Sum Calculation

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

Decoding the APs

The Determinant Dance

The Algebraic Revelation

Final Calculation

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