Q.If $a_1, a_2, a_3, \ldots, a_r$ are in G.P., then prove that the determinant $\begin{vmatrix} a_{r+1} & a_{r+5} & a_{r+9} \\ a_{r+7} & a_{r+11} & a_{r+15} \\ a_{r+11} & a_{r+17} & a_{r+21} \end{vmatrix}$ is independent of $r$.

Key idea: write each G.P. term as $a\,k^{\,n-1}$ and factor the powers of $k$ out of each row; the first two rows become identical, so the determinant is $0$ for every $r$.

Step 1 — with $a_n=a\,k^{\,n-1}$, factor $a k^{r}$ from $R_1$, $a k^{r+6}$ from $R_2$, $a k^{r+10}$ from $R_3$:

$\Delta = a^3k^{3r+16}\begin{vmatrix} 1 & k^4 & k^8 \\ 1 & k^4 & k^8 \\ 1 & k^6 & k^{10} \end{vmatrix}.$ …

Intuition

In a G.P. every term is $a\,k^{\,n-1}$. Inside the determinant that means each row is a power of $k$ times a fixed pattern. Once you pull those powers out, you can read off whether any two rows coincide — and here two of them do, which is what makes the value $0$ regardless of $r$.

Setting up

Let the first term be $a$ and the common ratio $k$, so $a_n=a\,k^{\,n-1}$:

Working the steps

1. Replace each entry. Using $a_{r+t}=a\,k^{\,r+t-1}$:

2. Factor a common power from each row. Row 1 has $a k^{r}$, row 2 has $a k^{r+6}$, row 3 has $a k^{r+10}$:

$\Delta = a^3 k^{3r+16}\begin{vmatrix} 1 & k^4 & k^8 \\ 1 & k^4 & k^8 \\ 1 & k^6 & k^{10} \end{vmatrix}.$ …

Method: Factoring Out G.P. Powers to Expose Equal (or Proportional) Rows

Use this method whenever a determinant's entries are consecutive terms of a G.P. (or any geometric-type sequence) and you're asked to prove the value doesn't depend on some index like $r$.

Steps

Step 1: Write every entry using the G.P.'s general term

If the sequence has first term $a$ and common ratio $k$, its $n$-th term is $a_n = a\,k^{\,n-1}$. Rewrite every entry of the determinant this way so all entries become powers of the same $a$ and $k$, with only the exponent changing from entry to entry.

Step 2: Factor the lowest common power out of each row

Each row shares a common factor $a\,k^{\,\text{(row's smallest exponent)}}$ — pull it out using the scaling property of determinants ($\det \to k\det$ when a row is scaled by $k$; pulling a factor OUT of a row is the same operation run in reverse, tracked explicitly as a multiplier in front of the determinant). What remains inside each row is a simple ratio-power pattern like $1,\ k^{p},\ k^{q}$.

Step 3: Compare the reduced rows …

Common Mistakes

Mistake 1: Misindexing the G.P. exponents when substituting $a_n = a\,k^{\,n-1}$

Why it's wrong: writing $a_{r+1}$ as $a\,k^{\,r}$ but then treating $a_{r+7}$ as $a\,k^{\,r+7}$ (forgetting the "$-1$" in the exponent) throws off every subsequent row and can hide or create a coincidence that isn't really there. Correct approach: substitute one general term first ($a_{r+t}=a\,k^{\,r+t-1}$), then plug in $t=1,5,9$ etc. for each row systematically, rather than guessing exponents by pattern.

Mistake 2: Jumping to direct cofactor expansion instead of factoring out powers of $k$ …