Question

The numbers $a_1,a_2,a_3,\ldots$ form an arithmetic sequence with $a_1\ne a_2$. The three numbers $a_1,a_2$ and $a_6$ form a geometric sequence in that order. Then the common difference of the arithmetic sequence is

• A. $a_{1}$
• B. $2a_{1}$
• C. $3 a_{1}$
• D. $4 a_{1}$
• E. $5a_{1}$

Hint:

Sequence Tip: Converting all given sequence terms into the universal variables of the first term ($a$ or $a_1$) and common difference ($d$) transforms any complex sequence problem into simple algebra.

Answer 1

The Correct Option is C

Concept:
In an Arithmetic Progression (A.P.), the $n$-th term is defined as $a_n = a_1 + (n-1)d$, where $a_1$ is the first term and $d$ is the common difference. If three terms $x, y, z$ form a Geometric Progression (G.P.), the square of the middle term equals the product of the extremes: $y^2 = xz$.

Step 1: Express the A.P. terms using $a_1$ and d.
Using the A.P. formula, we write the relevant terms: First term: $a_1 = a_1$ Second term: $a_2 = a_1 + d$ Sixth term: $a_6 = a_1 + 5d$

Step 2: Set up the Geometric Progression equation.
We are given that $a_1$, $a_2$, and $a_6$ form a G.P. Therefore, their relationship is: $$(a_2)^2 = a_1 \cdot a_6$$

Step 3: Substitute the A.P. expressions into the G.P. equation.
Replace $a_2$ and $a_6$ with their expressions in terms of $a_1$ and $d$: $$(a_1 + d)^2 = a_1(a_1 + 5d)$$

Step 4: Expand and simplify the algebraic equation.
Expand the binomial on the left and distribute on the right: $$a_1^2 + 2a_1d + d^2 = a_1^2 + 5a_1d$$ Subtract $a_1^2$ from both sides to cancel them out: $$2a_1d + d^2 = 5a_1d$$

Step 5: Solve for the common difference d.
Rearrange the equation to group terms with $d$: $$d^2 - 3a_1d = 0$$ Factor out $d$: $$d(d - 3a_1) = 0$$ This gives two possibilities: $d = 0$ or $d = 3a_1$. Since the problem states $a_1 \ne a_2$, the common difference cannot be zero ($d \ne 0$). Therefore, $d = 3a_1$. Hence the correct answer is (C) $3a_{1$}.