© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Express the given terms of the A.P.
Let the first term of the arithmetic progression (A.P.) be $A$ and its common difference be $d$. Then the 7th, 11th, and 13th terms (denoted by $a, b, c$ respectively) can be written as:
$a = T_7 = A + 6d$
$b = T_{11} = A + 10d$
$c = T_{13} = A + 12d$
Step 2: Use the condition that $a, b, c$ are in G.P.
Since $a, b, c$ are consecutive terms of a geometric progression (G.P.), they satisfy
$b^2 = a \cdot c$.
Substitute $a = A + 6d$, $b = A + 10d$, and $c = A + 12d$ into the above equation:
$(A + 10d)^2 = (A + 6d)(A + 12d)$.
Step 3: Simplify the equation to find a relation between $A$ and $d$
Expand both sides:
LHS: $(A + 10d)^2 = A^2 + 20Ad + 100d^2$.
RHS: $(A + 6d)(A + 12d) = A^2 + 18Ad + 72d^2$.
Equating the two expressions:
$A^2 + 20Ad + 100d^2 = A^2 + 18Ad + 72d^2.$
Subtract $A^2 + 18Ad + 72d^2$ from both sides:
$20Ad + 100d^2 - 18Ad - 72d^2 = 0,$
$2Ad + 28d^2 = 0,$
$2d(A + 14d) = 0.$
Since the progression is non-constant, $d \neq 0$, so we get
$A + 14d = 0 \quad \Rightarrow \quad A = -14d.$
Step 4: Compute $\displaystyle \frac{a}{c}$
Now, substitute $A = -14d$ back into $a$ and $c$:
$a = A + 6d = -14d + 6d = -8d,$
$c = A + 12d = -14d + 12d = -2d.$
Therefore,
$\displaystyle \frac{a}{c} = \frac{-8d}{-2d} = 4.$
Conclusion
The value of $\displaystyle \frac{a}{c}$ is $4$, which matches the correct answer.
