Q61 / 180 · COMEDK 2024 (Afternoon Shift) · Mathematics
$\text { If } 6^{\text {th }} \text { term of a geometric progression is }-\dfrac{1}{32} \text { and } 9^{\text {th }} \text { term is } \dfrac{1}{256} \text { then } r \text { is }$
- A$-2$
- B$\dfrac{1}{2}$
- C2
- Correct answerD$-\dfrac{1}{2}$
Solution
Let the first term of the geometric progression be $a$ and the common ratio be $r$. The $n^{th}$ term of a geometric progression is given by $T_n = ar^{n-1}$.
Given $T_6 = ar^5 = -\dfrac{1}{32}$ and $T_9 = ar^8 = \dfrac{1}{256}$.
Dividing the expression for $T_9$ by $T_6$, we get:
$\dfrac{ar^8}{ar^5} = \dfrac{1/256}{-1/32}$
$r^3 = \dfrac{1}{256} \times (-32) = -\dfrac{32}{256}$
$r^3 = -\dfrac{1}{8}$
Taking the cube root on both sides, we get $r = -\dfrac{1}{2}$.
Answer: $-\dfrac{1}{2}$
Given $T_6 = ar^5 = -\dfrac{1}{32}$ and $T_9 = ar^8 = \dfrac{1}{256}$.
Dividing the expression for $T_9$ by $T_6$, we get:
$\dfrac{ar^8}{ar^5} = \dfrac{1/256}{-1/32}$
$r^3 = \dfrac{1}{256} \times (-32) = -\dfrac{32}{256}$
$r^3 = -\dfrac{1}{8}$
Taking the cube root on both sides, we get $r = -\dfrac{1}{2}$.
Answer: $-\dfrac{1}{2}$
Hence, the correct answer is option: 'D'.