CAT 2025 Lesson : Progressions - GP Special Types

If the product of $4$ terms of an increasing GP is $1296$, and the product of the $2^{nd}$and $4^{th}$terms is $144$. Then, what is the $1^{st}$term?

Solution

Let the $4$ terms be$\dfrac{x}{y^{3}},\dfrac{x}{y},xy,xy^{3}$

Product of $4$ terms = $x^{4} = 1296 = 6^{4}$⇒ $x = 6$

Product of $2nd$ and $4th$ terms = $\dfrac{x}{y}\times xy^{3} = x^{2}y^{2} = 144$

⇒ $6^{2} \times y^{2} = 144$ ⇒ $y = 2$

$\therefore$ $1^{st}$ term = $\dfrac{x}{y^{3}} = \dfrac{6}{8} = \dfrac{3}{4}$

Answer: $\dfrac{3}{4}$

If $m$ and $n$ are the $2$ Geometric Means inserted between $6$ and $384$, then $m + n$ = ?

Solution

As the numbers along with the GMs will be in GP, let the terms be $a, ar,$ $ar^{2}$ and $ar^{3}$ respectively.

$1^{st}$ term = $\bm{a = 6}$

$4^{th}$term = $ar^{3}= 384$ ⇒ $r^{3}=\dfrac{384}{6}$

⇒ $r^{3}= 64$ ⇒ $\bm{r = 4}$

$m + n = ar + ar^{2} = ar( 1 + r ) = 6 \times 4 \times 5$ =$\bm{120}$

Answer: $120$