Question
Find the Geometric progression with $4^{th}$ term = $54$ and $7^{th}$ term = $1458$.
✓
Answer
Let the first term of the G.P. be a and its common ratio be r.
$4^{th}$ term = $54 \Rightarrow ar^3 = 54$
$7^{th}$ term = $1458 \Rightarrow ar^6 = 1458$
Now, $\frac{ ar ^6}{ ar ^3}=\frac{1458}{54}$
$\Rightarrow r^3 = 27$
$\Rightarrow r = 3$
$ar^3= 54$
$\Rightarrow a x (3)^3 = 54$
$\Rightarrow a =\frac{54}{27}=2$
$\therefore G.P. = a, ar, ar^2, ar^3,$ ........
$= 2, 2 \times 3, 2 \times (3)^2, 54,$...........
$= 2, 6, 18, 54$,............
$4^{th}$ term = $54 \Rightarrow ar^3 = 54$
$7^{th}$ term = $1458 \Rightarrow ar^6 = 1458$
Now, $\frac{ ar ^6}{ ar ^3}=\frac{1458}{54}$
$\Rightarrow r^3 = 27$
$\Rightarrow r = 3$
$ar^3= 54$
$\Rightarrow a x (3)^3 = 54$
$\Rightarrow a =\frac{54}{27}=2$
$\therefore G.P. = a, ar, ar^2, ar^3,$ ........
$= 2, 2 \times 3, 2 \times (3)^2, 54,$...........
$= 2, 6, 18, 54$,............
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