Find four numbers in G.P in which the sum of the extreme terms is $112$ and sum of middle terms is $48$ .

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Solution

Let $a, a r, a r^{2}, a r^{3}$ be 4 terms
Sum of Extreme $a + a r^{3} = 112$
sum of Middle $a r + a r^{2} = 48$
$⟹ \frac{1 + r^{3}}{r + r^{2}} = \frac{7}{3}$
$⟹ 3 r^{3} - 7 r^{2} - 7 r + 3 = 0$
$⟹ r = 3, - 1, \frac{1}{3}$
where $- 1$ doesnot fit to solution..
Let $r = 3$
$a (3) + a (3)^{2} = 48 ⟹ a = 4$
so the $4$ terms are $4, 12, 36, 108$

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Find four numbers in G.P in which the sum of the extreme terms is 112 and sum of middle terms is 48.

Let a,ar,ar2,ar3 be 4 terms Sum of Extreme a+ar3=112sum of Middle ar+ar2=48⟹1+r3r+r2=73⟹3r3−7r2−7r+3=0⟹r=3,−1,13where −1 doesnot fit to solution..Let r=3a(3)+a(3)2=48⟹a=4so the 4 terms are 4,12,36,108

Find four numbers in G.P in which the sum of the extreme terms is $112$ and sum of middle terms is $48$ .