Three numbers form a G.P. If the 3^{rd} term | vedclass

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(C) Let the three numbers in $G.P.$ be $a, ar, ar^2$.According to the first condition,$a, ar, ar^2 - 64$ are in $A.P.$Therefore,$2(ar) = a + (ar^2 - 6

Vedclass · Accepted Answer

$4, 20, 100$

Vedclass · Answer

(C) Let the three numbers in $G.P.$ be $a, ar, ar^2$.According to the first condition,$a, ar, ar^2 - 64$ are in $A.P.$Therefore,$2(ar) = a + (ar^2 - 64) \implies a(r^2 - 2r + 1) = 64 \implies a(r - 1)^2 = 64$ .....$(i)$According to the second condition,the second term of the $A.P.$ is decreased by $8$,so $a, ar - 8, ar^2 - 64$ are in $G.P.$Therefore,$(ar - 8)^2 = a(ar^2 - 64) \implies a^2r^2 - 16ar + 64 = a^2r^2 - 64a \implies 16ar - 64a = 64 \implies ar - 4a = 4 \implies a(r - 4) = 4$ .....(ii)Dividing $(i)$ by (ii): $\frac{a(r - 1)^2}{a(r - 4)} = \frac{64}{4} \implies \frac{(r - 1)^2}{r - 4} = 16 \implies r^2 - 2r + 1 = 16r - 64 \implies r^2 - 18r + 65 = 0$.Solving the quadratic equation: $(r - 5)(r - 13) = 0$,so $r = 5$ or $r = 13$.If $r = 5$,$a(5 - 4) = 4 \implies a = 4$. The numbers are $4, 4(5), 4(5^2) = 4, 20, 100$.If $r = 13$,$a(13 - 4) = 4 \implies 9a = 4 \implies a = 4/9$. The numbers are $4/9, 52/9, 676/9$.Checking option $(c)$: $4, 20, 100$. $3^{rd}$ term decreased by $64$ gives $4, 20, 36$,which is an $A.P.$ $(d=16)$. Decreasing the $2^{nd}$ term by $8$ gives $4, 12, 36$,which is a $G.P.$ $(r=3)$. Thus,the correct option is $(c)$.

Three numbers form a $G.P.$ If the $3^{rd}$ term is decreased by $64$,the three numbers thus obtained will constitute an $A.P.$ If the second term of this $A.P.$ is decreased by $8$,a $G.P.$ will be formed again. Find the numbers.

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