Question 1: Find three numbers in G.P. whose sum is $65$ and whose product is $3375$.

Let the terms of the G.P. be $\frac{a}{r}$ $, a, ar$

Given $\frac{a}{r}$ $\times a \times ar = 3375 \Rightarrow a^3 = 3375 \Rightarrow a = 15$

Also given $\frac{a}{r}$ $+ a + ar = 65$

$\Rightarrow$ $\frac{15}{r}$ $+ 15 + 15r = 65$

$\Rightarrow 15r^2 + 15r + 15 = 65r$

$\Rightarrow 15r^2 - 50r + 15 = 0$

$\Rightarrow 3r^2 - 10r + 3 = 0$

$\Rightarrow ( 3r-1) ( r-3) = 0$

$\Rightarrow r =$ $\frac{1}{3}$ or $r = 3$

Therefore when $a = 15$ and $r =$ $\frac{1}{3}$ , G.P. is $45, 15, 5, \ldots$

and when $a = 15$ and $r = 3$, G.P. is $5, 15, 45, \ldots$

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Question 2: Find three numbers in G.P. whose sum is $38$ and whose product is $1728$.

Let the terms of the G.P. be $\frac{a}{r}$ $, a, ar$

Given $\frac{a}{r}$ $\times a \times ar = 1728 \Rightarrow a^3 = 1728 \Rightarrow a = 12$

Also given $\frac{a}{r}$ $+ a + ar = 38$

$\Rightarrow$ $\frac{12}{r}$ $+ 12 + 12r = 38$

$\Rightarrow 12r^2 + 12r + 12 = 38r$

$\Rightarrow 12r^2 - 26r + 12 = 0$

$\Rightarrow 6r^2 - 13r + 6 = 0$

$\Rightarrow ( 3r-2) ( 2r-3) = 0$

$\Rightarrow r =$ $\frac{2}{3}$ or $r =$ $\frac{3}{2}$

Therefore when $a = 12$ and $r =$ $\frac{2}{3}$ , G.P. is $18, 12, 8 , \ldots$

and when $a = 12$ and $r =$ $\frac{3}{2}$, G.P. is $8, 12, 18, \ldots$

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Question 3: The sum of the first three terms of a G.P. is $\frac{13}{12}$ and their product is $-1$. Find the G.P.

Let the terms of the G.P. be $\frac{a}{r}$ $, a, ar$

Given $\frac{a}{r} \times a \times ar = -1 \Rightarrow a^3 = -1 \Rightarrow a = -1$

Also given $\frac{a}{r}$ $+ a + ar =$ $\frac{13}{12}$

$\Rightarrow$ $\frac{-1}{r}$ $-1 -1r =$ $\frac{13}{12}$

$\Rightarrow -12r^2 - 12r - 12 = 13r$

$\Rightarrow 12r^2 +25r + 12 = 0$

$\Rightarrow 4r(3r+4)+3(3r+4)=0$

$\Rightarrow (3r+4) ( 4r+3) = 0$

$\Rightarrow r =$ $\frac{-4}{3}$  or $r =$ $\frac{-3}{4}$

Therefore when $a = -1$ and $r =$ $\frac{-4}{3}$ , G.P. is $\frac{3}{4}$ $, -1,$ $\frac{4}{3} , \ldots$

and when $a = -1$ and $r =$ $\frac{-3}{4}$ , G.P. is $\frac{4}{3}$ $, -1,$ $\frac{3}{4}$ $, \ldots$

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Question 4: The product of the first three terms of a G.P. is $125$ and the sum of their products taken in pairs is $87$ $\frac{1}{2}$. Find them.

Let the terms of the G.P. be $\frac{a}{r}$ $, a, ar$

Given $\frac{a}{r}$ $\times a \times ar = 125 \Rightarrow a^3 = 125 \Rightarrow a = 5$

Also $\frac{a}{r}$ $\times a + a \times ar + ar \times$ $\frac{a}{r}$ $=$ $\frac{175}{2}$

$\Rightarrow$ $\frac{a^2}{r}$ $+ a^2 r + a^2 =$ $\frac{175}{2}$

$\Rightarrow$ $\frac{25}{r}$ $+ 25 r + 25 =$ $\frac{175}{2}$

$\Rightarrow$ $\frac{1}{r}$ $+ r + 1 =$ $\frac{7}{2}$

$\Rightarrow 2r^2 + 2r + 2 = 7r$

$\Rightarrow 2r^2 -5r + 2 = 0$

$\Rightarrow (2r-1)(r-2) = 0$

$\Rightarrow r =$ $\frac{1}{2}$ $\ or \ r = 2$

Therefore when $a = 5$ and $r =$ $\frac{1}{2}$ , G.P. is $10, 5,$ $\frac{5}{2}$ $, \ldots$

and when $a = 5$ and $r = 2$, G.P. is $\frac{5}{2}$ $, 5, 10, \ldots$

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Question 5: The sum of the first three terms of a G.P. is $\frac{39}{10}$ and their product is $1$. Find the common ratio and the terms.

Let the terms of the G.P. be $\frac{a}{r}$ $, a, ar$

Given $\frac{a}{r}$ $\times a \times ar = 1 \Rightarrow a^3 = 1 \Rightarrow a = 1$

Also given $\frac{a}{r}$ $+ a + ar =$ $\frac{39}{10}$

$\Rightarrow$ $\frac{1}{r}$ $+1 + r =$ $\frac{39}{10}$

$\Rightarrow 10r^2 +10r + 10 = 39r$

$\Rightarrow 10r^2-29r+10 = 0$

$\Rightarrow 5r(2r-5)-2 (2r-5)=0$

$\Rightarrow (2r-5) ( 5r-2) = 0$

$\Rightarrow r =$ $\frac{5}{2}$ or $r =$ $\frac{2}{5}$

Therefore when $a = 1$ and $r =$ $\frac{5}{2}$, G.P. is $\frac{5}{2}$ $, 1,$ $\frac{2}{5}$ $, \ldots$

and when $a = 1$ and $r =$ $\frac{2}{5}$, G.P. is $\frac{2}{5}$ $, 1,$ $\frac{5}{2}$ $, \ldots$

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Question 6: The sum of three numbers in G.P. is $14$. If the first two terms are each increase by $1$ and the third is decreased by $1$, the resulting numbers are in A.P. Find the numbers.

Let the terms of the G.P. be $\frac{a}{r}$ $, a, ar$

Therefore $\frac{a}{r}$ $+ a + ar = 14$

$\Rightarrow a ($ $\frac{1}{r}$ $+1+r) = 14$     … … … … … i)

Also,  new terms $\frac{a}{r}$ $+1 , a+1 , ar - 1$ are in A.P.

$\therefore 2 ( a+1) = ($ $\frac{a}{r}$ $+1) + (ar - 1)$

$\Rightarrow 2 ( a+1) =$ $\frac{a}{r}$ $+ ar$

Substituting from i) we get

$\Rightarrow 2 ( a+1) = 14 - a$

$\Rightarrow 3a = 12$

$\Rightarrow a = 4$

Substituting the value of a in i) we get

$\frac{4}{r} + 4 + 4r = 14$

$\Rightarrow$ $\frac{2}{r}$ $+ 2 + 2r = 7$

$\Rightarrow 2r^2+ 2r + 2 = 7r$

$\Rightarrow 2r^2 - 5r+2 = 0$

$\Rightarrow (r-2)(2r-1) = 0$

$\Rightarrow r = 2 \ or \ r =$ $\frac{1}{2}$

Therefore when $a = 4$ and $r = 2$, G.P. is $8, 4, 2 , \ldots$

and when $a = 4$ and $r =$ $\frac{1}{2}$, G.P. is $2, 4, 8, \ldots$

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Question 7: The product of three numbers in G.P. is $216$. If $2, 8, 6$ be added to them, the results are in A.P.. Find the numbers.

Let the terms of the G.P. be $\frac{a}{r}$ $, a, ar$

Therefore $\frac{a}{r}$ $\times a \times ar = 216 \Rightarrow a^3 = 216 \Rightarrow a = 6$

Also,  new terms $\frac{a}{r}$ $+2 , a+8 , ar +6$ are in A.P.

$\therefore 2 ( a+8) = ($ $\frac{a}{r}$ $+2) + (ar +6)$

$\Rightarrow 2 ( a+8) =$ $\frac{a}{r}$ $+ ar + 8$

Substituting for a we get

$\Rightarrow 2 ( 6+8) =$ $\frac{6}{r}$ $+ 6r + 8$

$\Rightarrow 20 =$ $\frac{6}{r}$ $+6r$

$\Rightarrow 6r^2 -20r + 6 = 0$

$\Rightarrow (r-3)(6r-2) = 0$

$\Rightarrow r = 3 \ or \ r =$ $\frac{1}{3}$

Therefore when $a = 6$ and $r = 3$, G.P. is $18, 6, 2 , \ldots$

and when $a = 6$ and $r =$ $\frac{1}{3}$, G.P. is $2, 6, 18, \ldots$

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Question 8: Find three numbers in G.P. whose product is $729$ and the sum of their product in pairs is $819$.

Let the terms of the G.P. be $\frac{a}{r}$ $, a, ar$

Therefore $\frac{a}{r} \times a \times ar = 729 \Rightarrow a^3 = 729 \Rightarrow a = 9$

Also $($ $\frac{a}{r}$ $\times a ) + ( a \times ar) + ( ar \times$ $\frac{a}{r}$ $) = 819$

$\Rightarrow ($ $\frac{9}{r}$ $\times 9 ) + ( 9 \times 9r) + ( 9r \times$ $\frac{9}{r}$ $) = 819$

$\Rightarrow$ $\frac{81}{r}$ $+ 81r + 81 = 819$

$\Rightarrow$ $\frac{9}{r}$ $+ 9r + 9 = 91$

$\Rightarrow 9r^2 + 9r + 9 = 91 r$

$\Rightarrow 9r^2 - 82r+9 = 0$

$\Rightarrow ( r-9) (9r-1) = 0$

$\Rightarrow r = 9 \ or \ r =$ $\frac{1}{9}$

Therefore when $a = 9$ and $r = 9$, G.P. is $1, 9, 81 , \ldots$

and when $a = 9$ and $r =$ $\frac{1}{9}$ , G.P. is $81, 9, 1, \ldots$

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Question 9: The sum of three numbers in G.P. is $21$ and the sum of their squares is $189$. Find the numbers.

Let the required numbers be $a, ar \text{ and } ar^2$

Therefore $a + ar + ar^2 = 21$

$\Rightarrow a ( 1 + r + r^2) = 21$     … … … … … i)

Also $a^2 + a^2r^2 + a^2 r^4 = 189$

$\Rightarrow a^2 ( 1 + r^2 + r^4) = 189$     … … … … … ii)

Squaring both sides of i) we get

$a^2 ( 1 + r + r^2)^2 = 441$

$\Rightarrow a^2 ( 1 + r^2 + r^4) + 2a^2r ( 1 + r + r^2)= 441$

$\Rightarrow 189+ 2ar [ a(1 + r + r^2)] = 441$

$\Rightarrow 189 + 2ar (21) = 441$

$\Rightarrow ar = 6$

$\Rightarrow a =$ $\frac{6}{r}$    … … … … … iii)

Substituting this in i) we get

$\frac{6}{r} ( 1 + r + r^2) = 21$

$\Rightarrow 6r^2 + 6r + 6 = 21 r$

$\Rightarrow 6r^2 - 15r + 6 = 0$

$\Rightarrow 2r^2 - 5r + 2 = 0$

$\Rightarrow (2r-1)(r-2) = 0$

$\Rightarrow r =$ $\frac{1}{2}$ $\ or \ r = 2$

When $r =$ $\frac{1}{2}$ $, a = 12$ and when $r = 2, a = 3$

Therefore when $a = 12$ and $r =$ $\frac{1}{2}$ , G.P. is $12, 6, 3 , \ldots$

and when $a = 2$ and $r = 3$, G.P. is $3, 6, 12, \ldots$