Question

$\huge\grey{\textbf{Question\: for\: mod}}$
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Find four numbers forming a geometric progession in which the third term is greater than the first term by 9, and the second term is greater than fourth by 18.
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Answer 1

Given

• Third term is greater than the first term by 9.
• Second term is greater than the fourth term by 18.

To find

• Four numbers forming a geometric progression.

Solution

$\sf\pink{⟶}$ Third term is greater than the first term by 9.

• ar² - a = 9⠀⠀⠀.....[1]

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★ Taking a as common

• a(r² - 1) = 9
• (r² - 1) = $\small\dfrac{9}{a}$⠀⠀....[2]

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$\sf\pink{⟶}$ Second term is greater than the fourth term by 18.

• ar - ar³ = 18

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★ Taking (-ar) as common

• -ar(r² - 1) = 18

$\tt\longmapsto{-\cancel{a}r (\dfrac{9}{\cancel{a}}) = 18}$⠀⠀[From (2)]

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$\tt\longmapsto{-9r = 18}$

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$\tt\longmapsto{r = \dfrac{18}{-9}}$

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$\tt\longmapsto{r = -2}$

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★ Putting the value of r in Equation [1]

$\tt:\implies\: \: \: \: \: \: \: \: {a(-2)^2 - a = 9}$

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$\tt:\implies\: \: \: \: \: \: \: \: {a (4) - a = 9}$

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$\tt:\implies\: \: \: \: \: \: \: \: {4a - a = 9}$

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$\tt:\implies\: \: \: \: \: \: \: \: {3a = 9}$

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$\tt:\implies\: \: \: \: \: \: \: \: {a = \dfrac{9}{3}}$

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$\tt:\implies\: \: \: \: \: \: \: \: {a = 3}$

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★ Now, we know that

$\: \: \: \: \: \: \: \: \: \: \: \: \boxed{\tt{\bigstar{G.P = a, ar, ar^2, ar^3{\bigstar}}}}$

⠀

$\tt\longmapsto{a = 3}$

$\tt\longmapsto{ar = 3 \times (-2) = -6}$

$\tt\longmapsto{ar^2 = 3 \times (-2)^2 = 3 \times 4 = 12}$

$\tt\longmapsto{ar^3 = 3 \times (-2)^3 = 3 \times (-8) = -24}$

⠀

※ Required G.P. $\sf{⟶}$ 3, +6, 12, -24.

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Answer 2

ɢɪᴠᴇɴ:-

• 3rd Term = 1st term + 9
• 2nd Term = 4th term + 18

ᴛᴏ ꜰɪɴᴅ:-

• The four numbers forming a geometric progession.

ꜱᴏʟᴜᴛɪᴏɴ:-

Let the GP be ( a , ar , ar² , ar³ )

$\begin{lgathered}\tt = t(3) = t(1) + 9 \\ \tt = ar {}^{(3 - 1)} = a {}^{(r - 1)} + 9 \\ \tt = ar = a + 9 \\ \sf = (ar {}^{2} - a) = 9 \;\;\;\;\;\;\;\;\;\;\bigg\lgroup\bf eq\;(1)\bigg\rgroup\end{lgathered}$

$\begin{lgathered}\tt = t(2) = t(4) + 18 \\ \tt = ar {}^{(2 - 1)} = ar {}^{(4 - 1)} + 19 \\ \tt = ar = ar {}^{3} + 18 \\ \sf = (ar + ar {}^{3} ) \;\;\;\;\;\;\;\;\;\;\bigg\lgroup\bf eq\;(2)\bigg\rgroup\end{lgathered}$

on dividing both the equations i.e. (1) & (2),

$\begin{lgathered}\tt = \frac{(ar {}^{2} - a) }{(ar - ar {}^{3)} } = \frac{9}{18} \\ \tt = \frac{(ar {}^{2} - 1)}{( - ar)(r {}^{2} - 1) } = \frac{1}{2} \\ \sf = r = ( - 2)\end{lgathered}$

On adding the value ( r = - 2 ),

$\begin{lgathered}\tt = ar {}^{2} - a = 9 \\ \tt = a(4) - a = 9 \\ \sf = a3\end{lgathered}$

Therefore , the required GP are,

• 3, -6 , 12 , -24

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