Question

Find four numbers forming a geometric progression in which the third term is greater than the first term by 9 and the second term is greater than by 4th by 18.​

Answer 1

$\color {red}\huge\bold\star\underline\mathcal{Question:-}$ we have to find GP series ....

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$\huge\underline\blue{\sf Given:}$

3rd Term = (1st Term +9)

2nd Term = (4th Term+18)

$\bold{\boxed{\huge{\boxed{\orange{\small{\boxed{\huge{\red{\bold{\:Answer}}}}}}}}}}$

Let the GP :---- a, ar, ar² ,ar³_________________

T(3) = T(1)+9

ar^(3-1) = a^(r-1) + 9

ar² = a + 9

(ar²-a) = 9 ---------------------- $\large\red{\boxed{\sf </strong><strong>Equation</strong><strong>(</strong><strong>1</strong><strong>)</strong><strong>}}$

Also,

T(2) = T(4)+18

ar^(2-1) = ar^(4-1) + 18

ar = ar³ + 18

(ar-ar³) = 18 --------------------$\large\red{\boxed{\sf Equation(</strong><strong>2</strong><strong>)}}$

Dividing Both equation we get,

(ar²-a)/(ar-ar³) = $\huge{</strong><strong>\</strong><strong>f</strong><strong>r</strong><strong>a</strong><strong>c</strong><strong>{</strong><strong>9</strong><strong>}</strong><strong>{</strong><strong>1</strong><strong>8</strong><strong>}</strong><strong>}</strong><strong>$

a(r²-1)/(-ar)(r²-1)= $\huge{\frac{</strong><strong>1</strong><strong>}{</strong><strong>2</strong><strong>}}$

$\large\red{\boxed{\sf </strong><strong>r</strong><strong>=</strong><strong>(</strong><strong>-</strong><strong>2</strong><strong>)</strong><strong>}}$

Now, putting value of r = (-2)

ar²-a = 9

a(4) -a = 9

$\large\red{\boxed{\sf </strong><strong>a</strong><strong>=</strong><strong>3</strong><strong>}}$

Hence,

3, 3(-2), 3(-2)² ,3(-2)³

3, -6 , 12 , -24 $\huge\blue{</strong><strong>Required</strong><strong>\:</strong><strong>GP</strong><strong>}$

Answer 2

$\huge\tt{AnsweR}$

Given:

3rd Term = 1st term + 9

2nd Term = 4th term + 18

Let the GP be ( a , ar , ar² , ar³ )

$\sf = t(3) = t(1) + 9 \\ \sf = ar {}^{(3 - 1)} = a {}^{(r - 1)} + 9 \\ \sf = ar = a + 9 \\ \sf = (ar {}^{2} - a) = 9 - - - equation(1)$

$\sf = t(2) = t(4) + 18 \\ \sf = ar {}^{(2 - 1)} = ar {}^{(4 - 1)} + 19 \\ \sf = ar = ar {}^{3} + 18 \\ \sf = (ar + ar {}^{3} ) - - - equation(2)$

Diving both equation (1) & (2)

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

Adding the value ( r = - 2 )

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

Therefore , the required GP are3,

• 3(-2), 3(-2)² ,3(-2)³
• 3, -6 , 12 , -24