Question

This one is really tough !! Help me


Q. The sum of first three terms of a G.P is 16/ sum of next three terms is 128. Find a , r and Sn

Answer 1

Heya !
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★Geometric Progressions ★
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Ques. → The sum of first three terms of a G.P is 16 and the sum of next three terms is 128. Find a , r and Sn.


Ans.→ Let the G.P be - a , ar , ar² , ar³.......... Where a is the first term and r is the common ratio.


°•° According to the Question we have ,


→ a + ar + ar² = 16 ---------------(Eq. 1 )


→ar³ + ar⁴ + ar^5 = 128----------(Eq.2 )



★Now divide Equation 2 & 1 so that a gets cancelled out !! We have ,


$= \frac{ar {}^{3} + ar {}^{4} + ar {}^{5} }{a + ar + ar {}^{2} } = \frac{128}{16}$

=> r³ = 8


•°• r = 2 ✔
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★Put value of r in the equation (1) , we have


→ a + 2a + 4a = 16

→7a = 16


•°• a = 16/7 ✔
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Now we have to find Sum of n terms , Sn .

•Here we see that r > 1


Thus, we have the formula ,


$= > sn = \frac{a(r {}^{n} - 1)}{r - 1} \\ \\ = > sn = \frac{ \frac{16}{7} (2 {}^{n} - 1)}{2 - 1} \\ \\ = > sn = \frac{16}{7} (2 {}^{n} - 1)$

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Answer 2

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