Question

for GP sum of first 3 terms is 125 and sum of next three terms is 27 find the value of r​

Answer 1

Answer:

The value of r is 3/5

Step-by-step explanation:

Let the first term is a and the common difference is r

then the first 3 terms of the GP are

$a, ar, ar^2$

sum of the first three terms = 125

$\implies a+ar+ar^2=125$

$\implies a(1+r+r^2)=125$   .......... (1)

the next three terms of the GP will be

$ar^3, ar^4, ar^5$

sum of these terms = 27

$\implies ar^3+ar^4+ar^5=27$

$\implies ar^3(1+r+r^2)=27$  .......... (2)

Dividing eq (2) by eq (1) we get

$r^3=\frac{27}{125}$

$\implies r^3=(\frac{3}{5})^3$

$\implies r=\frac{3}{5}$

Hope the answer is helpful.

Answer 2

Answer:

$r = \frac{3}{5}$

Step-by-step explanation:

Let a , r are first term and common ratio of a G.P.

Now,

$a,ar,ar^{2},ar^{3} \: are \: first\:three \:terms \\of \:G.P$

$sum \:of \:three \: terms = 125 \:(given)$

$\implies a+ar+ar^{2}=125$

$\implies a(1+r+r^{2})=125\:---(1)$

$ar^{3},ar^{4},ar^{5}, \: are \: next\:three \:terms \\of \:G.P$

$sum \:of\:next \:three \: terms = 27 \:(given)$

$\implies ar^{3}+ar^{4}+ar^{5}=27$

$\implies r^{3}[a(1+r+r^{2})]=27\:---(2)$

/* Substitute (1) in equation (2), we get

$\implies r^{3}\times 125= 27$

$\implies r^{3}\times =\frac{27}{125}\\=\frac{3^{3}}{5^{3}}$

$\implies r^{3}\times = \left(\frac{3}{5}\right)^{3}$

$\implies r = \frac{3}{5}$

Therefore,

$r = \frac{3}{5}$

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