Question

If for a G.P. first term is (27)2 and seventh
term is (8)2 , find Sum upto 8​

Answer 1

Answer:

Sum upto 8 terms as 65/2(3-√6)

Step-by-step explanation:

Let the first term be a and the common ratio be r

Thus, first term = a=27/2

7th term = ar⁶=8/2

(27/2)r⁶=8/2

r⁶=8/27

(r²)³=(2/3)³

r²=2/3

r=√2/3

Now, Sum in GP = a(1-rⁿ)/(1-r)

Therefore, Sum upto 8 terms = 27/2 [(1-(√2/3)⁸/(1-(√2/3))]

Solving the above equation, we have:

Sum upto 8 terms as 65/2(3-√6)

Answer 2

The sum upto 8 is 59.

Step-by-step explanation:

Given : If for a G.P. first term is $\frac{27}{2}$ and seventh  term is $\frac{8}{2}$.

To find : The sum upto 8 terms ?

Solution :

The nth term of GP is $a_n=ar^{n-1}$

Here, $a=\frac{27}{2}$

$a_7=\frac{8}{2}$

i.e. $ar^6=\frac{8}{2}$

$(\frac{27}{2})r^6=\frac{8}{2}\\\\r^6=\frac{8\times 2}{2\times 27}\\\\r^6=\frac{8}{27}\\\\(r^2)^3=(\frac{2}{3})^3\\\\r^2=\frac{2}{3}\\\\r=\sqrt{\frac{2}{3}}$

The sum of 8 terms of GP is

$S_n=\frac{a(1-r^n)}{1-r}\\\\S_n=\frac{(\frac{27}{2})(1-(\sqrt{\frac{2}{3}})^8)}{1-(\sqrt{\frac{2}{3}})}\\\\S_n=\frac{(\frac{27}{2})(1-(\frac{2}{3})^4)}{1-(\sqrt{\frac{2}{3}})}\\\\S_n\approx 59$

Therefore, the sum upto 8 is 59.

#Learn more

In Gp if a=1/2,r=1/3, then S6=?

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