Question

if for given G.P. a = 729 and th 7th term is 64 determine S7

Answer 1

First term, a = a1 = 729

7th term, a7 = 64

Sum of first 7 terms, S7 = ?
____________

we know that an = a × r^(n-1)

=> a7 = a × r^(7-1)

=> 64 = 729 × r^6

=> r^6 = 64/729 = (2/3)^6

=> r = 2/3
______________

Sum of 7 terms is given by

$sum = \frac{a( 1 - {r}^{n} )}{1 - r } \\ \\ sum = \frac{729 \times ( {1 - (\frac{2}{3} )}^{7} )}{1 - \frac{2}{3} }$

$sum = \frac{729 \times ( 1 - (\frac{128}{2187} ) )}{\frac{1}{3} } \\ \\ sum = 3 \times 729 \times \frac{2187 - 128}{2187}$

$sum = 2187 \times \frac{2059}{2187} \\ \\ sum = 2059$

$sum \: of \: 7 \: terms \: is \: 2059$

Answer 2

2059

The first term of a GP (a)

= 729

The seventh term of the GP (a7)

= 64

We can write the seventh term of the GP as:

$ar^{6} = 64$

Putting the value of a:

$729r^{6} = 64$

Taking 729 to the other side of the equation we get:

$r^{6} = \frac{64}{729}$

By Prime Factorization we can write 64 and 729 in exponential form as follows:

$r^{6} = \frac{2^{6} }{3^{6} }$

We get the common ratio as:

$r = \frac{2}{3}$

Formula used to find the sum of n terms of a GP:

$= \frac{a(1 - r^{n} )}{1 - r}$

Substituting all the values that are known to us in this formula we get:

= $\frac{729(1 - (\frac{2}{3}^{7} )}{1 - \frac{2}{3}}$

Rationalizing denominators and writing all the numbers in exponential form:

= $\frac{3^{6}\frac{(3^{7}- 2^{7})}{3^{7} } }{\frac{1}{3} }$

Simplifying we get:

= $3^{7} - 2^{7}$

= 2187-128

= 2059

Therefore, the sum of first seven terms of the GP is 2059.