Question

Find the sum of the first 5 terms of the geometric progression if the third term is 144 and the sixth term is 456. With solution​

Answer 1

Answer:

Step-by-step explanation:

486=144r^4-1

486=144r³

486/144=144r³/144

27/8=r³—use cuberoot here

√(27/8)=√(r³)

r=3/2

Answer 2

Given:

(Note:  there should be 486 in place of 456)

Find the sum of the first 5 terms of the geometric progression if the third term is 144 and the sixth term is 456.

Solution:

Know that,

The $\[{n^{th}}\]$ term of GP $\[{T_n} = a{r^{n - 1}}\]$

Sum of n terms of a GP $\[{S_n} = \frac{{a\left( {{r^{n - 1}} - 1} \right)}}{{r - 1}}\]$, when $r>1$

Using given information write the expression for third and sixth term,

$\[144 = a{r^2}\]$        ------(1)

$\[486 = a{r^5}\]$       ------(2)

Divide equation (1) by (2),

$\[ \Rightarrow \frac{{486}}{{144}} = {r^3}\]$

$\[ \Rightarrow {r^3} = \frac{{27}}{8}\]$

$\[ \Rightarrow r = \frac{3}{2}\]$

Find the first term of the GP using equation (1)

$\[ \Rightarrow 144 = a \times {\left( {\frac{3}{2}} \right)^2}\]$

$\[ \Rightarrow a = \frac{{144 \times 4}}{9}\]$

$\[ \Rightarrow a = 64\]$

Now, find the sum of the first five term of the GP,

$\[ \Rightarrow {S_5} = \frac{{64 \times \left\{ {{{\left( {\frac{3}{2}} \right)}^5} - 1} \right\}}}{{\frac{3}{2} - 1}}\]$

$\[ \Rightarrow {S_5} = \frac{{64 \times \left\{ {\frac{{243}}{{32}} - 1} \right\}}}{{\frac{1}{2}}}\]$

$\[ \Rightarrow {S_5} = 64 \times \frac{{211}}{{32}} \times \frac{2}{1}\]$

$\[ \Rightarrow {S_5} = 844\]$

Hence, the sum of the first 5 terms of the geometric progression is 844.