Question

The fourth term of a GP is 2/27 and 7th term is 2/729 then the series


7,7/3,7/9,7/27

5,5/3,5/7,5/27

2,2/3,2/9,2/27

1,1/3,1/9,1/27​

Answer 1

ANSWER:

Given:

• 4th term of GP = 2/27
• 7th term of GP = 2/729

To Find:

• The series

Solution:

$\text{We know that, the general term( a_n ) of a GP is:}\\\\:\hookrightarrow a_n=ar^{n-1}\\\\\text{where, a is first term, r is common ratio and n is no.of term.}\\\\\text{We are given that,}\\\\:\longrightarrow a_4=\dfrac{2}{27}\:\:and\:\:a_7=\dfrac{2}{729}\\\\\text{So,}\\\\:\implies a_4=ar^{4-1}=\dfrac{2}{27}\\\\:\implies a_4=ar^3=\dfrac{2}{27}- - - -(1)\\\\\text{And,}\\\\:\implies a_7=ar^{7-1}=\dfrac{2}{729}\\\\:\implies a_7=ar^6=\dfrac{2}{729}- - - -(2)$

$\text{Dividing (2) by (1),}\\\\:\implies\dfrac{a_7}{a_4}=\dfrac{a\!\!\!/\:r^{6\!\!\!/^{\:3}}}{a\!\!\!/\:r^3\!\!\!\!\!\!/}=\dfrac{\dfrac{2\!\!\!/}{729}}{\dfrac{2\!\!\!/}{27}}\\\\:\implies r^3=\dfrac{27\!\!\!\!/}{729\!\!\!\!\!/_{\:\:27}}\\\\:\implies r^3=\dfrac{1}{27}\\\\:\implies r=\sqrt[3]{\dfrac{1}{27}}\\\\:\implies r=\dfrac{1}{3}\\\\\text{Putting value of r in (1),}\\\\:\implies ar^3=\dfrac{2}{27}\\\\:\implies a\left(\dfrac{1}{3}\right)^3=\dfrac{2}{27}\\\\:\implies \dfrac{a}{27\!\!\!\!/}=\dfrac{2}{27\!\!\!\!/}\\\\:\implies a=4$

$\text{\bf{So, the series is 2, \: 2/3, \: 2/9, \: 2/27, . . . . . }}$

Formula Used:

• a_n = ar^(n-1)

Learn More:

• a_n = ar^(n-1)
• S_n = (ar^n- 1)/(r-1)