Question

the 4th and 10th term of a gp are 1/3 and 243 respectively. Find the 2nd term. ​

Answer 1

$t_2=\frac{1}{27}$

Step-by-step explanation:

Formula  of nth term of GP = $t_n=ar^{n-1}$

Where n is the no. of term

$t_n$ is the nth term

a = first term

r is the common ratio

Substitute n = 4

$t_4=ar^{4-1}$

$t_4=ar^3$

We are given that the 4 th term is $\frac{1}{3}$

$t_4=ar^3=\frac{1}{3}$ ---A

Substitute n = 10

$t_{10}=ar^{10-1}$

$t_{10}=ar^9$

We are given that the 10th term is 243

$t_{10}=ar^9=243$ --B

Substitute the value of a from A in B

$\frac{1}{3r^3}r^9=243$

$\frac{1}{3}r^6=243$

$r^6=243 \times 3$

$r=\sqrt[6]{243 \times 3}$

$r=3$

Substitute the value of r in A

$a(3)^3=\frac{1}{3}$

$a=\frac{1}{81}$

$t_n=ar^{n-1}$

Substitute n = 2

$t_2=\frac{1}{81} 3^{2-1}$

$t_2=\frac{1}{81} \times 3$

$t_2=\frac{1}{27}$

Hence the second term is $t_2=\frac{1}{27}$

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