Question

How do you find r and ${a_1}$ for the geometric sequence: ${a_3} = 5,\,{a_8} = \dfrac{1}{{625}}$ ?

Answer 1

In this question, we are given the 3rd and 8th term of a geometric sequence. We know the formula for finding the nth term of a geometric sequence, using that formula we will find the expression for 3rd and 8th term. Now, we have 2 unknown quantities that are the first term and the common ratio and we have exactly two equations involving these two unknown quantities. So we can easily find the value of the unknown quantities by solving the two equations.

Complete step-by-step solution:

We know that nth term of a G.P. is given as –

${a_n} = a{r^{n - 1}}$

We are given that ${a_3} = 5$ and ${a_8} = \dfrac{1}{{625}}$ , that is, we have –

$a{r^2} = 5$

And $a{r^7} = \dfrac{1}{{625}}$

Dividing these two equations, we get –

$\dfrac{{a{r^2}}}{{a{r^7}}} = \dfrac{5}{{\dfrac{1}{{625}}}}$

$\Rightarrow \dfrac{1}{{{r^5}}} = 5 \times 625$

Now, on prime factorization of 625 we get – $625 = 5 \times 5 \times 5 \times 5$

$\Rightarrow {r^5} = \dfrac{1}{{5 \times {{(5)}^4}}}$

We know that ${a^x} \times {a^y} = {a^{x + y}}$ , so $5 \times {5^4} = {5^{4 + 1}} = {5^5}$

$\Rightarrow {r^5} = \dfrac{1}{{{{(5)}^5}}}$

$\Rightarrow r = {\left(\dfrac{1}{{{5^5}}}\right)^{\dfrac{1}{5}}}$

$\Rightarrow r = \dfrac{1}{{{{({5^5})}^{\dfrac{1}{5}}}}}$

We also know that ${({a^x})^y} = {a^{xy}}$ -

$\Rightarrow r = \dfrac{1}{{{5^{5 \times \dfrac{1}{5}}}}}$

$\Rightarrow r = \dfrac{1}{5}$

Putting the value of r in one of the initial equations, we get –

$a{\left(\dfrac{1}{5}\right)^2} = 5$

$\Rightarrow a = 5 \times 25$

$\Rightarrow {a_1} = 125$

Hence, r and ${a_1}$ for the geometric sequence: ${a_3} = 5,\,{a_8} = \dfrac{1}{{625}}$ are $r = \dfrac{1}{5}$ and ${a_1} = 125$.

Note: A geometric progression is defined as a series or progression in which the ratio of any two consecutive terms of the sequence is constant, this constant value is known as the common ratio of the G.P. A geometric progression is of the form $a,ar,a{r^2}....$ and its explicit formula is found as follows –
Each term of the G.P. is given as $a{r^x}$ , where $x = n - 1$ . Thus ${a_n} = a{r^{n - 1}}$ is the explicit formula for any geometric sequence. In this question, we had to find the values of 2 unknown quantities so we formulated exactly 2 equations using the above formula.