Question

The sum of an infinite GP is 8, its second term is 2, find its first term.

Answer 1

Hint: First of all, consider the first term of the given infinite terms of GP as a variable and find the common ratio as we have given the value of the second term. Then equate the sum of infinite terms in GP to the given value to obtain the required answer.

Complete Step-by-Step solution:
We know that the sum of terms in infinite series of a GP with first term \[a\] and common ratio \[r\] is given by \[{S_\infty } = \dfrac{a}{{1 - r}}\].
Given the sum of infinite terms in GP \[{S_\infty } = 8\]
The second term in GP \[{a_2} = 2\]
Let the first term of the GP be \[a\]
The common ratio of the GP is given by \[\dfrac{{{a_2}}}{a} = \dfrac{2}{a}\]
By the above data and using formula of sum of terms in infinite GP, we have
\[
  {S_\infty } = \dfrac{a}{{1 - r}} = \dfrac{a}{{1 - \dfrac{2}{a}}} = 8 \\
  \dfrac{a}{{1 - \dfrac{2}{a}}} = 8 \\
  \dfrac{a}{{a - 2}} \times a = 8 \\
  \dfrac{{{a^2}}}{{a - 2}} = 8 \\
  {a^2} = 8\left( {a - 2} \right) \\
  {a^2} = 8a - 16 \\
  {a^2} - 8a + 16 = 0 \\
  {a^2} - 2\left( 4 \right)\left( a \right) + {\left( 4 \right)^2} = 0 \\
\]
By using the formula \[{a^2} - 2ab + {b^2} = {\left( {a - b} \right)^2}\], we have
\[
  {\left( {a - 4} \right)^2} = 0 \\
  \therefore a = 4 \\
\]
Thus, the value of the first term is 4.

Note: The common ratio of a series in GP is the ratio of any two consecutive terms in the GP. The sum of terms in an infinite series of a GP with first term \[a\] and common ratio \[r\] is given by \[{S_\infty } = \dfrac{a}{{1 - r}}\]. Remember that here the value of common ratio is always less than one i.e., \[r < 1\].