Question Video: Finding the First Term of an Infinite Geometric Sequence given Its Common Ratio and the Sum of the Terms | Nagwa Question Video: Finding the First Term of an Infinite Geometric Sequence given Its Common Ratio and the Sum of the Terms | Nagwa

# Question Video: Finding the First Term of an Infinite Geometric Sequence given Its Common Ratio and the Sum of the Terms Mathematics • Second Year of Secondary School

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Find the first term of the infinite geometric sequence where the common ratio is 1/4 and the sum is 98 6/7.

02:33

### Video Transcript

Find the first term of the infinite geometric sequence where the common ratio is one-quarter and the sum is 98 and six-sevenths.

Weβre given two pieces of information about this infinite geometric sequence. Firstly, its common ratio is one-quarter. This is the value we multiply each term by to give the following term. And secondly, weβre told that the sum of this infinite geometric sequence is 98 and six-sevenths. The formula for calculating the sum of an infinite geometric sequence, which we denote by π and then a subscript β, is π sub one over one minus π, where π sub one denotes the first term in the sequence and π denotes the common ratio.

Now, this formula is only valid if the absolute value of the common ratio is strictly less than one, as this is required for the series to be convergent. The value of π in this question is one-quarter. So the sum to β can indeed be found. We can answer this question then by substituting the value for the infinite sum, which is 98 and six-sevenths, and the value for the common ratio, one-quarter, and then solving the equation to give π sub one, which remember is the first term in the sequence. So we have 98 and six-sevenths is equal to π sub one over one minus a quarter. One minus a quarter is of course three-quarters. And at the same time, we may find it preferable to write the mixed number 98 and six-sevenths as the improper fraction 692 over seven.

To solve this equation for π sub one, we need to multiply both sides of the equation by three-quarters. So we have π sub one is equal to 692 over seven multiplied by three over four. Before we evaluate, we can cross cancel a factor of four from the numerator of the first fraction and the denominator of the second to give 173 over seven multiplied by three over one. 173 multiplied by three is 519. So we have π sub one is equal to 519 over seven.

So by recalling the formula for the sum of an infinite geometric sequence, which is valid in this case as the absolute value of the common ratio is strictly less than one, we were able to form and solve an equation to find that the first term of this sequence is 519 over seven.

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