Infinite Geometric Series Formula

Ans. An infinite geometric series (G.S.) is given as follows;

a + ar + ar² + ar³ +…,arⁿ+….\(\infty\)

Where a is called first term and r is the common ratio. It is known that this series will be convergent i.e., its sum will be definite and finite quantity, provided | r | < 1. So, taking | r | < 1 and denoting by S the sum of the series, that is;

S=a + ar + ar² + ar³ +…,arⁿ+…..(1) Then multiply it with r, we obtain

Sr=ar +  ar² + ar³ + ar⁴…,arⁿ….(2)

Now, (1) - (2) will give;

(1-r)S=a  which in turn implies

S=a + ar + ar² + ar³ +…\(\infty\)

Ans. Geometric series: a + ar + ar² + ar³ +…

The first term is equal to twice the sum of all the terms that follow it:

a = 2(ar + ar² + ar³ +…)

a=\(\frac{2ar}{1-r}\)

a(1-r)=2ar

(1-r)=2r (Assuming a≠0 )

1=3r 

r = \(\frac{1}{3}\)

Ans. An infinite geometric series has a total of 6 and a starting term of 2.

Let a be the first term of the geometric series and r be the common ratio.

The geometric series is therefore given by... 

a + ar + ar² + ar³ +…

Which converges to \(\frac{a}{1-r}\), r<1?

i.e., a+ar + ar² + ar³ +….\(\frac{a}{1-r}\)

⇒a(1+r+r²+r³+…)= 6 = \(\frac{a}{1-r}\)

\(\frac{2}{1-r}\)=6

⇒\(\frac{1 }{1-r}\)=3

⇒1-r = \(\frac{1}{3}\)

⇒r = 1 – \(\frac{1}{3}\)

⇒ r = \(\frac{2}{3}\)

Hence the required common ratio of the geometric series, whose sum 6 is \(\frac{2}{3}\).

Ans. Sum to Infinity = \(\frac{a}{1-r}\)

10 = \(\frac{2}{1-r}\)

Multiply both sides by (1-r) and divide both sides by 10

10 – 10r = 2 

10r = 10 – 2 

10r = 8 

r= \(\frac{8}{10}\)or \(\frac{4}{5}\)

Ans. \(\frac{1}{2}\) + \(\frac{3}{4}\) + \(\frac{9}{8}\) +….\(\infty\) 

a = \(\frac{1}{2}\)

Common ratio = \(\frac{\frac{3}{4}}{\frac{1}{2}}\)  = \(\frac{3}{4}\)x 2 = \(\frac{3}{2}\)

Common ratio = \(\frac{\frac{9}{8}}{\frac{3}{4}}\)  = \(\frac{9}{8}\)x \(\frac{4}{3}\) = \(\frac{3}{2}\)

So, the common ratio is = \(\frac{3}{2}\)

Sum till \(\infty\) =\(\frac{a }{1-r}\) r > 1

So, \(\frac{\frac{1}{2}}{1- \frac{3}{2}}\)

= \(\frac{\frac{1}{2}}{\frac{3-2}{2}}\) = \(\frac{1}{2}\)×2=1

Ans. From the question we have 

a + ar + ar² + ar³ +…= 15….. (1)

a² + a²r² + a²r⁴ + a²r⁶ +…= 45…... (2)

Solving the first equation 

a + ar + ar² + ar³ +…=15

\(\frac{a }{1-r}\) = 15……. (3)

Solving the second equation 

a² + a²r² + a²r⁴ + a²r⁶ +…=45

\(\frac{a^2 }{1-r^2}\) = 45

\(\frac{15^2(1-r)^2 }{1-r^2}\)=45, from equation 3 

5(1+r²-2r)  = 1-r²

5 + 5r² – 10r – 1 + r²=0

6r² - 10r + 4=0

3r² - 5r + 2=0

(r-1)(3r-2) = 0

But r ≠ 1

So, r = \(\frac{2}{3}\)

By substituting the value of r inequation (3) we get 

a = 15(1 – \(\frac{2}{3}\))

a = 5 

Thus, the required series is 

5 + \(\frac{10}{3}\) + \(\frac{20}{9}\) + \(\frac{40}{27}\) +….\(\infty\)

Ans. Since, \(\frac{a }{1-r}\)=3 

a=3(1 – r)

\(\frac{a^3 }{1-r^3}\) = \(\frac{27}{19}\)

\(\frac{27(1-r)^3 }{1-r^3}\) = \(\frac{27}{19}\) As long as the terms drop to zero, the total of a geometric series is finite; as the numbers approach zero, they turn non - significantly small, enabling a sum to be computed despite the series being infinite.

6r² – 13r + 6 = 0

r  = \(\frac{2}{3}\) as r<1

Ans. It is required (but not enough) for an infinite series to converge if the terms approach a limit of zero the terms of a geometric series with a common ratio larger than 1 keep on increasing, reaching positive infinity. The words swing back and forth between positive and negative if the common ratio is less than -1, but their absolute value keeps on growing and never goes close to zero.