Formulae for the

nth partial sum of an infinite geometric series are

π Question 1. How many terms of the geometric sequence 3, 3^{2}, 3^{3}, β¦ are needed to give the sum 120?

Answer

^{2}+3

^{3}+3

^{4}β¦120

3+9+27+81=120

Thus, four terms of the given geometric sequence are required to obtain the sum as 120.

π Q2. The first term of a geometric series is -1, and the common ratio is -3. How many terms are in the series if its sum is 182?

(A) 6 (B)7 (C) 8 (D) 9

β Solution:

The first term in the series is -1, the common ratio is -3, and the sum is 182. Use the formula for the sum of a finite geometric series to find the number of terms *n*.

Therefore, the correct answer is (A).

π Q3. How many terms of the geometric sequence 3, 3/2, ΒΎ, β¦ are needed to give the sum 3069/512?

β Solution: Let *n* be the number of terms needed. Given that *a*=3, *r*=Β½ and *S _{n}*=3069/512

which gives

*n*=10.

π Example 1. The sum of some terms of a geometric sequence is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.

β Solution:

Let the sum of *n* terms of the geometric sequence be 315.

It is known that,

It is given that the first term

*a*is 5 and common ratio

*r*is 2.

β΄ Last term of the geometric sequence = 6th term =

*aβ r*

^{(6-1)}=5β 2

^{5}=5β 32=160 Thus, the last term of the geometric sequence is 160.

π Activity 1.

1. Determine 3+6+12+24+β¦ to 10 terms

2. If 2+6+18+β¦=728, determine the value of *n*.

β Solutions:

π Q4. How many terms in the geometric sequence 1, 1.1, 1.21, 1.331, β¦ will be needed so that the sum of the first *n* terms is greater than 20?

β Solution:

The sequence is a geometric sequence with *a*=1 and *r*=1.1. We want to find the smallest value of *n* such that *S _{n}*>20. Now

If we now take logarithms of both sides, we get

*n*lnβ‘1.1>lnβ‘3

and as lnβ‘1.1>0 we obtain

*n*>lnβ‘3βlnβ‘1.1 =11.5267β¦

and therefore the smallest whole number value of

*n*is 12.

π Q5. The first term of a geometric series is 5, and the common ratio is -2. How many terms are in the series if its sum is -6825?

A 5 B 9 C 10 D 12

β Solution:

General sum Formula

Therefore, D is the correct answer.

Answer: D

π Ex2. If *S _{n}*=61/40, β
+Β½+β
+β―, find

*n*.

β Solution:

Find the common ratio.

β Γ·Β½=β

Substitute

*S*=61/40,

_{n}*a*

_{1}=β , and

*r*=β into the formula for the

*n*th partial sum of an infinite geometric series.

Because

*a*=β and the third term of the sequence is β ,

_{n}*n*=3.

π Ex3. Find *n* for 4.1+8.2+16.4+β― if *S _{n}*=61.5.

β Solution:

The common ratio is 2.

Substitute

*S*=61.5,

_{n}*a*

_{1}=4.1, and

*r*=2 into the formula for the

*n*th partial sum of an infinite geometric series.

Substitute

*a*=32.8,

_{n}*a*

_{1}=4.1, and

*r*=2 into the formula for the

*n*th term of a geometric sequence to find

*n*.