Formulae for the nth partial sum of an infinite geometric series are

📌 Question 1. How many terms of the geometric sequence 3, 3², 3³, … are needed to give the sum 120?
Answer

3+3²+3³+3⁴…120
3+9+27+81=120
Thus, four terms of the given geometric sequence are required to obtain the sum as 120.

📌 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
✍ 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).

📌 Q3. How many terms of the geometric sequence 3, 3/2, ¾, … are needed to give the sum 3069/512?
✍ Solution: Let n be the number of terms needed. Given that a=3, r=½ and S_n=3069/512

which gives n=10.

📌 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.
✍ 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.

∴ Last term of the geometric sequence = 6th term =a⋅r^(6-1)=5⋅2⁵=5⋅32=160 Thus, the last term of the geometric sequence is 160.

📌 Activity 1.
1. Determine 3+6+12+24+… to 10 terms
2. If 2+6+18+…=728, determine the value of n.
✍ Solutions:

📌 Q4. How many terms in the geometric sequence 1, 1.1, 1.21, 1.331, … will be needed so that the sum of the first n terms is greater than 20?
✍ 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 ln⁡1.1>ln⁡3
and as ln⁡1.1>0 we obtain
n>ln⁡3⁄ln⁡1.1 =11.5267…
and therefore the smallest whole number value of n is 12.

📌 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
✍ Solution:
General sum Formula

Therefore, D is the correct answer.
Answer: D

📌 Ex2. If S_n=61/40, ⅝+½+⅖+⋯, find n.
✍ Solution:
Find the common ratio.

½÷⅝=⅘
⅖÷½=⅘
Substitute S_n=61/40, a₁=⅝, and r=⅘ into the formula for the nth partial sum of an infinite geometric series.

Because a_n=⅖ and the third term of the sequence is ⅖, n=3.

📌 Ex3. Find n for 4.1+8.2+16.4+⋯ if S_n=61.5.
✍ Solution:
The common ratio is 2.
Substitute S_n=61.5, a₁=4.1, and r=2 into the formula for the nth partial sum of an infinite geometric series.

Substitute a_n=32.8, a₁=4.1, and r=2 into the formula for the nth term of a geometric sequence to find n.