💎 Examples 1 to 7. Write the first 3 terms of the infinite geometric series with the given characteristics.
1. S=12, r=½
✍ Solution:
Substitute S=12 and r=½ into the formula for the sum of an infinite geometric series to find a₁.

Use r=½ to find a₂ and a₃.
6⋅½=3
3⋅½=3/2.
Therefore, the first three terms of the sequence are 6, 3, and 3/2.

2. S=-25, r=0.2
✍ Solution:
Substitute S=-25 and r=0.2 into the formula for the sum of an infinite geometric series to find a₁.

Use r=0.2 to find a₂ and a₃.
-20⋅0.2=-4
-4⋅0.2=-⅘.
Therefore, the first three terms of the sequence are -20, -4, and -⅘.

3. S=-60, r=0.4
✍ Solution:
Substitute S=-60 and r=0.4 into the formula for the sum of an infinite geometric series to find a₁.

Use r=0.4 to find a₂ and a₃.
-36(0.4)=-14.4
-14.4(0.4)=-5.76.
Therefore, the first three terms of the sequence are -36, -14.4, and -5.76.

4. S=⅔, a₁=8/9
✍ Solution:
Substitute S=⅔ and a₁=8/9 into the formula for the sum of an infinite geometric series to find r.

Use r=⅓ to find a₂ and a₃.

Therefore, the first three terms of the sequence are 8/9, -8/27, and 8/81.

5. S=-115, a₁=-138
✍ Solution:
Substitute S=-115 and a₁=-138 into the formula for the sum of an infinite geometric series to find r.

Use r=-0.2 to find a₂ and a₃.
-138⋅(-0.2)=27.6
27.6⋅(-0.2)=-5.52.
Therefore, the first three terms of the sequence are -138, 27.6, and -5.52.

6. S=44.8, a₁=56
✍ Solution:
Substitute S=-25 and a₁=56 into the formula for the sum of an infinite geometric series to find r.

Use r=-0.25 to find a₂ and a₃.
56(-0.25)=-14
-14(-0.25)=3.5.
Therefore, the first three terms of the sequence are 56, -14, and 3.5.

7. S=-126.25, a₁=-50.5
✍ Solution:
Substitute S=-126.25 and a₁=-50.5 into the formula for the sum of an infinite geometric series to find r.

Use r=0.6 to find a₂ and a₃.
-50.5⋅0.6=-30.3
-30.3⋅0.6=-18.18.
Therefore, the first three terms of the sequence are -50.5, -30.3, and -18.18.

📌 Ex8. The sum to infinity of a geometric series is 36 and the second term of the series is 8. Find two possible series.
✍ Solution:
Let the series be a+ar+a⋅r²+ …
Given: S_∞=36

Given: u₂=8.∴a⋅r=8.
We now solve between (1) and (2):
a⋅r=8 … (2)
36(1-r)r=8 [replace a with 36(1-r)]
(36-36r)r=8
36r-36r²=8
-36r²+36r-8=0
36r²-36r+8=0
9r²-9r+2=0
(3r-2)(3r-1)=0
3r-2=0 or 3r-1=0
3r=2 or 3r=1
r=⅔ or r=⅓.
Put r=⅔ and r=⅓ into (1) or (2) to find the value of a.

Thus, we have two series which obey the two given conditions:
(i) a=12, r=⅔, the series is 12+8+5⅓+ …
(ii) a=24, r=⅓, the series is 24+8+2⅔+ …