Correct Answer - Option 4 : 9 or 3
Concept:
The sum of an infinite geometric progression \(\rm S = \dfrac {a}{1-r}\), Where a is the first term and r is common ratio of geometric progression.
Calculations:
Given, the sum of an infinite geometric progression is 6.
The sum of an infinite geometric progression \(\rm S = \dfrac {a}{1-r}\)
⇒ \(\rm \dfrac {a}{1-r} = 6\)
⇒ a = 6 - 6r.
Also, given the sum of the first two terms is 9/2.
⇒ a + ar = \(\dfrac 92\)
⇒ a(1 + r) = \(\dfrac 92\)
Put the value of a in above equation, we get
⇒ (6 - 6r)(1 + r) = \(\dfrac 92\)
⇒ 6(1 - r)(1 + r) = \(\dfrac 92\)
⇒ 1 - r2 = \(\dfrac 23\)
⇒ r2 = \(\dfrac 14\)
⇒ r = \(\pm \dfrac 12\)
When r = \(\dfrac 12\), a = 6 - 6 (\(\dfrac 12\)) = 6 - 3 = 3
and when r = \(\dfrac {-1}2\), a = 6 - (6) \(\dfrac {-1}2\) = 6 + 3 = 9
Hence, the sum of an infinite geometric progression is 6. If the sum of the first two terms is 9/2, then the first term is 3 or 9.
