3. The difference between the first and sixth terms of a geometric
progression is 1,210 and q = 3. Find Ss.
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Answer:
Given that: T
3
=12 and T
6
=96
T
3
=12
⇒ar
2
=12 ......... (1)
T
6
=96
ar
5
=96 ........ (2)
divide (2) by (1):
ar
2
ar
5
=
12
96
∴r
3
=8
r
3
=2
3
∴r=2
Consider, ar
2
=12
a(2)
2
=12∴a×4=12∴a=3
Now, let us find S
9
.
S
n
=
r−1
a(r
n
−1)
S
9
=
2−1
3(2
9
−1)
=
1
3(512−1)
=3(511)
∴S
9
=1533
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