the sum of first terms of a gp is 30 and the sum of infinite number of terms of the same gp is 40 the common ratio of this gp is
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Sum of 1
st
4 terms=30
⇒
r−1
a
1
(r
4
−1)
=30−(i)
Sum of 1
st
8 terms=30+480
⇒
r−1
a
1
(r
8
−1)
=510
⇒
r−1
a
1
(r
4
−1)(r
4
+1)
=510−(ii)
Dividing (ii) by (i) we get
r−1
a
1
(r
4
−1)
r−1
a
1
(r
4
−1)(r
4
+1)(r−1)
=
30
510
⇒
a
1
(r
4
−1)(r−1)
a
1
(r
4
−1)(r
4
+1)(r−1)
=
30
510
⇒r
4
+1=17
⇒r
4
=16
⇒r=2
∴ From (i) replacing r=2
⇒
2−1
a
1
(2
4
−1)
=30
⇒a
1
×(16−1)=30×1
⇒a
1
×15=30⇒a
1
=2
∴ Sum of first 12 terms
S
12
=
r−1
a
1
(r
12
−1)
=
2−1
2×(2
12
−1)
=2×(2
12
−1)
=2×4095
=8190
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