the 4th term of a g. p is 2/3 and 7th term is 16/81. find the geometric series
Answers
Step-by-step explanation:
a.r3=2/3
a.r6=16/81
a.r3/a.r6=2/3*81/16
1/r3=27/8
r=2/3
a.r3=2/3
a=9/4
geometric series=9/4,3/2,1.....
Answer:
t(1)=ar^(n-1)=9/4×r^(1-1)=9/4
t(2)=ar^(n-1)=9/4×r^(2-1)=9/4×2/3=18/12=3/2
t(3)=ar^(n-1)=9/4×r^(3-1)=9/4×4/9=36/36=1
t(4)=ar^(n-1)=9/4×r^(4-1)=9/4×8/27=72/108=2/3
and soon
Step-by-step explanation:
The nth term of a GP series is Tn = ar^(n-1), where a = first term and r = common ratio = Tn/Tn-1) . The sum of infinite terms of a GP series S∞= a/(1-r) where 0< r<1. If a is the first term, r is the common ratio of a finite G.P. consisting of m terms, then the nth term from the end will be = ar^m-n.
Case-i
t(n)=ar^(n-1)
2/3=ar^(4-1)
2/3=ar^3
2/3r^3=a----------i
Case-ii
t(n)=ar^(n-1)
16/81=ar^(7-1)
16/81=ar^6
16/81=ar^6
16/81r^6=a -----------ii
By equation i and ii we get,
2/3r^3=16/81r^6
By cross multiplying we get,
48r^3=162r^6
48/162=r^3
8/27=r^3
r=2/3
Substitute r value in equation i
2/3r^3=a
2/(3×(2/3)^3)=a
2/(3×(8/27))=a
2/(8/9)=a
a=9/4
t(1)=ar^(n-1)=9/4×r^(1-1)=9/4
t(2)=ar^(n-1)=9/4×r^(2-1)=9/4×2/3=18/12=3/2
t(3)=ar^(n-1)=9/4×r^(3-1)=9/4×4/9=36/36=1
t(4)=ar^(n-1)=9/4×r^(4-1)=9/4×8/27=72/108=2/3
and soon
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