The 4th term of a geometric progression is 2/3 and the 7th term is 16/81 find the geometric series
Answers
Answer:
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Step-by-step explanation:
Answer:
The required GP is
\frac{9}{4},\frac{3}{2},1,...49,23,1,...
Step-by-step explanation:
It is given that fourth term of a GP is 2/3 and 7th term is 16/81.
Let the first term of the GP be a and the common ratio is r.
The nth term of a GP is
a_n=ar^{n-1}an=arn−1
4th term of the GP is 2/3.
a_4=ar^{4-1}a4=ar4−1
\frac{2}{3}=ar^{3}32=ar3 .... (1)
7th term of the GP is 16/81..
a_7=ar^{7-1}a7=ar7−1
\frac{16}{81}=ar^{6}8116=ar6 .... (2)
Divide equation (2) by equation (1).
\frac{3}{2}\times \frac{16}{81}=r^{3}23×8116=r3
\frac{8}{27}=r^{3}278=r3
\frac{2}{3}=r32=r
The common ratio is 2/3.
Put this value in equation (1).
\frac{2}{3}=a(\frac{2}{3})^332=a(32)3
\frac{2}{3}=a(\frac{8}{27})32=a(278)
a=\frac{2}{3}\times \frac{27}{8}=\frac{9}{4}a=32×827=49
The first term is 9/4.
The required GP is
\frac{9}{4},\frac{9}{4}\times \frac{2}{3},\frac{9}{4}\times \frac{2}{3},...49,49×32,49×32,...
\frac{9}{4},\frac{3}{2},1,...49,23,1,...