Math, asked by saaksravani, 4 months ago

The 4th term of a geometric progression is 2/3 and the 7th term is 16/81 find the geometric series

Answers

Answered by sitatriptibvks7
0

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,...

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