Question

4th and 10th terms of a geometric sequence are 1/8 and 1/128 respectively. Find the 5th term

Answer 1

@ganeshie8 @Whitemonsterbunny17 @samhang @.Sam. @EclipsedStar @sweetburger

Answer 2

You know the GP formula right?

Answer 3

whats the gp formula?

Answer 4

\[a _{n}=a _{1}r ^{n-1} \]
where r is common ration, a1 is the first term of the gp, n is the number of terms

Answer 5

What do we do now?

Answer 6

We dont know a1 nor r, so we have to find them.
Now you use the given value and put it in the formula, you will get two equations, using which you can eliminate a1 to find r.

Answer 7

\[a_4=a_1r^{4-1}=\frac{1}{8} \\ \\ a_{10}=a_1r^{10-9}=\frac{1}{128}\]
is this it?

Answer 8

check your a10 again?

Answer 9

oops

Answer 10

@qtpi3.14 and can you also check the question itself? I feel there might be some mistake... just check and if it is correct we will proceed with the same.

Answer 11

not sure

Answer 12

i got \[a_1=128^2\]

Answer 13

How did you get that? :O
No that is wrong.

Answer 14

\[a_5=128^2(^6\sqrt{\frac{1}{128})}^4\]

Answer 15

how did you find a1?

Answer 16

um
I made a mistake

Answer 17

Can you check your mistake?

Answer 18

hold on

Answer 19

from a4 i'm able to find a1=1/8r^3

Answer 20

once a1 is 1/8r^3

Answer 21

now substituting a1 into a10
\[a_{10}=(\frac{1}{8r^3})(r^9)=\frac{1}{128}\]

Answer 22

Yes that is correct

Answer 23

\[r=~^6\sqrt{\frac{1}{16}}\]

Answer 24

Perfect, that is correct.

Answer 25

\[8a_1=\frac{1}{[(\frac{1}{16})^{\frac{1}{6}}]^3}\]

Answer 26

where did 8a1 come from?

Answer 27

from a4

Answer 28

i got a1=1/2

Answer 29

Yeah now you know r,a1,n. Substitute in the general formula and that is the answer

Answer 30

thank you

Answer 31

Anytime :)

Answer 32

@qtpi3.14 still here? lol