find a2 and a3 for each geometric sequence in 8, a_2, a_3,27
Answers
Answer:
We know the terms in geometric series are in the form ar,ar2,ar3,ar4,ar5,....
Given, a1=10,r=−1
a2=a1r=10(−1)=−10
a3=a1r2=10(−1)2=10
a4=a1r3=10(−1)3=−10
a5=a1r4=10(−1)4=10
Step-by-step explanation:
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Answer:
The value of
a
10
=
19683
Explanation:
a
3
=
−
9
,
a
6
=
243
;
a
10
=
?
n
th term in G.P series is
a
n
=
a
1
⋅
r
n
−
1
;
a
1
and
r
are
first term and common ratio of G.P. series.
a
3
=
a
1
⋅
r
3
−
1
or
a
1
⋅
r
2
=
−
9
;
(
1
)
similarly ,
a
6
=
a
1
⋅
r
6
−
1
or
a
1
⋅
r
5
=
243
;
(
2
)
Dividing
equation (2) by equation (1) we get,
r
5
r
2
=
−
243
9
or
r
3
=
−
27
or
r
=
−
3
Putting
r
=
−
3
in equation (1) we
get ,
a
1
⋅
(
−
3
)
2
=
−
9
or
a
1
=
−
9
9
or
a
1
=
−
1
∴
a
10
=
a
1
⋅
r
10
−
1
=
−
1
⋅
(
−
3
)
9
=
19683
The value of
a
10
=
19683
[Ans]
Step-by-step explanation:
Explanation:
The general term for a GP is
a
n
=
a
1
r
n
−
1
where
a
1
is the first term and
r
is the common ratio.
You are given the values of two terms in a GP.
Divide the two terms: The formula and the values
a
6
a
3
=
a
r
5
a
r
2
=
243
−
9
This gives:
r
3
=
−
27
←
find the cube root.
×
×
×
×
r
=
−
3
Now find
a
1
from
a
3
a
1
r
2
=
a
1
(
−
3
)
2
=
−
9
a
1
=
−
9
9
=
−
1
Now you can find the value of
a
10
a
10
=
−
1
(
−
3
)
9
=
19
,
683