the third term of an GP is 18 and the 7th term is 32/9 find the 10th term
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In detail:
Let 1st term be 'a'.
So, ar^2=18.......(1)...(given)
Also ar^6=32/9..(2)...(given)
Dividing (2) by (1), we get,
ar^6/ar^2 = 32/162
r^4=16/81.
r=2/3.
Putting r in eqn.(1)
a=81/2.
So, 10th term = ar^9
= (81/2) ×(512/19683)
= 256/243.
...HOPE IT IS CLEAR...
Let 1st term be 'a'.
So, ar^2=18.......(1)...(given)
Also ar^6=32/9..(2)...(given)
Dividing (2) by (1), we get,
ar^6/ar^2 = 32/162
r^4=16/81.
r=2/3.
Putting r in eqn.(1)
a=81/2.
So, 10th term = ar^9
= (81/2) ×(512/19683)
= 256/243.
...HOPE IT IS CLEAR...
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Answer:
10th term = 256/243
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The third term of an GP is 18 and the 7th term is 32/9 find the 10th term
Report by Aka1e8 31.12.2017
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In detail:
Let 1st term be 'a'.
So, ar^2=18.......(1)...(given)
Also ar^6=32/9..(2)...(given)
Dividing (2) by (1), we get,
ar^6/ar^2 = 32/162
r^4=16/81.
r=2/3.
Putting r in eqn.(1)
a=81/2.
So, 10th term = ar^9
= (81/2) ×(512/19683)
= 256/243.
...HOPE IT IS CLEAR...
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