find the 10th term of the sequence root2,root 8,root18.......
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Hi ,
Given sequence is √2 , √8, √18,...
a1 = √2 ,
a2 = √8 = √ ( 2 × 2 ) ×2 = 2√2
a3 = √18 = √ ( 3 × 3 ) × 2 = 3 √2
First term (a ) = a1 = √2
a2 - a1 = 2√2 - √2 = √2
a3 - a2 = 3√2 - 2√2 = √2
Therefore ,
a2 - a1 = a3 - a2 = √2
Common difference (d) = √2
The given sequence is in A.P .
nth term = an = a + ( n-1 ) d
10th term of the A.P = a10
a10 = a + ( 10 - 1 ) d
= a + 9d
= √2 + 9 × √2
= √2 + 9√2
= 10√2
10 th term of the given A.P = 10√2
Or
a10 = √200
I hope this helps you.
:)
Given sequence is √2 , √8, √18,...
a1 = √2 ,
a2 = √8 = √ ( 2 × 2 ) ×2 = 2√2
a3 = √18 = √ ( 3 × 3 ) × 2 = 3 √2
First term (a ) = a1 = √2
a2 - a1 = 2√2 - √2 = √2
a3 - a2 = 3√2 - 2√2 = √2
Therefore ,
a2 - a1 = a3 - a2 = √2
Common difference (d) = √2
The given sequence is in A.P .
nth term = an = a + ( n-1 ) d
10th term of the A.P = a10
a10 = a + ( 10 - 1 ) d
= a + 9d
= √2 + 9 × √2
= √2 + 9√2
= 10√2
10 th term of the given A.P = 10√2
Or
a10 = √200
I hope this helps you.
:)
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