Question

Find the 10th term of the sequence root2,root 6,root18.......

Answer 1

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.

:)

Answer 2

The sequence is in GP
a=root 2. r=root 6/root 2=3
T10=a+9d
=root 2 +9*3
=root 2+27

Hope it helped u