Given, a₅ = 27 and a₈ = 729.
=>ar⁴ = 27 and ar⁷ = 729
On dividing we get, r³ = 27 => r=3
=> a=27 / (3⁴) = 1/3
=>a₁₁ = ar¹⁰ = (1/3) (3¹⁰) = 39 = 19683.

Given, n^th term is n(n-2)
So, a_k = k(k-2)
Taking summation from k=1 to k=n on both sides, we get
\(\sum_{i=0}^n a_k = \sum_{i=0}^n k^2 ? 2 \sum_{i=0}^n k = \frac{n(n+1)(2n+1)}{6} – 2\frac{n(n+1)}{2} = \frac{n(n+1)(2n-5)}{6}\).

Given series is G.P. with first term 1 and common ratio 1/2.
We know, S_n = a(1-r_n)/(1-r) for r<1.
S₆ = 1(1-(1/2)⁶) / (1-1/2) = (1-1/64) / (1/2) = 63*2/64 = 63/32.

Sum of series 2+3+5+7 is finite because given series has finite number of terms. The sum of given 4 terms i.e. 17.

If a is first term of G.P., n G.P. look like a, a, a, a, ????
Then sum to n terms becomes n*a.

Given, a_n = 3n.
We know, d = a_n-a_n-1 = 3n ? 3(n-1) = 3.

a₁=1 and a₂=1.
a_n = a_n-1 + a_n-2, n>2.
This is a recurrence relation which gives Fibonacci sequence.

Given, n^th term is n² + 3ⁿ
So, a_k = k² + 3^k
Taking summation from k=1 to k=n on both sides, we get
\(\sum_{i=0}^na_k = \sum_{i=0}^nk^2 + \sum_{i=0}^n3^k\)
\(\sum_{i=0}^nk^2 = n(n+1) (2n+1)/6\)
\(\sum_{i=0}^n3^k = 3*(3^n-1)/ (3-1) = (3/2) (3^n-1)\)
So, \(\sum_{i=0}^na_k = n(n+1) (2n+1)/6 + (3/2) (3^n-1)\)
Sum up to 6^th term = 6*7*13/6 + (3/2) (3⁶-1) = 91+1092 = 1183.

The isometric series is not a series. Rest all are series i.e. arithmetic series, geometric series and harmonic series.

We know, G.M. of two numbers a and b is ?ab.
So, a + b = 4 ?ab
Squaring we get, a²+b² = 16ab
=>(a/b) + (b/a) = 16
Let x = a/b.
So, x + 1/x = 16 => x² ? 16x + 1 = 0
=>x = \(\frac{16±\sqrt{256-4}}{2} = \frac{16±\sqrt{252}}{2} = \frac{16±6\sqrt{7}}{2} = \frac{8±3\sqrt{7}}{1}\).