In given G.P., a=25 and r = 125/25 = 5.
Given, a_n = 390625 => ar^n-1 = 390625
=>25*5^n-1 = 390625
=> 5^n-1 = 390625/25 = 15625 = 5⁶
=> n-1 = 6 => n=7.

Given, a=20, a₁₂= 120, n=12.
S_n = \(\frac{n}{2}\) (a+l) => S₁₂ = \(\frac{12}{2} (20+120)\) = 6*140 = 840.

Since 2,3,5 and 7 all are consecutive prime numbers so, it is a sequence of prime numbers. Prime number next to 7 is 11. So, 2,3,5,7,11.

We know, sum of squares of first n terms is given by \(\frac{(n(n+1)(2n+1))}{6}\).
Here, n=10. So, sum = 10*11*21/6 = 385.

When we use addition between terms of sequence, it is said to be series.
We know that addition can also be written in form of sigma so, series can also be denoted by ∑a_n.

Let three terms be a/r, a, a*r.
Product = 27 => (a/r) ( a ) (a*r) = 27 => a³ = 27
=>a = 3.
Sum = 21/2 => (a / r + a + a*r) = 21/2 => a (1 / r + 1 + 1*r) = 21/2
=> (1 / r + 1 + 1*r) = (21/2)/3 = 7/2
=> (r² + r + 1) = (7/2) r => r² ? (5/2) r + 1 = 0
=> r = 2 and 1/2.

We know that arithmetic mean of two numbers is given by average of two numbers i.e. A.M. = (2+6)/2=8/2 = 4.

Sum of squares of first n terms = 1²+2²+3²+?????+n²
k³?(k ? 1)³=3k²?3k + 1
On substituting k = 1, 2, 3, ??, n and adding we get,
n³ = 3 \(\sum_{i=0}^n k^2 – 3 \sum_{i=0}^n k + n\)
n³ = 3 \(\sum_{i=0}^n k^2 – 3 \frac{n(n+1)}{2}\) + n
\(\sum_{i=0}^n k^2 = \frac{n(n+1)(2n+1)}{6}\).

a_n = 4n+6 and n=15
=>a₁₅ = 4*15 + 6 = 60+6 = 66.

a=171 and d=162-171 = -9.
a_n<0
=>171+(n-1) (-9) < 0
=>180-9n < 0
=>9n > 180
=>n > 20 => n=21 for first negative term.
First negative term is 171+(20) (-9) = 171-180 = -9