Prove that the nth term of the GP with first term a and common ratio r is given by `a_n=ar^(n-1)`

Prove that the sum of n terms of GP with first term a and common ratio r is given by S_(n)=a(r^(n)-1)/(r-1)

Write three terms of the GP when the first term : a 'and the common ratio r are given a=4, r=3

Theorem: The sum of nth terms of an AP with first term a and common difference d is S_(n)=(n)/(2)(2a+(n-1)d)

if S_(n) denotes the sum of n terms of a G.P whose first term is a and common ratio is r then find the sum of S_(1),S_(3),S_(5).........,S_(2)n-1

Sum to infinity of a GP is 15 and the sum to infinity of their squares is 45 .If a is the first term and r is the common ratio then the sum of the first 5 terms of the AP with first term a and common difference 3r is

If S_(n) denotes the sum fo n terms of a G.P. whose first term and common ratio are a and r respectively then show that: S_(1)+S_(3)+S_(5)+……..+S_(2n)-1=(an)/(1-r)-(ar(1-r^(2n)))/((1-r^(2))

Find the sum of the first 25 terms of an A.P. whose nth term is given by a_(n)=2-3n

Find the sum of the first 25 terms of an A.P. whose nth term is given by a_(n)=7-3n