Show that the linear function in n i.e. `f(n)=an+b` determines an arithmetic progression, where a,b are constants.

If a, b and c are in arithmetic progression, then b+ c, c + a and a + b are in

If a, b and c are three positive numbers in an arithmetic progression, then:

Sum of 'n' arithmetic means between a and b is

The nth term of a series is given by t_(n)=(n^(5)+n^(3))/(n^(4)+n^(2)+1) and if sum of its n terms can be expressed as S_(n)=a_(n)^(2)+a+(1)/(b_(n)^(2)+b) where a_(n) and b_(n) are the nth terms of some arithmetic progressions and a, b are some constants, prove that (b_(n))/(a_(n)) is a costant.

If the sum of 'n' terms of an arithmetic progression is n^(2)-2n , then what is the n^(th) term?

Show that the function f:R rarr given by f(x)=xa+b, where a,b in R,a!=0 is a bijection.

(V)A Sequence is an AP ; if its nth term is linear expressions in n i.e.a_(n)=An+B where A and B are constants.In such a case the cofficients of n is a_(n) is the common difference of AP.

Show that the function f:N rarr N given by,f(n)=n-(-1)^(n) for all n in N is a bijection.

let a < b , if the numbers a , b and 12 form a geometric progression and the numbers a , b and 9 form an arithmetic progression , then (a+b) is equal to: