Question

Prove that no matter what the real numbers a and b are, the sequences with nth term a+nb is always in A.P. what is the common difference?​

Answer 1

Step-by-step explanation:

ANSWER

Given the sequence which is defined by.

a

n

=a+nb.....eq(1)

and we know that nth term of an A.p is given by

a

n

=a

1

+(n−1)d

which can also be written as

a

n

=a

1

−d+nd.....eq(2)

comparing eq(1) and eq(2) we get

common difference is

d=b (by comparing coefficients of n)

and

a

1

−d=a

by putting value of d=b in above equation we get

a

1

−b=a

⟹a

1

=a+b (first term of A.P)

hence if a and b are real numbers this sequence form an A.P with first term a

1

=a+b

and have a common difference d=b

Answer 2

Answer:

let the nth term of a given profession be given by

.tn=an+b. 【where a and b is constant 】

=)then ,tn-1=a(n-1)+b

=)tn-1 =an+b-a.

now ,there is two term let tn-1 is first term and tn is first term of an Ap.

【like t2-t1=d 】simularly here

so,tn-tn-1=d

so,an+b-an-b+a

=)a is common difference ,which is constant ,

hence the given progression is in an Ap