Question

prove that no matter what the real numbers a and b are, the sequence with nth term a +nb is always an A.P. what is the common difference help me guys

Answer 1

Heya friends!!
.
I think it's too easy ☺..

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 progession is in an Ap


hope it help you 1☺

@rajukumar☺

Answer 2

Answer:Yes it is an AP

Step-by-step explanation: 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 progession is in an Ap

@divyanshugaur555

#Brainliest