Question

if a+1, 2a+1, 4a-1 are in A-P fibd the value of a

Answer 1

Answer:

A=2

Step-by-step explanation:

Let the first be x = a+1

2nd term be y = 2a+1

3rd term be z= 4a-1

y-x=z-y

2y=x+z

2(2a+1) = a+1+4a-1

4a+2 = 5a

a= 2

Answer 2

We must recall that:

An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.

Given: $a+1,2a+1,4a-1$ are in AP

$a_2-a_1=a_3-a_2$

$2a+1-a-1\\=4a-1-2a-1$

$a=2a-2$

$a-2a=-2$

$-a=-2$

$a=2$

Hence $a=2.$