Question

Take any Arithmetic progression ,Now multiply each term by a fixed number and write the resulting numbers as a list.Now, observe whether the new list is also in A.P.?Why?​

Answer 1

Answer:

Given statement is correct.

Step-by-step explanation:

Taking a sequence( in AP ) :

Let that sequence be 1 , 2 , 3 , 4 ...   : where 1 is the first term and 1 is the common difference.

In this AP :

               1st term = 1

              2nd term = 2

             3rd term = 3

A rule which will confirm if it is an AP : 2b = a + c, it means twice the second term is equal to the sum of other terms.

= > 2 x 2 = 3 + 1       = > 4 = 4        which is true, so this is an AP.

Now, multiplying this AP by a fix number( say n ) : Then

            1st term = 1n

            2nd term = 2n

            3rd term = 3n

Checking whether it is AP :

        2 x second term = 2 x 2n = 4n

       Sum of extremes = 1n + 3n = 4n

Both result the same, thus it forms an AP.

Answer 2

Step-by-step explanation:

let us consider an AP ,

a, a+d, a+2d, ...

on multiplying by a constant say n, the terms of the AP changes to

na, na+nd, na+2nd

as we know that AP is a progression in which the difference between consecutive terms is a constant so wet can observe that in the new progression the condition is satisfied so we can say that it is also an AP.

$(na+nd)-na=nd=(na+2nd)-(na+nd)$

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