Question

is 121 a term of AP 6,10,14...?​

Answer 1

Here, a=6 , d=10-6=4

We want to see if 121 is a term of following A.P.

an=121

an= a+(n-1)d

121 = 6 + (n-1)4

115 = (n-1)4

115/4 = n-1

115/4 + 1 =n

119/4 = n


But n is a natural number


Therefore n is not equal to 119/4


And therefore we can say that 121 is not a term of the AP

Answer 2

Answer:

No,

121 is not a term of AP 6,10,14.

Step-by-step explanation:

Let me start off by explaining what is AP or Arithmetic Progression.

It is a sequence of numbers in which the difference between the preceding and succeeding numbers are the same as the next preceding and succeeding number.

For example, here

the numbers given are 6, 10, 14

So, the difference between 6 and 10 is 4, also the difference between 10 and 4 is 4.

I hope, you,re getting it now.

Now coming to whether 121 is a term of this arithmetic progression or not.

In this question they have used all the even numbers to get a difference as 4 and 121 is not a even number

The number should have been 120 and 116, we can guess this by the numbers given in the question.

Thus, 121 is not a term of this particular AP

The general formula for AP to find out how many terms can be there is

Last term= a+(n-1)d

where a= first term

           n= number of terms

           d= difference between the terms

#SPJ3