Calculating the value of ith position in a given Arithmetic Series.

Question

I'm doing some self-study on Arithmetic series and I'm confused about why some variables are necessary to find the nth value in a given series.

Let $A = {3, 6, 9, 12, 15}$

To find the value of 3rd element the formula is $a_n = a_1 + (n-m)d$.

So if:

$n=3$

$a_1=1$

$m=1$

$d=3$

Then $a_3 = 3 + (3-1)3$ which is $9$. But then why not just say $n*d = a_n$? Because in the above case $3 * 3 = 9$. Why do i need the other variables? Is there a case where this fails?

Answer 1

Your Case only works becuase $a=d$ . Hence ,

$$a_n= a+(n-1)d = a+n*d -d = (n*d) +(a-d) = n*d$$

If we consider another A.P with $a\ne d$ , For ex . $3,7,11,15,19,.....$

then , $$a_n = 3+(n-1)4 = 4n-1$$

While according to your formula , $$a_n = 4n$$

Answer 2

There is no need for $m$ in the formula.

$$a_1, a_1+d, a_1+2d, a_1+3d, ..., a_1+ (n-1)d,...$$

So the n_th term is $$a_n = a_1 + (n-1)d$$

For example in $$5,9,13, 17,...$$

we have $$a_{20} = 5+19(4) = 81 $$