Arithmetic Progression generation with sum of all elements , first and last elements provided

Question

Is there a method to find the common difference between elements of a progression with the first element, last element and the sum of all elements provided? The sum of all elements of the progression must match the value provided.

Answer 1

HINT:

We know $$S_n=\dfrac n2\{2a+(n-1)d\}=\dfrac n2(a+l)$$

where $n$ is the number of elements, $a$ is the first term, $l$ is the last term, $d$ being the common difference.

Answer 2

Yes you can find the difference using first term, last term and sum.

I am taking example to show.

Let first term a=2, last term $a_n$=16 and sum S=72.

Then,

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

$16=2+(n-1)d$

$d = \frac{14}{(n-1)} .....(1)

Also,

S = $\frac n2[2a+(n-1)d$

$72 = \frac n2 [4+nd-d]$

$144 = 4n + n^2d - nd$

$144 = 4n + n^2 \cdot \frac{14}{(n-1)} - n \cdot \frac{14}{(n-1)}$

$144(n-1) = 4n(n-1) + 14n^2 - 14n$

$144n-144=4n^2-4n+14n^2-14n$

$18n^2-162n+144=0$

$n^2-9n+8=0$

You can find value of n. Then from equation (1) you can find d.

In this example we got two series. Maybe in your question its only one series.