An algebra problem by RMNDK4life

Call $a_1;a_2;a_3;a_4;a_5$ are that sequence. Cause it follow an arithmetic progression, $a_1+a_2+a_3+a_4+a_5$ $=(a_3-2d)+(a_3-d)+a_3+(a_3+d)+(a_3+2d)$ $=5a_3+(2d-2d)+(d-d)$ $=5a_3$ $=5 \times 12$ $=\boxed{60}$ I don't think you need to show the $1^{st}$ term.

$1st$ term of given arithmetic progression, $a_1 = - 5$
$3rd$ term of given arithmetic progression, $a_3 = 12$

Using the general formula for $n^{th}$ term of an A.P. , we can write,
$a_3 = a_1 + (3-1)d$
$\implies 12 = -5 + 2d$
$\implies 17 = 2d$
$\implies \boxed {d = \frac{17}{2}}$

Now, using value of $d$ , we can calculate $5th$ term of the A.P. as follows:
$a_5 = a_1 + 4d$
$\implies a_5 = -5 + (4 \times \frac{17}{2})$
$\implies a_5 = -5 + 34$
$\implies \boxed {a_5 = 29}$

Applying the formula for sum of A.P. , $S_n = \frac{n}{2}(a_1 + a_n)$

Putting in the corresponding values gives us, $S_5 = \frac{5}{2}(-5 + 29)$
$\implies S_5 = \frac{5}{2} \times 24$
$\implies \boxed {S_5 = 60}$

ALTERNATIVE METHOD:

Find the common difference as shown in previous method and apply the direct formula given below for summation of A.P. : $S_n = \frac{n}{2}(2 a_1 + (n-1)d)$

Putting $n = 5$ , $a_1 = -5$ and $d = \frac{17}{2}$ , we get,
$S_5 = \frac{5}{2}{({2(-5)} + (5-1) \frac{17}{2})}$
$\implies S_5 = \frac{5}{2}{(-10 + 34)}$
$\implies S_5 = \frac{5}{2} \times 24$
$\implies \boxed{S_5 = 60}$

Find common difference:

Since $a_{n} = a_{1} + (n-1)d$

Therefore $a_{3} = -5 + (3-1)d$

$12 = -5 + 2d$

Hence d = 8.5

Find Sum:

Sum of an arithmetic series is $S_{n} = \frac{n}{2}[2a + (n-1)d]$

Therefore sum of first five terms is

$S_{5} = \frac{5}{2}[2(-5) + (5-1)(8.5)]$

$S_{5} = \frac{5}{2}[-10 + 34]$

$S_{5} = 60$

Alternatively you can find all first five terms and sum them

First five terms should be -5, 3.5, 12, 20.5, 29, 37.5. (Continue adding 8.5 to first term)

-5 + 3.5 +12 + 20.5 + 29 + 37.5 = 60

Notice that:

$T_5 - T_3 = T_3 - T_1\\ T_5 = 2T_3 - T_1\\ =2(12) - (-5)\\ =29$

Therefore,

$S_5 = \dfrac{5}{2}(-5+29)\\ =\dfrac{5}{2}(24)\\ =5(12)\\ =\boxed{60}$