[tex] \large\underline \bold{{SOLUTION\:1}}[/tex]
We noticed that the given questions are arithmetic sequence. Hence , We will use the sum of arithmetic sequence formula.
First , Finding for the common difference by subtracting the terms.
[tex]\sf{A_2-A_1=5-1 = 4}[/tex]
[tex]\sf{A_3-A_2=9-5=4}[/tex]
By the first 2 tries , We knew that the common difference is 4. And Going back to our equation:
Given that: [tex]\textsf{$A_1=1$ , n=10 , d=4}[/tex]
[tex]\longmapsto\sf{S_n = \frac{n}{2}[2(A_1)+(n-1)d]}[/tex]
[tex]\longmapsto\sf{S_{10} = \frac{10}{2}[2(1)+(10-1)4]}[/tex]
[tex]\longmapsto\sf{S_{10} = 5[2+(9)4]]}[/tex]
[tex]\longmapsto\sf{S_{10} = 5[2+36]}[/tex]
[tex]\longmapsto\sf{S_{10} = 5(38)}[/tex]
[tex]\longmapsto\boxed{\sf{S_{10} = 190}}\longleftarrow\rm{Answer}[/tex]
[tex] \large\underline \bold{{SOLUTION\:2}}[/tex]
Given that the terms are -15 to 40. By subtracting we will get a total of 56 terms including 0. Then , We knew that their common difference is 1 because -15,-14,-13...,40.
[tex]\sf{A_2-A_1=-14-(-15) = 1}[/tex]
[tex]\sf{A_3-A_2=-13-(-14) = 1}[/tex]
Now , Going back to the formula:
Given that: [tex]\textsf{$A_1=-15$ , n=56 , d=1}[/tex]
[tex]\longmapsto\sf{S_n = \frac{n}{2}[2A_1+(n-1)d]}[/tex]
[tex]\longmapsto\sf{S_{56} = \frac{56}{2}[2(-15)+(56-1)1]}[/tex]
[tex]\longmapsto\sf{S_{56} = 28[2(-15)+(55)1]}[/tex]
[tex]\longmapsto\sf{S_{56} = 28[-30+55]}[/tex]
[tex]\longmapsto\sf{S_{56} = 28(25)}[/tex]
[tex]\longmapsto\boxed{\sf{S_{56} = 700}}\longleftarrow\rm{Answer}[/tex]
