Dose anyone know how to do it?

Question

find the sum of the first 43 terms of an arithmetic sequence in which the 11th term is 83 and the 62nd term is 440

Please explaine

Answer 1

Im not sure but try this (it will take a while)

You are dealing with a difference of 357 numbers between 83 and 440. Considering the 11th number is 83 you need to find the numbers 1-10 and 12-43. There seems to be a pattern to follow:

-9-8-8-9-8-8-9-8-8-9, 83, +8+8+9+8+8+9+8+8+9...

There should be 8.3 numbers between each number so accounting for the .3 on every 3rd number you add +9 of cours this is open to interpretation.

Answer 2

a11=83, a62=440 or

a1=83, a52 =440

440 = 83 +(52-1)d

357 =51d

d=7

a11 =a1 +(11-1)7 =83

83 =a1 +70

a1 =13

a43 =13 +(43-1)7 =13 +294 =307

S43 =43(13 +307)/2 =6880

Answer 3

298

62-11=51......440-83=357.........51 goes into 357   7 times..........4+7x42=298!