MATH SOLVE

4 months ago

Q:
# 1. [4 marks]In an arithmetic sequence, the sum of the 3rd and 8th terms is 1.Given that the sum of the first seven terms is 35, determine the first term and the common difference.β

Accepted Solution

A:

An arithmetic sequence [tex]a_n[/tex] has a fixed difference [tex]d[/tex] between consecutive terms, so that they are recursively described by[tex]a_n=a_{n-1}+d[/tex]We're told that the sum of the 3rd and 8th terms is 1, so[tex]a_3+a_8=1[/tex]Using the recursive rule above, we have[tex]a_8=a_7+d[/tex][tex]a_8=(a_6+d)+d=a_6+2d[/tex][tex]a_8=(a_5+d)+2d=a_5+3d[/tex]and so on down to[tex]a_8=a_3+5d[/tex]which means[tex]a_3+(a_3+5d)=\boxed{2a_3+5d=1}[/tex]More generally, we can do the same manipulation with the recursive rule to find the explicit rule:[tex]a_n=(a_{n-2}+d)+d=a_{n-2}+2d[/tex][tex]a_n=(a_{n-3}+d)+2d=a_{n-3}+3d[/tex]and so on down to[tex]a_n=a_3+(n-3)d[/tex]We're also told that the sum of the first 7 terms is 35:[tex]a_1+a_2+a_3+a_4+a_5+a_6+a_7=35[/tex]and using the explicit rule above, this is the same as[tex](a_3-2d)+(a_3-d)+a_3+(a_3+d)+(a_3+2d)+(a_3+3d)+(a_3+4d)=35[/tex][tex]\implies7a_3+7d=35\implies\boxed{a_3+d=5}[/tex]Now solve for [tex]a_3[/tex] and [tex]d[/tex]:[tex]a_3+d=5\implies d=5-a_3[/tex][tex]2a_3+5d=1\implies2a_3+5(5-a_3)=1[/tex][tex]\implies25-3a_3=1[/tex][tex]\implies3a_3=24[/tex][tex]\implies a_3=8[/tex]The common difference is then[tex]d=5-8\implies\boxed{d=-3}[/tex]and in turn, the first term is[tex]a_1=a_3-2d=8-2(-3)\implies\boxed{a_1=14}[/tex]