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.​

Accepted Solution

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]