Question

1. [4 marks]

Answer 1

An arithmetic sequence $a_n$ has a fixed difference $d$ between consecutive terms, so that they are recursively described by

$a_n=a_{n-1}+d$

We're told that the sum of the 3rd and 8th terms is 1, so

$a_3+a_8=1$

Using the recursive rule above, we have

$a_8=a_7+d$

$a_8=(a_6+d)+d=a_6+2d$

$a_8=(a_5+d)+2d=a_5+3d$

and so on down to

$a_8=a_3+5d$

which means

$a_3+(a_3+5d)=\boxed{2a_3+5d=1}$

More generally, we can do the same manipulation with the recursive rule to find the explicit rule:

$a_n=(a_{n-2}+d)+d=a_{n-2}+2d$

$a_n=(a_{n-3}+d)+2d=a_{n-3}+3d$

and so on down to

$a_n=a_3+(n-3)d$

We're also told that the sum of the first 7 terms is 35:

$a_1+a_2+a_3+a_4+a_5+a_6+a_7=35$

and using the explicit rule above, this is the same as

$(a_3-2d)+(a_3-d)+a_3+(a_3+d)+(a_3+2d)+(a_3+3d)+(a_3+4d)=35$

$\implies7a_3+7d=35\implies\boxed{a_3+d=5}$

Now solve for $a_3$ and $d$:

$a_3+d=5\implies d=5-a_3$

$2a_3+5d=1\implies2a_3+5(5-a_3)=1$

$\implies25-3a_3=1$

$\implies3a_3=24$

$\implies a_3=8$

The common difference is then

$d=5-8\implies\boxed{d=-3}$

and in turn, the first term is

$a_1=a_3-2d=8-2(-3)\implies\boxed{a_1=14}$