\[k-2, 3k-5, 8k-7\]

The problem is to find the first term and the common differences.

We need to find an equation for the common difference, which is equal to the 2nd term minus the first term, or the 3rd term minus the second term.

\[8k-7)-(3k-5)=(3k-5)-(k-2)\]

\[5k-2=2k-3\]

\[5k-2k=2-3\]

\[3k=-1 \rightarrow k=- \frac{1}{3}\]

The 1st term is

\[k-2=- \frac{1}{3}-2= - \frac{7}{3}\]

The common difference is

\[2k-2 = 2 \times - \frac{1}{3}- 3 = - \frac{11}{3}\]

An arithmetic sequence has third term 5 and seventh term 13. Find the sum of the first ten terms.

The formula for the nth term is

\[a_n=a+(n-1)d\]

where \[a, \: d\]

are the first term and the difference of successive terms respectively. Hence\[a_3=5=a+(3-1)d=a+2d \rightarrow 5=a+2d\]

(1)\[a_7=5=a+(7-1)d=a+6d \rightarrow 13=a+6d\]

(2)(2)-(1) gives

\[8=4d \rightarrow d=2\]

From (1) then

\[5=a+2 \times 2=a+4 \rightarrow a=1\]

.