a6 = 45, a14 = -43, n=10

an = a1 + (n-1)d
a6 = a1 + 5d
a14 = a1 + 13d

a14 = a6 + 8d => -43 = 45 + 8d => -88 = 8d => d = -11.
a6 = a1 + 5d => a1 = a6 - 5d = 45 - (-11)*5 = 45+55 = 100.
a10 = a1 + 9d = 100 + 9(-11) = 1.

Sn = (a1 + an)/2 * n = (1 + 100)/2 * 10 = 1010/2 = 505.
Ответ: а) 505

[tex]\displaystyle\bf a_{n} = a_{1} + (n - 1)d \\ \displaystyle\bf\\\left \{ {{ a_{6} = 45} \atop {a_{14} = - 43}} \right. \\ \displaystyle\bf\\\left \{ {{a_{1} + 5d = 45 \: \: | \times ( - 1)} \atop {a_{1} + 13d = - 43}} \right. \\ \displaystyle\bf\\ + \left \{ {{ - a_{1} - 5d = - 45} \atop { a_{1}   + 1 3d =  - 43}} \right. \\ \\ - 5d + 13d = - 45 - 43 \\ 8d = - 88 \\ d = - 88 \div 8 \\ d = - 11\\  \\ a_{1} + 5 \times ( - 11) = 45 \\ a_{1} - 55= 45 \\ a_{1} =45 + 55 \\ a_{1} = 100 \\  \\ a_{10} = a_{1} + 9d = 100+ 9 \times ( - 11)= 100 - 99 = 1 \\  \\ S_{n} =  \frac{a_{1} + a_{n}}{2} n \\ S_{10} =  \frac{a_{1} + a_{10}}{2}  \times   10 = 5(a_{1} + a_{10}) = \\ = 5 \times (100 + 1) = 5 \times 101 = 505[/tex]

Ответ: а) 505