Calculating standard combustion enthalpy of dodecane from that of heptane & octane
Heptane is $\ce{CH3-CH2(5)-CH3}$

Octane is $\ce{CH3-CH2(6)-CH3}$

Given combustion enthalpies for each of $-4816$ and $-5470~\mathrm{kJ~mol^{-1}}$  respectively, we now have a simple set of algeraic equations:

5A + 2B = $-4816~\mathrm{kJ~mol^{-1}}$

6A + 2B = $-5470~\mathrm{kJ~mol^{-1}}$

First we can solve for A as the combustion enthaply of a CH2 in a linear hydrocarbon and we get the value $-654~\mathrm{kJ~mol^{-1}})$.

We can now solve for B,

In the first equation we get 2B = ($-4816 - (654 * 5)~\mathrm{kJ~mol^{-1}}$, A = $-773~\mathrm{kJ~mol^{-1}}$

In the second equation we get 2B = ($-5470 - (654 * 6)~\mathrm{kJ~mol^{-1}}$ , A = $-773~\mathrm{kJ~mol^{-1}}$

dodecane  has CH3-CH2(10)-CH3 , for 10A + 2B.

10 * A  + 2 * B  = ($-6540$ + $-1546$)$~\mathrm{kJ~mol^{-1}}$ = $-8086~\mathrm{kJ~mol^{-1}}$