已知等差数列\(\left\{ {{a_n}} \right\}\)的首项为\(a\)，公差为\(d\)；等差数列\(\left\{ {{b_n}} \right\}\)的首项为\(b\)，公差为\(e\)．如果\({{c}_{n}}={{a}_{n}}+{{b}_{n}}\left ( {n\geq 1} \right )\)，且\({c_1} = 4\)，\({c_2} = 8\)，则\({c_{10}} = \)___．

因为\({c_n} - {c_{n - 1}} = {a_n} + {b_n} - \left( {{a_{n - 1}} + {b_{n - 1}}} \right) = {a_n} - {a_{n - 1}} + {b_n} - {b_{n - 1}} = d + e\)，

所以数列\(\{ {c_n}\} \)是以\(a + b\)为首项，\(d+e\)为公差的等差数列，

因为\({c_1} = 4\)，\({c_2} = 8\)，所以\(d + e = {c_2} - {c_1} = 8 - 4 = 4\)

所以\({c_{10}} = 4 + 4 \times 9 = 40\).

故答案为：40

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