If $$\{a_n\}$$ is a geometric sequence with all terms greater than 0, and common ration $$q
e 1$$, then

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Uniteasy Admin · Accepted Answer

$$a_8 = a_1q^7$$
$$a_4 = a_1q^3$$
$$a_5 = a_1q^4$$
$$a_1+a_8 = a_1(1+q^7) $$
$$a_4+a_5 = a_1(q^3+q^4)$$
$$a_1+a_8-(a_4+a_5 ) = a_1(1+q^7-q^3-q^4)$$
$$=a_1(1-q^3)(1-q^4)$$
$$=a_1(1+q^2)(1-q)^2(1+q)(q^2+q+1)>0$$

Therefore, $$a_1+a_8 > a_4+a_5$$

Multiple Choice Question (MCQ)

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