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Algebra Difficulty 6.9 National olympiad Find the answer

Example 1 Given that the sequences {an}\left\{a_{n}\right\} and {bn}\left\{b_{n}\right\} are both arithmetic sequences, SnS_{n} and TnT_{n} are their respective sums of the first nn terms, and SnTn=7n+1n+3\frac{S_{n}}{T_{n}}=\frac{7 n+1}{n+3}, find the value of a2+a5+a17+a22b8+b10+b12+b16\frac{a_{2}+a_{5}+a_{17}+a_{22}}{b_{8}+b_{10}+b_{12}+b_{16}}.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Analysis: Students generally can use the property of the general term of an arithmetic sequence: "If m,n,k,lNm, n, k, l \in \mathbf{N}^{*} and m+n=k+lm+n=k+l, then am+an=ak+ala_{m}+a_{n}=a_{k}+a_{l}" to simplify the expression a2+a5+a17+a22b8+b10+b12+b16\frac{a_{2}+a_{5}+a_{17}+a_{22}}{b_{8}+b_{10}+b_{12}+b_{16}}. However, students often fail to establish the relationship between the sum of the first nn terms and the general term formula of an arithmetic sequence, leading to cumbersome calculations. In fact, it is only necessary to constructively use Sn=(a1+an)n2S_{n}=\frac{\left(a_{1}+a_{n}\right) n}{2}.
a2+a5+a17+a22b8+b10+b12+b16=(a5+a17)+(a2+a22)(b10+b12)+(b8+b16)=2a11+2a122b11+2b12=22(a12+a12)222(b11+b12)2=22(a1+a22)222(b1+b22)2=S22T22,a2+a5+a17+a22b8+b10+b12+b16=S22T22=7×22+122+3=315\begin{array}{l} \because \frac{a_{2}+a_{5}+a_{17}+a_{22}}{b_{8}+b_{10}+b_{12}+b_{16}}=\frac{\left(a_{5}+a_{17}\right)+\left(a_{2}+a_{22}\right)}{\left(b_{10}+b_{12}\right)+\left(b_{8}+b_{16}\right)} \\ =\frac{2 a_{11}+2 a_{12}}{2 b_{11}+2 b_{12}}=\frac{\frac{22\left(a_{12}+a_{12}\right)}{2}}{\frac{22\left(b_{11}+b_{12}\right)}{2}}=\frac{\frac{22\left(a_{1}+a_{22}\right)}{2}}{\frac{22\left(b_{1}+b_{22}\right)}{2}}=\frac{S_{22}}{T_{22}}, \\ \therefore \frac{a_{2}+a_{5}+a_{17}+a_{22}}{b_{8}+b_{10}+b_{12}+b_{16}}=\frac{S_{22}}{T_{22}}=\frac{7 \times 22+1}{22+3}=\frac{31}{5} \text{. } \end{array}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.