If two arithmetic sequences {an} {BN} satisfy a1 + A2 + a3 +. + an / B1 + B2 + B3 +. + BN = 7n + 2 / N + 3, find A5 / B5 I use two methods to calculate, the results are different... Help me see what's wrong with the second one The first type: (a1 + A2 + a3 +. + an) / (B1 + B2 + B3 +. + BN) = 7n + 2 / N + 3 San/Tbn=(7n+2)/(n+3) a5/b5=[1/2*(a1+a9)]/[1/2*(b1+b9)] =9*[1/2*(a1+a9)]/9*[1/2*(b1+b9)] =S9/T9 =(7*9+2)/(9+3) =65/12 The second type: a1 + A2 +... + an = 7 * n square + 2n; B1 + B2 +... + BN = n square + 3N Then A5 / B5 = 4.625

If two arithmetic sequences {an} {BN} satisfy a1 + A2 + a3 +. + an / B1 + B2 + B3 +. + BN = 7n + 2 / N + 3, find A5 / B5 I use two methods to calculate, the results are different... Help me see what's wrong with the second one The first type: (a1 + A2 + a3 +. + an) / (B1 + B2 + B3 +. + BN) = 7n + 2 / N + 3 San/Tbn=(7n+2)/(n+3) a5/b5=[1/2*(a1+a9)]/[1/2*(b1+b9)] =9*[1/2*(a1+a9)]/9*[1/2*(b1+b9)] =S9/T9 =(7*9+2)/(9+3) =65/12 The second type: a1 + A2 +... + an = 7 * n square + 2n; B1 + B2 +... + BN = n square + 3N Then A5 / B5 = 4.625

The first one is right. Instead of finding A5 and B5, it can be found directly by conditions;
The second method is right at the beginning, but we can use the summation formula of arithmetic sequence to further find A1 and D, A1 = 9, d = 14, A5 = 65;
b1=4,d=2,b5=12；a5/b5=65/12