The first problem is that you directly list two equations about d 1 and d2, solve these two equations, and then substitute them into the formula to find a 10.

The second question recommends the reduction to absurdity: a2n = 2+(n-1) d2a (2n-1) =1+(n-1) d1.

So according to the meaning of the question: A2n+1> A2n & gtA2n- 1 holds, so we get:

1+nd 1 & gt； 2+(n- 1)D2 & gt； 1+(n- 1)d 1

There are three relationships between d 1 and d2: d1> d2，d 1=d2，d 1 & lt； D2, as long as we exclude their inequality, then the rest must be equal.

Therefore, it is appropriate to set d 1; 2+(n- 1)d2 is easy to get: n (D2-D 1)

Similarly, let's say d1>; D2, at this time, we solve 2+(n-1) D2 >1+(n-1) d1and get: n (d1-D2) < d1-D2+.

Therefore, only d 1=d2 can satisfy the condition: An < An+ 1 holds.

Since d 1=d2, then we bring him into S 15= 15a8, which is a good solution: d 1=d2=2,

But at this time, we didn't get the proof that {an} is arithmetic progression, so the last step is still:

a2n = 2+(n- 1)D2 a(2n- 1)= 1+(n- 1)d 1 a2n+ 1 = 1+nd 1

a2n-a2n- 1= 1

a2n+ 1-a2n=d- 1

At this time, the previously solved d=2 is brought in, and it is proved that A2n-A2n-1= A2n+1-A2n =1.

At this point, the card has been completed.