The first term of the infinite arithmetic sequence {an} is A1 = 93, and the tolerance D1 = - 7. The first term of the infinite arithmetic sequence {BN} is B1 = 17, and the tolerance D2 = 12, Then, how many items are there in these two sequences with equal values?

The first term of the infinite arithmetic sequence {an} is A1 = 93, and the tolerance D1 = - 7. The first term of the infinite arithmetic sequence {BN} is B1 = 17, and the tolerance D2 = 12, Then, how many items are there in these two sequences with equal values?


By title
93-7x>=17①
17+12y=0)
The solution has X



When the sequence {an} and {BN} are two infinite arithmetic sequences, the tolerance is D1 and D2 respectively. This paper proves that the sequence {an + BN} is arithmetic sequence, and finds its tolerance


An + BN - (an-1 + bn-1) = (an-an-1) + (bn-bn-1) = D1 + D2, so {an + BN} is an arithmetic sequence, and the tolerance is D1 + D2



If the sequence {an}, {BN} is an arithmetic sequence, and the tolerances are D1 and D2 respectively, is the sequence {A2N}, {an, 2bn) an arithmetic sequence? If so, what is the tolerance


If the sequence {an}, {BN} is an arithmetic sequence and the tolerance is D1 and D2 respectively, then {A2N}, {an, 2bn) is an arithmetic sequence? If so, what is the tolerance {A2N} is. A (2n) - A (2 (n-1)) = 2 * D1, {an ± 2bn} is. A (n + 1) ± 2B (n + 1) - (an ± 2bn) = (a (n + 1) - an)) ± (2B (n + 1) - (2bn) = D1 ± 2 * D2



The sequence {an}, {BN} are all arithmetic sequence, and the tolerance is D1 and D2 respectively. Then the tolerance of {an + qbn} (q is constant) can be reduced


The tolerance of {an + qbn} (q is constant) is: D1 + QD2
an=a1 + (n-1)d1
bn=b1+ (n-1)d2,qbn=qb1 + (n-1)qb2
Let CN = an + qbn
Then, CN = a1 + QB1 + (n-1) (D1 + QD2)
Therefore, for CN, it can be equivalent to C1 = a1 + QB1,
Tolerance d = D1 + QD2
Then, the tolerance of {an + qbn} (q is a constant) is: D1 + QD2