As shown in the figure, C and D are two points on the line ab. given BC = 14ab, ad = 13ab, ab = 12cm, find the length of CD and BD

As shown in the figure, C and D are two points on the line ab. given BC = 14ab, ad = 13ab, ab = 12cm, find the length of CD and BD

∵ BC = 14ab, ad = 13ab, ab = 12cm, ∵ BC = 3cm, ad = 4cm, ∵ CD = ab-ad-bc = 12cm-4cm-3cm = 5cm, BD = BC + CD = 3cm + 5cm = 8cm

Am * an = AP * AQ if m * n = P * q?

No
Should be
m+n=p+q
Then am * an = AP * aq

In the arithmetic sequence {an}, if AP = q, AQ = P, (P, Q ∈ n, and P ≠ q), then AP + Q=______ ．

Let the first term be A1 and the tolerance be D, then AP = a1 + (p-1) d = q, AQ = a1 + (Q-1) d = P, subtracting the two expressions to get (P-Q) d = Q-P, so the solution is & nbsp; d = - 1, substituting into A1 = P + Q-1, so AP + q = a1 + (P + Q-1) d = (P + Q-1) + (P + Q-1) * - 1 = 0

Given that the sum of the first n terms of two arithmetic sequences {an} and {BN} is an and BN respectively, and anbn = 7n + 45N + 3, then the number of positive integers n with anbn as an integer is ()
A. 2B. 3C. 4D. 5

From the properties and summation formula of arithmetic sequence, we can get: anbn = 2an2bn = a1 + A2N − 1B1 + b2n − 1 = (2n − 1) (a1 + A2N − 1) 2 (2n − 1) (B1 + b2n − 1) 2 = A2N − 1b2n − 1 = 7 (2n − 1) + 45 (2n − 1) + 3 = 7 + 12n + 1. We know that anbn is an integer when n = 1, 2, 3, 5, 11

Arithmetic sequence {an}, {BN}, an / BN = (7n + 45) / (n + 3), find the value of n where an / BN is an integer
Maybe what I said is not clear. An and BN refer to the sum of two sequences respectively, while an and BN refer to each term. See clearly

According to the summation formula Sn = (a1 + an) * n / 2An / BN = [(a1 + an) * n / 2] / [(B1 + BN) * n / 2] = (a1 + an) / (B1 + BN), the arithmetic sequence has a1 + an = 2 * a [(1 + n) / 2], where (1 + n) / 2 in square brackets is the subscript, which indicates the position in the sequence. Similarly, B1 + BN = 2 * B [(1 + n) / 2] = (a1 + an) / (B1 + BN) = a

Given that the sum of the first n terms of two arithmetic sequences {an} and {BN} is an and BN respectively, and anbn = 7n + 45N + 3, then the number of positive integers n with anbn as an integer is ()
A. 2B. 3C. 4D. 5

From the properties and summation formula of arithmetic sequence, we can get: anbn = 2an2bn = a1 + A2N − 1B1 + b2n − 1 = (2n − 1) (a1 + A2N − 1) 2 (2n − 1) (B1 + b2n − 1) 2 = A2N − 1b2n − 1 = 7 (2n − 1) + 45 (2n − 1) + 3 = 7 + 12n + 1. We know that anbn is an integer when n = 1, 2, 3, 5, 11

Let the sum of the first n terms of the arithmetic sequence {an} be Sn, if SM-1 = - 2, SM = 0, SM + 1 = 3, then M=
M-1, M + 1 are subscripts

Am = SM - SM-1 = 2 similarly am + 1 = 3 tolerance q = 1 sm = 0 recursively get am-1 = 1 AM-2 = 0 AM-3 = - 1 am-4 = - 2 it is easy to know that this is the first term, so the result of M = 5 listed by summation formula is the same, the number of items is not much, you can write it directly

In the arithmetic sequence {an}, Sn = m, SM = n, find Sn + m?

Sn=(A1+An)n/2=(A1+A1+(n-1)d)n/2=m 2A1+(n-1)d=2m/n
Sm=(A1+Am)m/2=(A1+A1+(m-1)d)m/2=n 2A1+(m-1)d=2n/m
Subtraction of two formulas
(n-m)d=2m/n-2n/m=2(m^2-n^2)/(mn)=2(m+n)(m-n)/(mn)
d=-2(m+n)/(mn) mnd=-2(m+n)
A(n+1)=A1+nd
A(n+2)=A2+nd
……
A(n+m)=Am+nd
The above formula has m terms in total and is added
A(n+1)+A(n+2)+…… +A(n+m)=(A1+A2+…… +Am)+mnd=Sm-2(m+n)
S(n+m)=Sn+(A(n+1)+A(n+2)+…… +A(n+m))
=Sn+Sm-2(m+n)
=n+m-2(n+m)
=-(n+m)

Let the sum of the first n terms of the arithmetic sequence {an} be Sn, if SM-1 = - 2, SM = 0, SM + 1 = 3, then M = ()
A. 3B. 4C. 5D. 6

Am = sm-sm-1 = 2, am + 1 = SM + 1-sm = 3, so tolerance d = am + 1-am = 1, SM = m (a1 + AM) 2 = 0, get A1 = - 2, so am = - 2 + (m-1) · 1 = 2, get m = 5, so choose C

If Sn = 93, an = 48, q = 2, then n = 1=______ ．

By substituting the first formula into the second one, A1 = 3 ∵ 48 = 3 ∵ 2N-1 ∵ n = 5, so the answer is 5