The sequence an = 6n-3, BN = 5n-4, if an is less than or equal to 1000, BN is less than or equal to 1000, a new sequence cn is composed of both the terms in an and BN. How many terms does an have in total, and what is the sum of all these terms

The sequence an = 6n-3, BN = 5n-4, if an is less than or equal to 1000, BN is less than or equal to 1000, a new sequence cn is composed of both the terms in an and BN. How many terms does an have in total, and what is the sum of all these terms

Let the nth term in {an} be the m-th term in {BN}. Then: 5m-6n = 1. By solving the Diophantine equation of two variables, n = 5k-1, M = 6k-1. (k is any integer)
So: CN = a (5n-1) = B (6n-1) = 30n-9

In known sequence an, A1 = 2, an + 1 = an + 2 (n belongs to n *) sequence BN satisfies BN = 1 / Anan + 1, find the sum of the first ten terms of BN

a(n+1)=an+2a(n+1)-an =2an -a1 =2(n-1)an = 2nbn=1/[an.a(n+1)]= (1/4)( 1/n - 1/(n+1)] Sn =b1+b2+...+bn= (1/4)( 1 - 1/(n+1)] = n/[4(n+1)]S10 = 10/44 = 5/22

If a1 + A2 + a3 = 7, a1a2a3 = 8, find the general term formula of sequence an
An is an equal ratio sequence

a1+a1q+a1q^2=7a1^3q^3=8a1q=2a1+2+a1q^2=7a1+a1q^2=5a1=2/q2/q+2/q*q^2=52/q+2q=52+2q^2=5q2q^2-5q+2=0(2q-1)(q-2)=0q1=1/2,q2=21.q1=1/2,a1=2/1/2=4an=a1q^(n-1)=4*(1/2)^(n-1)=2^(2-n+1)=2^(3-n)(n:N*)2.q2=2,a1=...