百度试题 结果1 题目Find the nth term of each arithmetic sequence described.a_1=10,d=-5,n=21 相关知识点: 试题来源: 解析 —90 反馈 收藏
a_2-a_1=5-(5+k) or -k find the difference between pairs of consecutive termsa_3-a_2=5-k-5 or -k to verify the common difference.The common difference is -k.Add -k to the third term to get the fourth term, and so on.a_4=5-k-k or 5-2ka_5=5-2k-k or 5-3ka_6=...
Number of terms, n = 15 First term, a = 1 Common difference, d = 2 – 1 Now, we know that finding the nth term of an arithmetic sequence is given byan= a + (n − 1) × d. Substituting the given values in this equation, we have, ...
百度试题 结果1 题目Find the sum of the arithmetic sequence.∑_(i=1)^∑(2i-3) 相关知识点: 试题来源: 解析 =3601a_1=2(1)-3=-1 a_(20)=2(20)-3=37 20 反馈 收藏
Question: Find the number of terms in the finite arithmeticsequence:{-5,-3,-1,1,3,dots,127}n= Find the number of terms in the finite arithmetic -3,-1 There are 2 steps to solve this one.
Find the number of terms in the sequence. a_1 = 4, a_n = 42, d = 2 Find the number of terms in the sequence: 2, -6, 18, -54, . . ., -4374. Find the first five terms of the sequence given that A_1=9 and A_n=A_{n-1}+n. ...
百度试题 结果1 题目 12. The first four terms of an arithmetic sequence are21 17 13 9Find, in terms of n, an expression for the nth term of this sequence.-4n+25(Total 2 marks) 相关知识点: 试题来源: 解析 -4n+25 反馈 收藏 ...
Find the sum of the first 20 terms of the arithmetic sequence where the first term is 3 and the common difference is 4. A. 780 B. 790 C. 800 D. 810 相关知识点: 试题来源: 解析 C。解析:使用等差数列求和公式 Sn = n(a1 + an)/2,先求出第 20 项为 79,再计算可得和为 800。
Problem 26: In an arithmetic sequence a =2an-1 +n. Find an in the explicit form in terms of n. 相关知识点: 试题来源: 解析 Problem 26: Solution:-n-2. Let the common difference of the arithmetic se- quence an =2an-1 +n be d. We know that n is a positive integer greater ...
The sum, s, of the arithmetic series a1+a2+⋯+an is given by the formula s=n(a1+an2) where n is the number of terms, a1 is the first term of the series, and an is last term of the series. Answer and Explanation: Let's expand the given series. {eq}\...