an = a1 + d(n - 1)
by the general formula ,a+(n-1)*d * * * * * That assumes that it is an arithmetic sequence. The sequence cound by geometric ( t(n) = a*rn ) or power ( t(n) = n2 ) or something else.
The nth term of an arithmetic sequence = a + [(n - 1) X d]
The sequence 3, 7, 11 is an arithmetic sequence where the first term is 3 and the common difference is 4. The nth term formula for an arithmetic sequence can be expressed as ( a_n = a_1 + (n - 1)d ), where ( a_1 ) is the first term and ( d ) is the common difference. Substituting the values, the nth term formula for this sequence is ( a_n = 3 + (n - 1) \cdot 4 ), which simplifies to ( a_n = 4n - 1 ).
An arithmetic sequence can be represented using a formula, typically in the form (a_n = a_1 + (n-1)d), where (a_n) is the nth term, (a_1) is the first term, (d) is the common difference, and (n) is the term number. Additionally, an arithmetic sequence can be represented as a list of its terms, such as {a1, a2, a3, ..., an}, where each term increases by the common difference (d).
an = a1 + d(n - 1)
-7
arithmetic sequence * * * * * A recursive formula can produce arithmetic, geometric or other sequences. For example, for n = 1, 2, 3, ...: u0 = 2, un = un-1 + 5 is an arithmetic sequence. u0 = 2, un = un-1 * 5 is a geometric sequence. u0 = 0, un = un-1 + n is the sequence of triangular numbers. u0 = 0, un = un-1 + n(n+1)/2 is the sequence of perfect squares. u0 = 1, u1 = 1, un+1 = un-1 + un is the Fibonacci sequence.
Yes, it can both arithmetic and geometric.The formula for an arithmetic sequence is: a(n)=a(1)+d(n-1)The formula for a geometric sequence is: a(n)=a(1)*r^(n-1)Now, when d is zero and r is one, a sequence is both geometric and arithmetic. This is because it becomes a(n)=a(1)1 =a(1). Note that a(n) is often written anIt can easily observed that this makes the sequence a constant.Example:a(1)=a(2)=(i) for i= 3,4,5...if a(1)=3 then for a geometric sequence a(n)=3+0(n-1)=3,3,3,3,3,3,3and the geometric sequence a(n)=3r0 =3 also so the sequence is 3,3,3,3...In fact, we could do this for any constant sequence such as 1,1,1,1,1,1,1...or e,e,e,e,e,e,e,e...In general, let k be a constant, the sequence an =a1 (r)1 (n-1)(0) with a1 =kis the constant sequence k, k, k,... and is both geometric and arithmetic.
by the general formula ,a+(n-1)*d * * * * * That assumes that it is an arithmetic sequence. The sequence cound by geometric ( t(n) = a*rn ) or power ( t(n) = n2 ) or something else.
t(n) = 12*n + 5
The following formula generalizes this pattern and can be used to find ANY term in an arithmetic sequence. a'n = a'1+ (n-1)d.
The nth term of an arithmetic sequence = a + [(n - 1) X d]
The set of odd numbers is an arithmetic sequence. Let say that the sequence has n odd numbers where the first term is a1 and the last one is n. The formula to find the sum on nth terms for an arithmetic sequence is: Sn = (n/2)(a1 + an) or Sn = (n/2)[2a1 + (n - 1)d] where d is the common difference that for odd numbers is 2. Sn = (n/2)(2a1 + 2n - 2)
Oh, dude, you're hitting me with the math questions, huh? So, the formula for finding the nth term of an arithmetic sequence is a + (n-1)d, where a is the first term and d is the common difference. In this sequence, the common difference is 8 (because each term increases by 8), and the first term is 14. So, the formula for the nth term would be 14 + 8(n-1). You're welcome.
The nth term is referring to any term in the arithmetic sequence. You would figure out the formula an = a1+(n-1)d-10where an is your y-value, a1 is your first term in a number sequence (your x-value), n is the term you're trying to find, and d is the amount you're increasing by.
7 - 4n where n denotes the nth term and n starting with 0