an = a1 + d(n - 1)
To find the sum of the first 28 terms of an arithmetic sequence, you need the first term (a) and the common difference (d). The formula for the sum of the first n terms (S_n) of an arithmetic sequence is S_n = n/2 * (2a + (n - 1)d). Once you have the values of a and d, plug them into the formula along with n = 28 to calculate the sum.
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 = 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.
t(n) = 12*n + 5
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 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)
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 ).
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.