t(n) = a + 8*d = 54 .. .. .. .. .. .. .. (A)
s(12) = 12*a + 66*d = 438 .. .. .. .. (B)
12*(A) - (B) => 12*A + 96*d -12*A - 66*D = 648 - 438
=> 30*d = 210 = d = 7
Then substituting this value in (A) gives a + 54 = 54 => a = -2
So the first term is -2 and the common difference is 7.
It is an Arithmetic Progression with a constant difference of 11 and first term 15.
In an arithmetic progression the difference between each term (except the first) and the one before is a constant. In a geometric progression, their ratio is a constant. That is, Arithmetic progression U(n) - U(n-1) = d, where d, the common difference, is a constant and n = 2, 3, 4, ... Equivalently, U(n) = U(n-1) + d = U(1) + (n-1)*d Geometric progression U(n) / U(n-1) = r, where r, the common ratio is a non-zero constant and n = 2, 3, 4, ... Equivalently, U(n) = U(n-1)*r = U(1)*r^(n-1).
-4 is the first negative term. The progression is 24,20,16,12,8,4,0,-4,...
What is the 14th term in the arithmetic sequence in which the first is 100 and the common difference is -4? a14= a + 13d = 100 + 13(-4) = 48
16
For an Arithmetic Progression, Sum = 15[a + 7d].{a = first term and d = common difference} For a Geometric Progression, Sum = a[1-r^15]/(r-1).{r = common ratio }.
It is an Arithmetic Progression with a constant difference of 11 and first term 15.
In an arithmetic progression the difference between each term (except the first) and the one before is a constant. In a geometric progression, their ratio is a constant. That is, Arithmetic progression U(n) - U(n-1) = d, where d, the common difference, is a constant and n = 2, 3, 4, ... Equivalently, U(n) = U(n-1) + d = U(1) + (n-1)*d Geometric progression U(n) / U(n-1) = r, where r, the common ratio is a non-zero constant and n = 2, 3, 4, ... Equivalently, U(n) = U(n-1)*r = U(1)*r^(n-1).
In an arithmetic progression the difference between each term (except the first) and the one before is a constant. In a geometric progression, their ratio is a constant.That is,Arithmetic progressionU(n) - U(n-1) = d, where d, the common difference, is a constant and n = 2, 3, 4, ...Equivalently,U(n) = U(n-1) + d = U(1) + (n-1)*dGeometric progressionU(n) / U(n-1) = r, where r, the common ratio is a non-zero constant and n = 2, 3, 4, ...Equivalently,U(n) = U(n-1)*r = U(1)*r^(n-1).
35 minus 4 differences, ie 4 x 6 so first term is 11 and progression runs 11,17,23,29,35...
-4 is the first negative term. The progression is 24,20,16,12,8,4,0,-4,...
What is the 14th term in the arithmetic sequence in which the first is 100 and the common difference is -4? a14= a + 13d = 100 + 13(-4) = 48
It is a + 8d where a is the first term and d is the common difference.
16
2
RAMANUJANRAMANUJAN
If a is the first term and r the common difference, then the nth term is tn = a * (n-1)r So t16 = a + 15r Then 6*t16 = 6(a + 15r) or 6a + 90r No further simplifiaction is possible.