For any index n (>1) calculate D(n) = U(n) - U(n-1). If this is the same for all integers n (>1) then D is the common difference. The sign of D determines whether the common difference is positive or negative.
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 a sequence of numbers, a(1), a(2), a(3), ... , a(n), a(n+1), ... he first differences are a(2) - a(1), a(3) - a(2), ... , a(n+1) - a(n) , ... Alternatively, d the sequence of first differences is given by d(n) = a(n+1) - a(n), n = 1, 2, 3, ...
An arithmetic sequence is a sequence of numbers such that the difference between successive terms is a constant. This constant is called the common difference and is usually denoted by d. If the first term is a, then the iterative definition of the sequence is U(1) = a, and U(n+1) = U(n) + d for n = 1, 2, 3, ... Equivalently, the position-to-term rule which defines the sequence is U(n) = a + (n-1)*d for n = 1, 2, 3, ...
a + (n-1)d = last number where a is the first number d is the common difference.
For any index n (>1) calculate D(n) = U(n) - U(n-1). If this is the same for all integers n (>1) then D is the common difference. The sign of D determines whether the common difference is positive or negative.
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 a sequence of numbers, a(1), a(2), a(3), ... , a(n), a(n+1), ... he first differences are a(2) - a(1), a(3) - a(2), ... , a(n+1) - a(n) , ... Alternatively, d the sequence of first differences is given by d(n) = a(n+1) - a(n), n = 1, 2, 3, ...
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).
An arithmetic sequence is a sequence of numbers such that the difference between successive terms is a constant. This constant is called the common difference and is usually denoted by d. If the first term is a, then the iterative definition of the sequence is U(1) = a, and U(n+1) = U(n) + d for n = 1, 2, 3, ... Equivalently, the position-to-term rule which defines the sequence is U(n) = a + (n-1)*d for n = 1, 2, 3, ...
a + (n-1)d = last number where a is the first number d is the common difference.
clearly the given series is an arithmetic progression with a common difference of -11,that is every term is obtained by subtracting 11 from the previous term for any A.P, n-th term is a(n)=a(1)+ (n-1)d where a(1)=first term and d=common difference here a(1)=100, and d= -11 so, a(n)=100+(n-1)x(-11) or, a(n)=111-11n
Sum = n/2[2Xa1+(n-1)d] where n is last number, a1 is the first number & d is the common difference between the numbers, here d=2 for the even /odd numbers. Sum = n/2 [2Xa1+(n-1)2]
The general (or nth) term is given by the equation t(n) = a + (n-1)d where a is the first term and d is the common difference between successive terms.
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tn = a + (n - 1)d where a is the first term and d is the difference between each term.
I am assuming this is an arithmetic series. Use the formula nth term = a + (n-1)d where a is the 1st term, and d is the difference between each term. 10th term = 8 + (10-1)d ==>8 +9d=53 ==> 9d = 53-8 = 45 ==> d = 5 The difference is 5.