Q: How do you find the product of n terms in an progression?

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It is not possible to answer this question without information on whether the terms are of an arithmetic or geometric (or other) progression, and what the starting term is.

The sum of the 1st n terms is : N(3N-1)/2 Explanation : The sum from 1 to N of (3m-2) = 3 * sumFrom1toN(m) - sumFrom1toN(2) = 3 * (N*(N+1)/2) -2*N = N(3N-1)/2 For N=10 => 145

Find the Sum to n terms of the series 5 5+55+555+ +n Terms

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).

You.... have to apply this formula! n(n+1)/2 and n is the no. of terms

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It is not possible to answer this question without information on whether the terms are of an arithmetic or geometric (or other) progression, and what the starting term is.

RAMANUJANRAMANUJAN

The sum of n terms in a harmonic progression is given by the formula ( S_n = \frac{n}{a_1+ \frac{ (n-1)d}{2}} ) where ( S_n ) is the sum of n terms, ( a_1 ) is the first term, d is the common difference.

The formula for the sum of the first n terms of an arithmetic progression is Sn = n/2 * (a + l), where Sn is the sum, n is the number of terms, a is the first term, and l is the last term.

The sum of the 1st n terms is : N(3N-1)/2 Explanation : The sum from 1 to N of (3m-2) = 3 * sumFrom1toN(m) - sumFrom1toN(2) = 3 * (N*(N+1)/2) -2*N = N(3N-1)/2 For N=10 => 145

The nth term of the series is [ 4/2(n-1) ].

A product of 3 and N would be 3 and N multiplied together, so the product would be 3N. To get a numeric answer, you would first need to find what the value of N is.

The series given is an arithmetic progression consisting of 5 terms with a common difference of 5 and first term 5 → sum{n} = (n/2)(2×5 + (n - 1)×5) = n(5n + 5)/2 = 5n(n + 1)/2 As no terms have been given beyond the 5th term, and the series is not stated to be an arithmetic progression, the above formula only holds for n = 1, 2, ..., 5.

Find the Sum to n terms of the series 5 5+55+555+ +n Terms

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).

You.... have to apply this formula! n(n+1)/2 and n is the no. of terms

The product of any nonzero real number and its reciprocal is the number 1. This can be mathematically given as n multiplied by 1/n, where n represents the nonzero real number. The product of these two terms is 1.