Find the Sum of n terms of Harmonic Sequence a = 2, n=6, and d=2

Steps to find sum of n terms of harmonic sequence:

sum of n terms of harmonic sequence formula:-

S_n = `(1/d)*ln( (2*a+(2*n-1)*d)/(2*a-d) )`

where:

• s_n is the sum of n terms
• a is first term
• n is total number of terms
• d is common difference

Input values are:-

a = 2

n = 6

d = 2

Put values into formula

S₆ = `(1/d)*ln( (2*a+(2*n-1)*d)/(2*a-d) )`

S₆ = `(1/2)*ln( (2*2+(2*6-1)*2)/(2*2-2) )`

S₆ = 1.28247

Following are the step by step process to find the sum of N terms in a harmonic sequence easily by hand.

• Firstly, Write down the values that were given in the problem, such as a = 2, n=6, and d=2.
• As we know that arithmetic progression is the reciprocal of harmonic progression.
• Apply the formula of sum of n term of Arithmetic Progression, i.e., Sn = n/2[2a + (n − 1) × d]
• Substitute the values in the formula, i.e., Sn = 6/2[2*2 + (6 − 1) × 2].
• Simplify the equation, i.e., Sn = 6/2[2*2 + (6-1) × 2] = 42.0
• At last , the sum of n terms of Arithmetic progression is 42.0.
• Finally, the result of sum of n terms of Harmonic sequence is 1/42.0

Example for Finding Sum of n terms of Harmonic Sequence