Concept:

We believe that the sequence a1, a2, a3 …. an is an arithmetic series.

• Common difference “d”= a2 – a1 = a3 – a2 = …. = an – an – 1
• The n^th term of the arithmetic series is given by an = a + (n – 1) d
• The n^th term from the end is given by an = l – (n – 1) d
• The sum of the first n terms = S = n/2[2a + (n − 1) × d]

Or the sum of the first n terms = n/2(a + l)

Where, a = first term, d = common difference, n = number of terms and an = n^th term

Calculation:

Given: n^th term of the arithmetic series = a_n = \(\frac{{3 + {\rm{n}}}}{4}\)

For the first term, let n = 1

a₁ = a = (3 + 1)/4 = 4/4 = 1

For the second term, let n = 2

a₂ = (3 + 2)/4 = 5/4

Common difference “d”= a₂ - a₁ = (5/4) - 1 = 1/4

We need to find the sum of the first 105 terms,

\({{\rm{s}}_{105}} = \;\frac{{105}}{2}\left[ {2 \times 1 + \left( {105 - 1} \right) \times \frac{1}{4}} \right] = 1470\)         (∵S = n/2[2a + (n - 1) × d])