Question

The third term of an arithmetic series is 108, and the twelfth term is 54. Find:

a) The common difference of the series.
b) The first term of the series.
c) The sum of the first n terms of the series.
d) The possible values of n given Sn = 1200.

Answer 1

a) The common difference (\(d\)) is -6.

b) The first term (\(a\)) is 120.

c) The sum of the first \(n\) terms
`(\(S_n\)) is \(n(60 + 60n)\).`

d) The possible values of \(n\) are 10 and -20.

Step-by-step explanation:

a) In an arithmetic series, the n-th term is given by
`\(a_n = a + (n-1)d\)`, where \(a\) is the first term and \(d\) is the common difference. Given \(a_3 = 108\) and \(a_{12} = 54\), we can set up equations to solve for \(d\), which results in \(d = -6\).

b) Once \(d\) is known, we can find the first term \(a\) using the formula
`\(a_n = a + (n-1)d\)`, with
`\(a_3 = 108\)`. Solving for \(a\), we get a = 120.

c) The sum of the first
`\(n\) terms (\(S_n\))` of an arithmetic series is given by
`\(S_n = (n)/(2)[2a + (n-1)d]\)`. Substituting the known values a = 120and d = -6, we obtain
`\(S_n = n(60 + 60n)\).`

d) To find the possible values of n given
`\(S_n = 1200\)`, we set the expression for
`\(S_n\)` equal to 1200 and solve for n , resulting in the possible values n = 10 and n = -20.