Finding the Sum of the First 140 Terms of a Sequence

The question asks for the sum of the first 140 terms of a sequence whose nth term is given by the formula \(a_n = \frac{2n+5}{7}\).

Identifying the Type of Sequence

To find the sum of a sequence, it's helpful to know what type of sequence it is. Let's calculate the first few terms:

• For \(n=1\), the first term is \(a_1 = \frac{2(1)+5}{7} = \frac{2+5}{7} = \frac{7}{7} = 1\).
• For \(n=2\), the second term is \(a_2 = \frac{2(2)+5}{7} = \frac{4+5}{7} = \frac{9}{7}\).
• For \(n=3\), the third term is \(a_3 = \frac{2(3)+5}{7} = \frac{6+5}{7} = \frac{11}{7}\).

Now, let's look at the difference between consecutive terms:

• \(a_2 - a_1 = \frac{9}{7} - 1 = \frac{9}{7} - \frac{7}{7} = \frac{9-7}{7} = \frac{2}{7}\).
• \(a_3 - a_2 = \frac{11}{7} - \frac{9}{7} = \frac{11-9}{7} = \frac{2}{7}\).

Since the difference between consecutive terms is constant, this sequence is an Arithmetic Progression (AP). The first term is \(a = 1\) and the common difference is \(d = \frac{2}{7}\).

Calculating the 140th Term

We need the sum of the first 140 terms. For the sum of an AP, we can use the formula \(S_n = \frac{n}{2}(a_1 + a_n)\), which requires the first term (\(a_1\)) and the nth term (\(a_n\)). We already have \(a_1 = 1\). We need to find the 140th term (\(a_{140}\)). We can use the given formula for the nth term directly:

\(a_{140} = \frac{2(140)+5}{7}\)

\(a_{140} = \frac{280+5}{7}\)

\(a_{140} = \frac{285}{7}\)

Calculating the Sum of the First 140 Terms

Now we can find the sum of the first 140 terms using the sum formula \(S_n = \frac{n}{2}(a_1 + a_n)\) with \(n=140\), \(a_1 = 1\), and \(a_{140} = \frac{285}{7}\):

\(S_{140} = \frac{140}{2}(a_1 + a_{140})\)

\(S_{140} = 70(1 + \frac{285}{7})\)

To add 1 and \(\frac{285}{7}\), we find a common denominator:

\(1 + \frac{285}{7} = \frac{7}{7} + \frac{285}{7} = \frac{7+285}{7} = \frac{292}{7}\)

Substitute this back into the sum formula:

\(S_{140} = 70\left(\frac{292}{7}\right)\)

Now, we can simplify by dividing 70 by 7:

\(S_{140} = 10 \times 292\)

\(S_{140} = 2920\)

So, the sum of the first 140 terms of the sequence is 2920.

Conclusion

The sum of the first 140 terms of the sequence with \(a_n = \frac{2n+5}{7}\) is 2920.

Revision Table: Sequence Sum Calculation

Additional Information: Arithmetic Progressions

An Arithmetic Progression (AP) is a sequence where the difference between consecutive terms is constant. This constant difference is called the common difference, denoted by \(d\).

• The general form of an AP is \(a, a+d, a+2d, a+3d, \dots\)
• The nth term of an AP is given by \(a_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference. In the given problem, \(a = 1\) and \(d = \frac{2}{7}\), so \(a_n = 1 + (n-1)\frac{2}{7} = 1 + \frac{2n-2}{7} = \frac{7+2n-2}{7} = \frac{2n+5}{7}\), which matches the given formula, confirming it's an AP.
• The sum of the first \(n\) terms of an AP can be calculated using two main formulas:
  • \(S_n = \frac{n}{2}(a_1 + a_n)\), where \(a_1\) is the first term and \(a_n\) is the nth term.
  • \(S_n = \frac{n}{2}(2a + (n-1)d)\), where \(a\) is the first term and \(d\) is the common difference.
  Both formulas yield the same result and can be used depending on what information is readily available. In this solution, we used the first formula after calculating \(a_{140}\).