Finding the First Term of an Arithmetic Progression Using Summation

When dealing with the terms of an arithmetic progression (AP), it is not uncommon to face situations where the sum of certain terms is given, and we need to find the first term of the sequence. In this article, we will explore a specific problem where the sum of the first 40 terms is 200, and the sum of the next 40 terms is 3500. We will walk through the steps to determine the first term of the AP.

Understanding the Problem and the Formula

The sum of the first n terms of an AP is given by the formula:

Sn (n/2) [2a (n-1)d]

where:

a is the first term of the AP, d is the common difference between the terms, n is the number of terms.

Given Information and Initial Calculations

We are given the sum of the first 40 terms, S40, and the sum of the first 80 terms, S80. The values are:

S40 200 S80 3700

Using the summation formula for the first 40 terms:

S40 (40/2) [2a (40-1)d] 200

Let's simplify this equation:

200 20[2a 39d]

10 2a 39d … (1)

Similarly, for the first 80 terms:

S80 (80/2) [2a (80-1)d] 3700

2a 79d 92.5 … (2)

Now, we can equate equations (1) and (2) to solve for d and a.

Solving for the Common Difference d

From equation (1):

2a 39d 10

From equation (2):

2a 79d 92.5

By subtracting equation (1) from equation (2):

40d 82.5

d 82.5 / 40 2.0625

Having found d, we can substitute this value back into one of the equations to solve for a. Using equation (1):

2a 39(2.0625) 10

2a 80.4375 10

2a -70.4375

a -35.21875

Verification

To verify our solution, we can substitute the values of a and d back into the sum formulas for the first 40 and 80 terms.

Sum of the First 40 Terms

S40 (40/2) [-2(-35.21875) 39(2.0625)]

20 [-70.4375 80.4375]

20 (10) 200

This matches our given condition, confirming that the calculation for the first term is correct.

Sum of the First 80 Terms

S80 (80/2) [-2(-35.21875) 79(2.0625)]

40 [-70.4375 162.9375]

40 (92.5) 3700

This also confirms that our solution is correct.

Conclusion

The first term of the arithmetic progression is -35.21875. This solution demonstrates the power of algebraic manipulation in solving real-world problems involving sequences and series.