Math Problem Statement

The thirteenth term of an arithmetic progression is 12 and the sum of the first 30 terms is−15. Find the sum of the first 50 terms of the progression.

Solution

Let's denote the first term of the arithmetic progression (AP) as aa, and the common difference as dd.

Step 1: Use the information about the 13th term

The general formula for the nn-th term of an arithmetic progression is:

Tn=a+(n1)dT_n = a + (n-1) d

For the 13th term, we know that:

T13=a+(131)d=a+12dT_{13} = a + (13-1) d = a + 12d

We are told that T13=12T_{13} = 12, so:

a+12d=12(Equation 1)a + 12d = 12 \quad \text{(Equation 1)}

Step 2: Use the information about the sum of the first 30 terms

The sum of the first nn terms of an arithmetic progression is given by the formula:

Sn=n2(2a+(n1)d)S_n = \frac{n}{2} \left( 2a + (n-1) d \right)

For n=30n = 30, we are told that S30=15S_{30} = -15, so:

S30=302(2a+(301)d)=15(2a+29d)=15S_{30} = \frac{30}{2} \left( 2a + (30-1) d \right) = 15 \left( 2a + 29d \right) = -15

Simplifying:

2a+29d=1(Equation 2)2a + 29d = -1 \quad \text{(Equation 2)}

Step 3: Solve the system of equations

We now have the system of two equations:

a+12d=12(Equation 1)a + 12d = 12 \quad \text{(Equation 1)} 2a+29d=1(Equation 2)2a + 29d = -1 \quad \text{(Equation 2)}

Solving Equation 1 for aa:

From Equation 1:

a=1212da = 12 - 12d

Substitute into Equation 2:

Substitute a=1212da = 12 - 12d into Equation 2:

2(1212d)+29d=12(12 - 12d) + 29d = -1 2424d+29d=124 - 24d + 29d = -1 24+5d=124 + 5d = -1 5d=255d = -25 d=5d = -5

Find aa:

Now that we know d=5d = -5, substitute this into a=1212da = 12 - 12d:

a=1212(5)=12+60=72a = 12 - 12(-5) = 12 + 60 = 72

Step 4: Find the sum of the first 50 terms

Now we need to find the sum of the first 50 terms, S50S_{50}. Using the sum formula:

S50=502(2a+(501)d)S_{50} = \frac{50}{2} \left( 2a + (50-1) d \right) S50=25(2(72)+49(5))S_{50} = 25 \left( 2(72) + 49(-5) \right) S50=25(144245)S_{50} = 25 \left( 144 - 245 \right) S50=25×(101)S_{50} = 25 \times (-101) S50=2525S_{50} = -2525

Final Answer:

The sum of the first 50 terms is 2525\boxed{-2525}.


Would you like further clarification or have any questions?

Here are 5 related questions to explore further:

  1. How would the sum change if the common difference was positive?
  2. How do you find the sum of an infinite arithmetic progression?
  3. What happens to the sum of an arithmetic progression if the common difference is 0?
  4. How do you find the sum of an arithmetic progression when the last term is known, instead of the total number of terms?
  5. What is the general method to derive the formula for the nn-th term in an arithmetic progression?

Tip: Always check if the arithmetic progression's sum involves a negative or positive value, as this can significantly impact the behavior of the sequence, especially in practical applications!

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Math Problem Analysis

Mathematical Concepts

Arithmetic Progression
Algebra
Sequences and Series

Formulas

n-th term of an arithmetic progression: T_n = a + (n - 1)d
Sum of first n terms of an arithmetic progression: S_n = n/2 * (2a + (n - 1)d)

Theorems

Sum of terms in an arithmetic progression

Suitable Grade Level

Grades 8-10