Math Problem Statement

Another method

The sum of a certain number of terms of the Arithmetic Progression (A.P.) 20, 17, 14, ... is 65. Find the: (a) number of terms, (b) last term.

Solution

The question you've provided is related to an Arithmetic Progression (A.P.):

The A.P. given is:
20, 17, 14, ...

The sum of a certain number of terms of this A.P. is 65. We are asked to find:

  • (a) The number of terms.
  • (b) The last term.

Step 1: Formula for Sum of an Arithmetic Progression

The formula for the sum SnS_n of the first nn terms of an arithmetic progression is: Sn=n2×(2a+(n1)d)S_n = \frac{n}{2} \times (2a + (n - 1) d) Where:

  • aa is the first term.
  • dd is the common difference.
  • nn is the number of terms.
  • SnS_n is the sum of the first nn terms.

Step 2: Identifying Given Values

From the sequence: 20, 17, 14, ..., we have:

  • a=20a = 20 (first term)
  • d=1720=3d = 17 - 20 = -3 (common difference)
  • Sn=65S_n = 65 (the given sum)

We need to find the number of terms nn.

Step 3: Substituting Values into the Formula

Substitute the known values into the sum formula: 65=n2×(2×20+(n1)×(3))65 = \frac{n}{2} \times (2 \times 20 + (n - 1) \times (-3)) 65=n2×(403n+3)65 = \frac{n}{2} \times (40 - 3n + 3) 65=n2×(433n)65 = \frac{n}{2} \times (43 - 3n) 130=n×(433n)130 = n \times (43 - 3n) 130=43n3n2130 = 43n - 3n^2

Step 4: Solving the Quadratic Equation

Rearrange the equation: 3n243n+130=03n^2 - 43n + 130 = 0

Use the quadratic formula: n=b±b24ac2an = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} Where a=3a = 3, b=43b = -43, and c=130c = 130.

n=43±(43)24(3)(130)2(3)n = \frac{43 \pm \sqrt{(-43)^2 - 4(3)(130)}}{2(3)} n=43±184915606n = \frac{43 \pm \sqrt{1849 - 1560}}{6} n=43±2896n = \frac{43 \pm \sqrt{289}}{6} n=43±176n = \frac{43 \pm 17}{6}

This gives two solutions: n=43+176=606=10n = \frac{43 + 17}{6} = \frac{60}{6} = 10 or n = \frac{43 - 17}{6} = \frac{26}{6} = 4.33 \quad (\text{which is not possible since n must be an integer})

Thus, n=10n = 10.

Step 5: Finding the Last Term

The last term lnl_n of an arithmetic progression is given by: ln=a+(n1)dl_n = a + (n - 1)d Substitute n=10n = 10, a=20a = 20, and d=3d = -3: l10=20+(101)×(3)l_{10} = 20 + (10 - 1) \times (-3) l10=20+9×(3)=2027=7l_{10} = 20 + 9 \times (-3) = 20 - 27 = -7

Final Answers:

  • (a) The number of terms is 1010.
  • (b) The last term is 7-7.

Would you like further details or clarification? Here are five related questions to consider:

  1. How do we derive the formula for the sum of an arithmetic progression?
  2. What are other types of sequences that use similar methods?
  3. How can you solve for the number of terms in a geometric progression?
  4. Why is the quadratic formula necessary when dealing with A.P. sums?
  5. What are some real-world applications of arithmetic progressions?

Tip: Always ensure that the common difference is correctly identified, as it directly affects the sum and general term of the A.P.

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Math Problem Analysis

Mathematical Concepts

Arithmetic Progression
Algebra
Quadratic Equations

Formulas

Sum of arithmetic progression: S_n = (n/2) × (2a + (n-1)d)
Quadratic equation: ax^2 + bx + c = 0
Last term of an A.P.: l_n = a + (n-1)d

Theorems

Quadratic Formula

Suitable Grade Level

Grades 8-10