Math Problem Statement

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 problem asks about an arithmetic progression (A.P.) with first term 20, second term 17, and third term 14. The sum of these terms is given as 65, and we are required to find:

(a) the number of terms,
(b) the last term.

Solution Outline:

  1. Identify the given values:

    • First term, a=20a = 20,
    • Common difference, d=1720=3d = 17 - 20 = -3,
    • Sum of the progression, Sn=65S_n = 65.
  2. Sum formula for an arithmetic progression (A.P.): Sn=n2×(2a+(n1)d)S_n = \frac{n}{2} \times (2a + (n-1)d) where SnS_n is the sum of nn terms, aa is the first term, dd is the common difference, and nn is the number of terms.

  3. Use the given information to find the number of terms:

    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)) Simplify the equation: 65=n2×(403n+3)65 = \frac{n}{2} \times (40 - 3n + 3) 65=n2×(433n)65 = \frac{n}{2} \times (43 - 3n) Multiply both sides by 2 to eliminate the denominator: 130=n×(433n)130 = n \times (43 - 3n) Expand: 130=43n3n2130 = 43n - 3n^2 Rearrange into a standard quadratic form: 3n243n+130=03n^2 - 43n + 130 = 0

  4. Solve the quadratic equation:

    Using the quadratic formula: n=(43)±(43)24×3×1302×3n = \frac{-(-43) \pm \sqrt{(-43)^2 - 4 \times 3 \times 130}}{2 \times 3} Simplify: 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 possible values for nn: n=43+176=606=10n = \frac{43 + 17}{6} = \frac{60}{6} = 10 or n=43176=2664.33n = \frac{43 - 17}{6} = \frac{26}{6} \approx 4.33

    Since nn must be an integer, n=10n = 10.

  5. Find the last term:

    The last term ll of an arithmetic progression can be calculated using the formula: l=a+(n1)×dl = a + (n-1) \times d Substituting the known values: l=20+(101)×(3)l = 20 + (10-1) \times (-3) l=2027=7l = 20 - 27 = -7

Final Answers:

  • (a) The number of terms, nn, is 10.
  • (b) The last term, ll, is 7-7.

Would you like a more detailed explanation, or do you have any questions?


Here are 5 related questions:

  1. How do you derive the sum formula for an arithmetic progression?
  2. What happens if the common difference is positive instead of negative?
  3. Can you solve the quadratic equation using factoring instead of the quadratic formula?
  4. How does the sum of an arithmetic progression change if more terms are added?
  5. What are the applications of arithmetic progressions in real-world problems?

Tip: Always double-check the discriminant in the quadratic formula to ensure the solution is valid and real!

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Math Problem Analysis

Mathematical Concepts

Arithmetic Progression
Algebra
Quadratic Equations

Formulas

Sum of an arithmetic progression: S_n = (n/2) × [2a + (n-1)d]
Quadratic formula: n = [-b ± sqrt(b^2 - 4ac)] / 2a
Last term of arithmetic progression: l = a + (n-1)d

Theorems

Quadratic Formula
Properties of Arithmetic Progressions

Suitable Grade Level

Grades 9-10