Arithmetic Mean and Selection of Terms in an AP for CBSE Class 10

When solving algebraic word problems involving Arithmetic Progressions—such as finding three numbers in an AP whose sum and product are given, or determining the interior angles of a quadrilateral—choosing your variables wisely can transform an intimidating system of equations into a simple, single-variable calculation.

In CBSE Class 10 Mathematics, mastering the selection of terms in an AP and understanding the concept of the Arithmetic Mean (AM) provides students with powerful algebraic shortcuts that eliminate redundant variables and guarantee full marks on 3-mark and 4-mark board exam questions.

What You Will Learn

• Concept and mathematical definition of the Arithmetic Mean (AM)
• Property of terms equidistant from the extremes in a finite AP
• Strategic selection of 3, 4, and 5 terms in an AP using symmetry
• Why symmetric selection simplifies the sum to a single variable
• Solved CBSE board examination problems (dividing numbers into parts, angles of triangles)
• Common algebraic traps and error-prevention tips

1. What is the Arithmetic Mean?

When three numbers form an Arithmetic Progression, the middle number is the average of the two outer numbers.

Definition of Arithmetic Mean

If three numbers $a, A,$ and $b$ are in an Arithmetic Progression, then the middle term $A$ is called the Arithmetic Mean (AM) of $a$ and $b$.

Since $a, A, b$ are in AP, the common difference between consecutive terms is equal: $A - a = b - A$ Transposing variables: $A + A = a + b$ $2A = a + b \implies A = \frac{a + b}{2}$

$\text{Arithmetic Mean of } a \text{ and } b = \frac{a + b}{2}$

Example:

The Arithmetic Mean of $8$ and $20$ is: $A = \frac{8 + 20}{2} = \frac{28}{2} = 14$ Notice that $8, 14, 20$ forms an AP with common difference $d = 6$.

2. Property of Equidistant Terms in a Finite AP

In any finite Arithmetic Progression, an elegant balance exists between the beginning and the end:

In a finite AP, the sum of any two terms that are equidistant from the beginning and the end is always constant, and is equal to the sum of the first and the last terms.

$a_1 + a_n = a_2 + a_{n-1} = a_3 + a_{n-2} = \dots$

Verification:

Let the AP be $a, a+d, a+2d, \dots, a+(n-1)d$:

• Sum of 1st and $n$-th terms: $a_1 + a_n = a + [a + (n-1)d] = 2a + (n-1)d$
• Sum of 2nd and $(n-1)$-th terms: $a_2 + a_{n-1} = (a + d) + [a + (n-2)d] = 2a + (n-1)d$ The sums are identical!

3. Strategic Selection of Terms in an AP

When an exam question gives the sum of consecutive terms, do NOT set the terms as $a, a+d, a+2d$. Setting terms this way forces you to solve an equation with both $a$ and $d$ simultaneously.

Instead, take advantage of algebraic symmetry so that the common difference $d$ cancels out when the terms are added!

                      Symmetric Selection of Terms in an AP
                                        |
       +--------------------------------+--------------------------------+
       |                                |                                |
    3 Terms                          4 Terms                          5 Terms
  (a - d), a, (a + d)          (a - 3d), (a - d),               (a - 2d), (a - d), a,
  Common diff = d              (a + d), (a + 3d)                 (a + d), (a + 2d)
  Sum = 3a (d vanishes!)       Common diff = 2d                  Common diff = d
                               Sum = 4a (d vanishes!)            Sum = 5a (d vanishes!)

1. When Three Terms Are in AP:

Choose the terms as: $\mathbf{a - d, \quad a, \quad a + d}$

• Notice that their common difference is $d$.
• Their sum is: $(a - d) + a + (a + d) = 3a$. The variable $d$ completely cancels out, immediately giving the value of $a$!

2. When Four Terms Are in AP:

Choose the terms as: $\mathbf{a - 3d, \quad a - d, \quad a + d, \quad a + 3d}$

• Notice that their common difference is $2d$ (not $d$!).
• Their sum is: $(a - 3d) + (a - d) + (a + d) + (a + 3d) = 4a$. The variable $d$ cancels out, immediately giving the value of $a$!

3. When Five Terms Are in AP:

Choose the terms as: $\mathbf{a - 2d, \quad a - d, \quad a, \quad a + d, \quad a + 2d}$

• Their common difference is $d$.
• Their sum is: $(a - 2d) + (a - d) + a + (a + d) + (a + 2d) = 5a$.

4. Solved CBSE Board Examination Problems

Solved Example 1: Three Numbers with Sum and Product

Problem: The sum of three numbers in an AP is 27 and their product is 405. Find the numbers.

Solution:

1. Let the three numbers in AP be $a - d, a, a + d$.
2. Condition 1 (Sum is 27): $(a - d) + a + (a + d) = 27$ $3a = 27 \implies a = 9$
3. Condition 2 (Product is 405): $(a - d) \times a \times (a + d) = 405$ Substitute $a = 9$: $(9 - d) \times 9 \times (9 + d) = 405$ Divide by 9: $(9 - d)(9 + d) = \frac{405}{9} = 45$
4. Apply the algebraic identity $(a - b)(a + b) = a^2 - b^2$: $81 - d^2 = 45$ $d^2 = 81 - 45 = 36 \implies d = \pm 6$
5. Form the Numbers:
   • When $a = 9$ and $d = +6$: $\text{Numbers are: } (9 - 6), 9, (9 + 6) \implies \mathbf{3, 9, 15}$
   • When $a = 9$ and $d = -6$: $\text{Numbers are: } (9 - (-6)), 9, (9 + (-6)) \implies \mathbf{15, 9, 3}$
6. In both cases, the collection of three numbers is identical: <u>$3, 9, 15$</u>.

Solved Example 2: Angles of a Triangle in AP

Problem: The angles of a triangle are in an AP. The greatest angle is twice the least. Find all the angles of the triangle.

Solution:

1. Let the three interior angles of the triangle in AP be $(a - d)^\circ, a^\circ, (a + d)^\circ$.
2. We know that the sum of the angles of a triangle is always $180^\circ$: $(a - d) + a + (a + d) = 180^\circ$ $3a = 180^\circ \implies a = 60^\circ$
3. The angles are now $(60 - d)^\circ, 60^\circ, (60 + d)^\circ$. Here the least angle is $(60 - d)^\circ$ and the greatest angle is $(60 + d)^\circ$.
4. Given Condition (Greatest angle = 2 times least angle): $60 + d = 2(60 - d)$ $60 + d = 120 - 2d$ $3d = 60 \implies d = 20^\circ$
5. Compute the angles:
   • Smallest angle: $60^\circ - 20^\circ = 40^\circ$
   • Middle angle: $60^\circ$
   • Largest angle: $60^\circ + 20^\circ = 80^\circ$
6. Check: $40^\circ + 60^\circ + 80^\circ = 180^\circ$, and $80^\circ = 2 \times 40^\circ$.
7. Therefore, <u>the angles of the triangle are $40^\circ, 60^\circ, 80^\circ$</u>.

5. Summary and Examination Tips

Exam Tip: If a question involves the angles of a triangle in AP, you can IMMEDIATELY deduce that the middle angle is $60^\circ$, because $(a-d) + a + (a+d) = 180^\circ \implies 3a = 180^\circ \implies a = 60^\circ$!

Common Mistake: When using the 4-term selection $(a - 3d, a - d, a + d, a + 3d)$, students often forget that the common difference is $2d$, not $d$. Once you find $d$, multiply it by $2$ to state the common difference of the sequence!