Choosing AP Terms: Split a Number into Parts | Class 10

Quick answer: When a problem asks you to split a number into terms of an AP, choose the terms symmetrically so that $d$ cancels when you add them. Use $a - d,\ a,\ a + d$ for three terms (sum $3a$), and $a - 3d,\ a - d,\ a + d,\ a + 3d$ for four terms (sum $4a$, common difference $2d$). For example, dividing $32$ into four parts in AP, with the product of the extremes to the product of the means in the ratio $7 : 15$, gives $2, 6, 10, 14$.

Suppose a question says: the sum of three numbers in an AP is $15$ and their product is $105$; find the numbers. If you write the numbers as $a$, $a + d$, $a + 2d$, the sum is $3a + 3d$, and you are left with two unknowns in a complicated equation.

A smarter choice is $a - d$, $a$, $a + d$. Their sum is just $3a$, so $a$ comes out immediately, and the product becomes much simpler.

In CBSE Class 10 Mathematics, Chapter 5 (Arithmetic Progressions), choosing terms this way is a useful technique for word problems on the sum and product of AP terms.

What You Will Learn

• How to choose 3, 4 or 5 terms of an AP symmetrically
• Why this makes $d$ cancel in the sum
• Example 1: three numbers with sum $15$ and product $105$
• Example 2: three numbers with sum $21$ where the first and third multiply to $45$
• Example 3: dividing $32$ into four parts in AP with a given ratio

1. Choosing the Terms

Key point: In each case, the $d$ terms cancel when you add, so the sum gives $a$ directly.

Why $a - 3d$ for four terms? The four terms must have equal gaps. In $a - 3d,\ a - d,\ a + d,\ a + 3d$, each gap is $2d$, so they form an AP.

2. Example 1: Sum 15, Product 105

Question

The sum of three numbers in AP is $15$ and their product is $105$. Find the numbers.

Solution

Step 1: Choose the terms. Let the numbers be $a - d$, $a$, $a + d$.

Step 2: Use the sum.

$3a = 15 \implies a = 5$

Step 3: Use the product.

$(a - d) \times a \times (a + d) = 105 \implies 5(25 - d^2) = 105$

$25 - d^2 = 21 \implies d^2 = 4 \implies d = \pm 2$

Answer: The numbers are $3, 5, 7$ (or $7, 5, 3$ when $d = -2$).

Check: $3 + 5 + 7 = 15$ and $3 \times 5 \times 7 = 105$. ✓

3. Example 2: Sum 21, First × Third = 45

Question

The sum of three numbers in AP is $21$, and the product of the first and third numbers is $45$. Find the numbers.

Solution

Step 1: Let the numbers be $a - d$, $a$, $a + d$. Then $3a = 21$, so $a = 7$.

Step 2: $(a - d)(a + d) = 45 \implies 49 - d^2 = 45 \implies d = \pm 2$.

Answer: The numbers are $5, 7, 9$.

Check: $5 + 7 + 9 = 21$ and $5 \times 9 = 45$. ✓

4. Example 3: Dividing 32 into Four Parts

Question

Divide $32$ into four parts that are in AP, such that the ratio of the product of the extremes (first and last) to the product of the means (middle two) is $7 : 15$.

Solution

Step 1: Choose the terms. Let the parts be $a - 3d$, $a - d$, $a + d$, $a + 3d$.

Step 2: Use the sum.

$4a = 32 \implies a = 8$

Step 3: Write the products.

$\text{Extremes: } (a - 3d)(a + 3d) = a^2 - 9d^2 = 64 - 9d^2$

$\text{Means: } (a - d)(a + d) = a^2 - d^2 = 64 - d^2$

Step 4: Use the ratio.

$\frac{64 - 9d^2}{64 - d^2} = \frac{7}{15}$

$15(64 - 9d^2) = 7(64 - d^2) \implies 960 - 135d^2 = 448 - 7d^2$

$512 = 128d^2 \implies d^2 = 4 \implies d = \pm 2$

Step 5: Find the parts. With $a = 8$ and $d = 2$: $8 - 6 = 2$, $8 - 2 = 6$, $8 + 2 = 10$, $8 + 6 = 14$.

Answer: The four parts are $2, 6, 10, 14$ (the same numbers in reverse order when $d = -2$).

Check: $2 + 6 + 10 + 14 = 32$; the common difference is $4$; and $\frac{2 \times 14}{6 \times 10} = \frac{28}{60} = \frac{7}{15}$. ✓

5. Summary Table

6. Exam Tips and Common Mistakes

Exam Tip: For four terms, the common difference is $2d$, not $d$. With $d = 2$, the terms go up by $4$: $2, 6, 10, 14$.

Exam Tip: Always check your answer against every condition in the question.

Common Mistake: Choosing four terms as $a - 2d,\ a - d,\ a + d,\ a + 2d$. The gaps are $d$, $2d$ and $d$, which are not equal, so these are not in AP.

Common Mistake: Forgetting that $d$ can be negative. $d = -2$ gives the same numbers in reverse order.

Frequently Asked Questions

How do you choose three terms in an AP?

Take them as a − d, a, a + d. Their sum is 3a, so a can be found straight from the sum.

How do you choose four terms in an AP?

Take them as a − 3d, a − d, a + d, a + 3d. Their sum is 4a and the common difference is 2d.

How do you divide 32 into four parts in AP?

Using a − 3d, a − d, a + d, a + 3d gives a = 8. With the ratio condition 7 : 15, d = ±2, so the parts are 2, 6, 10 and 14.

Why do we choose the terms symmetrically?

So that the d terms cancel when we add them, which gives the value of a immediately and makes the equations easier to solve.