Choosing Three and Four Terms in an AP | Class 10 Maths

Quick answer: When the sum of some numbers in an AP is given, choose the terms symmetrically so that $d$ cancels when you add them: 3 terms → $a - d,\ a,\ a + d$; 4 terms → $a - 3d,\ a - d,\ a + d,\ a + 3d$ (common difference $2d$); 5 terms → $a - 2d$ to $a + 2d$. For example, three numbers in AP with sum 21 and product 231 are 3, 7, 11.

Why choose terms symmetrically?

If you take three terms as $a$, $a + d$, $a + 2d$, the sum is $3a + 3d$ and the product is a messy expression in two unknowns. If you take them symmetrically about the middle term, the $d$'s cancel when you add, and you find $a$ at once.

$(a - d) + a + (a + d) = 3a$

Symmetric choice of terms

Watch out for 4 terms: Consecutive terms differ by $(a - d) - (a - 3d) = 2d$. So the common difference is $2d$, not $d$.

This choice is most useful when the sum is given. If it is not, the usual $a$, $a + d$, $a + 2d$, … may be simpler.

Solved example 1: three numbers in AP

The sum of three numbers in an AP is 21 and their product is 231. Find the numbers.

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

Step 2: Use the sum.

$(a - d) + a + (a + d) = 21$

$3a = 21 \quad \Rightarrow \quad a = 7$

Step 3: Use the product.

$(7 - d) \times 7 \times (7 + d) = 231$

Divide both sides by 7:

$(7 - d)(7 + d) = 33$

$49 - d^2 = 33$

$d^2 = 16 \quad \Rightarrow \quad d = \pm 4$

Step 4: Write the numbers.

• If $d = 4$: the numbers are 3, 7, 11.
• If $d = -4$: the numbers are 11, 7, 3.

Answer: The numbers are 3, 7 and 11. Check: 3 + 7 + 11 = 21 and 3 × 7 × 11 = 231. ✓

Solved example 2: four terms in AP

The sum of four consecutive terms of an AP is 32, and the ratio of the product of the first and last terms to the product of the two middle terms is 7 : 15. Find the terms.

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

Step 2: Use the sum.

$4a = 32 \quad \Rightarrow \quad a = 8$

Step 3: Use the ratio.

$\frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{7}{15}$

With $a = 8$:

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

Step 4: Cross-multiply and solve.

$15(64 - 9d^2) = 7(64 - d^2)$

$960 - 135d^2 = 448 - 7d^2$

$512 = 128d^2$

$d^2 = 4 \quad \Rightarrow \quad d = \pm 2$

Step 5: Write the terms.

• If $d = 2$: the terms are 2, 6, 10, 14.
• If $d = -2$: the terms are 14, 10, 6, 2.

Answer: The terms are 2, 6, 10 and 14. Check: sum = 32; first × last = 28, middle two = 60, and 28 : 60 = 7 : 15. ✓ The common difference is $2d = 4$.

Summary table

Exam tips

• Say clearly which terms you have chosen before you start the algebra.
• Use the identity $(x - y)(x + y) = x^2 - y^2$ to simplify products quickly.
• Write both values of $d$ (positive and negative) and note that they give the same numbers in reverse order.
• Always check your answer against both conditions in the question.

Common mistakes

• Writing $(a - 3d)(a + 3d) = a^2 - 3d^2$ instead of $a^2 - 9d^2$.
• Taking the common difference as $d$ in the four-term case; it is $2d$.
• Forgetting the negative value of $d$.
• Using the symmetric form when the question gives the first term, not the sum, which can make it harder.

Frequently Asked Questions

How do you choose three terms in an AP?

Take them as a − d, a and a + d. Their sum is 3a, so the sum condition gives a directly.

How do you choose four terms in an AP?

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

What are three numbers in AP whose sum is 21 and product is 231?

The numbers are 3, 7 and 11. Here a = 7 from the sum, and 49 − d² = 33 gives d = ±4.

Why is the common difference 2d when choosing four terms?

Consecutive terms such as a − 3d and a − d differ by 2d, so the common difference of the AP is 2d.

When should I use symmetric terms in an AP?

Use them when the sum of the terms is given, because the d terms cancel and you can find a in one step.