CAT 2025

Slot 3

QA

Question & SolutionQues & Sol

In an arithmetic progression, if the sum of fourth, seventh and tenth terms is 99, and the sum of the first fourteen terms is 497, then the sum of first five terms is

1. Concept Used

• Topic: Arithmetic Progression — General Term and Sum Formula
• Formula: $$T_n = a + (n-1)d \quad \text{and} \quad S_n = \frac{n}{2}(2a + (n-1)d)$$

2. Calculation

Let the first term of the AP be $a$ and the common difference be $d$.

The fourth, seventh, and tenth terms are: $$T_4 = a + 3d, \quad T_7 = a + 6d, \quad T_{10} = a + 9d$$

Their sum is given as 99: $$(a + 3d) + (a + 6d) + (a + 9d) = 99$$ $$3a + 18d = 99$$ $$a + 6d = 33 \quad \cdots (1)$$

Note that $T_7 = a + 6d$, so equation (1) simply tells us the 7th term equals 33 — a clean insight that the middle term of a symmetric trio equals the average.

Next, the sum of the first 14 terms is 497: $$S_{14} = \frac{14}{2}(2a + 13d) = 497$$ $$7(2a + 13d) = 497$$ $$2a + 13d = 71 \quad \cdots (2)$$

From equation (1): $a = 33 - 6d$

Substituting into equation (2): $$2(33 - 6d) + 13d = 71$$ $$66 - 12d + 13d = 71$$ $$d = 5$$

Substituting back into equation (1): $$a + 6(5) = 33 \Rightarrow a = 33 - 30 = 3$$

So the AP starts at $a = 3$ with common difference $d = 5$.

The first five terms are: $3, 8, 13, 18, 23$

Sum of first five terms: $$S_5 = \frac{5}{2}(2(3) + (5-1)(5)) = \frac{5}{2}(6 + 20) = \frac{5}{2} \times 26 = 65$$

3. Solution

Answer = 65 ✅

The sum of the first five terms of the arithmetic progression is 65.