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CAT 2025 Slot 3 QA Question 1

Type-in-the-answer (no negative marking) · Algebra · Progression & Series · Try it, then check the answer and solution below.

CAT 2025 Slot 3QAAlgebra • Progression & SeriesModerate
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
TITA Answer:
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Answer and solution

Answer: 65

📌 Core Concept & Formula
In an arithmetic progression (AP), the nth term is given by:
an=a+(n−1)da_n = a + (n-1)d
where:
aa is the first term,
dd is the common difference.
The sum of the first nn terms of an AP is: Sn=n2[2a+(n−1)d]S_n = \frac{n}{2} [2a + (n-1)d]$
🔢 Step-by-Step Solution
1
Given:
Sum of 4th, 7th, and 10th terms is 99.
Sum of first 14 terms is 497.
2
Express the terms:
4th term: a4=a+3da_4 = a + 3d
7th term: a7=a+6da_7 = a + 6d
10th term: a10=a+9da_{10} = a + 9d
3
Sum of these terms:
(a+3d)+(a+6d)+(a+9d)=993a+18d=99a+6d=33(Equation 1)(a + 3d) + (a + 6d) + (a + 9d) = 99 3a + 18d = 99 a + 6d = 33 \quad \text{(Equation 1)}
4
Sum of first 14 terms:
S14=142[2a+13d]=4977(2a+13d)=4972a+13d=71(Equation 2)S_{14} = \frac{14}{2} [2a + 13d] = 497 7(2a + 13d) = 497 2a + 13d = 71 \quad \text{(Equation 2)}
5
Solve Equations 1 and 2:
Multiply Equation 1 by 2:
2a+12d=66−SubtractfromEquation2:2a + 12d = 66 - Subtract from Equation 2: (2a + 13d) - (2a + 12d) = 71 - 66 d = 5 $
6
Find aa:
a+6(5)=33a=3a + 6(5) = 33 a = 3
7
Sum of first 5 terms:
S5=52[2(3)+4(5)]S5=52[6+20]S5=52×26S5=5×13S5=65S_5 = \frac{5}{2} [2(3) + 4(5)] S_5 = \frac{5}{2} [6 + 20] S_5 = \frac{5}{2} \times 26 S_5 = 5 \times 13 S_5 = 65
⚡ 30-Second Shortcut
Recognize the pattern in the terms (4th, 7th, 10th) which are spaced 3 terms apart, simplifying to 3a+18d=993a + 18d = 99.
Use the sum formula for 14 terms to set up a second equation.
Solve the system of equations efficiently by elimination or substitution.
Plug values back into the sum formula for the first 5 terms.
🎯 Final Answer
Correct Answer: 65

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