In an A.P., the sum of three consecutive terms is 24 and the sum of their squares is 194. Find the numbers.
Generated by claude-sonnet-4-6 · 2026-06-15 10:31 · grounding rag
Model Answer
Let the three consecutive terms of the A.P. be $(a-d)$, $a$, $(a+d)$.
Condition 1: Sum = 24
$$(a-d) + a + (a+d) = 24$$
$$3a = 24 \implies a = 8$$
Condition 2: Sum of squares = 194
$$(a-d)^2 + a^2 + (a+d)^2 = 194$$
$$3a^2 + 2d^2 = 194$$
$$3(64) + 2d^2 = 194$$
$$2d^2 = 194 - 192 = 2 \implies d^2 = 1 \implies d = \pm 1$$
When $d = 1$: terms are 7, 8, 9
When $d = -1$: terms are 9, 8, 7
∴ The three numbers are 7, 8, 9.
Source: Chapter 5, Arithmetic Progressions
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Explanation
- Always assume consecutive A.P. terms as $(a-d), a, (a+d)$ — this makes the sum condition simple (middle term directly gives $a$).
- Expand $(a-d)^2 + a^2 + (a+d)^2 = 3a^2 + 2d^2$ — a standard identity to remember.
- Both $d = +1$ and $d = -1$ give the same set of numbers, so either answer is accepted; state both for full marks.