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Quarterly Examination, Mathematics, Grade 10
Any school test works the same way — quarterly, half-yearly, pre-board.
Chapter · Arithmetic Progressions
Concept · Sum and product of three AP terms
A 3-mark question
Find three consecutive terms of an A.P. such that their sum equals 30 and their product equals 910.
What the student wrote
“Let the three terms be a−d, a, a+d … a−d+a+a+d = 30 → 3a = 30 ∴ a = 10 … 10(a² + d²) = 910 … 100 + d² = 91 … d² = −9 … no real value of d”
2 of 3 marks lost, step by step
CBSE marking scheme — ½ + ½ + 1 + 1
Take the three terms as a−d, a, a+d
½The board's standard assumption for three terms in A.P. — half a mark for setting the question up this way.
Use the sum: 3a = 30, so a = 10
½Correct method and correct value — and it is carried into the next step.
Use the product: a(a² − d²) = 910, giving d² = 9
0 / 1Wrote a(a² + d²) — the minus became a plus, so 100 + d² = 91 gives d² = −9. The product mark is lost here, once.
d = ±3, so the terms are 7, 10, 13
0 / 1d² came out negative, so no terms were ever named. The paper stops at d² = −9.
Marks awarded
1/ 3
The assumption and a = 10 were right — that is the 1 mark. The flipped sign in the product step ends the paper.
Next step
Sign before expanding: a² − d², not a² + d². Then 100 − d² = 91 → d² = 9 → d = ±3 → 7, 10, 13.
A real Grade 10 paper — student details withheld
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