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In this Class 10 Mathematics topic from Arithmetic Progressions (AP), students learn how to find the sum of the first n terms of an arithmetic progression. They identify the first term, common difference, and number of terms, then apply the formulas Sₙ = n/2 [2a + (n−1)d] and Sₙ = n/2(a + l) when the last term is known. Examples help learners solve numerical problems, verify results, and understand the pattern behind sums in an AP.
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Medium · Level 69 · arithmetic progression,even terms sum,Finding the sum of the first $n$ terms of an AP,finding the sum of the first n terms of an ap,Arithmetic Progressions (AP),arithmetic progressions ap,Mathematics,Class 10 MCQView options
Medium · Level 69 · arithmetic progression,sum formula,Finding the sum of the first $n$ terms of an AP,finding the sum of the first n terms of an ap,Arithmetic Progressions (AP),arithmetic progressions ap,Mathematics,Class 10 MCQView options
In an arithmetic progression, S₂₀ = 1160 and t₁₀ = 56. What is t₁₁?
Correct answer: B
The governing idea is the paired-term property of an arithmetic progression. For an even number of terms, the sum can be written as S₂ₘ = m(tₘ + tₘ₊₁), because the first and last terms, second and second-last terms, and so on have equal pair sums. Here 20 = 2 × 10, so S₂₀ = 10(t₁₀ + t₁₁). Substituting the given values gives 1160 = 10(56 + t₁₁). Dividing by 10 gives 116 = 56 + t₁₁, hence t₁₁ = 60. Therefore option B is correct. The other options result from an incorrect pairing or arithmetic error.
The sum of the first 24 terms of the arithmetic progression x, x+5, x+10, ... is 1788. What is x?
Correct answer: C
Use the sum formula Sₙ = n/2[2a + (n−1)d] for an arithmetic progression. In this sequence, the first term is a = x, the common difference is d = 5, and n = 24. Therefore 1788 = 24/2[2x + 23(5)] = 12(2x + 115). Dividing by 12 gives 149 = 2x + 115. Subtracting 115 gives 2x = 34, so x = 17. Thus option C is correct. Option 15 would produce a sum of 1740, while 16 and 18 would produce different sums; they arise from mishandling the 23d term or dividing incorrectly.
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