Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>If <mathjax>a_1, a_2, a_3, \ldots</mathjax> are in AP, and <mathjax>a_1 + a_{30} = a_6 + a_{25} = a_{10} + a_{21} = S</mathjax> (say), with <mathjax>a_1 + a_6 + a_{10} + a_{21} + a_{25} + a_{30} = 120</mathjax>, then find <mathjax>\sum_{i=1}^{30} a_i</mathjax>.</p>
Step-by-Step Solution
Key Concept: In an AP, terms equidistant from the ends have the same sum. Use this property to find S, then apply the AP sum formula.
<p><strong>Step 1:</strong> Since <mathjax>a_1, a_2, a_3, \ldots</mathjax> are in AP:</p><p><mathjax>a_1 + a_{30} = a_6 + a_{25} = a_{10} + a_{21} = S</mathjax></p><p><strong>Step 2:</strong> Given: <mathjax>a_1 + a_6 + a_{10} + a_{21} + a_{25} + a_{30} = 120</mathjax></p><p><strong>Step 3:</strong> Rearrange as: <mathjax>(a_1 + a_{30}) + (a_6 + a_{25}) + (a_{10} + a_{21}) = 120</mathjax></p><p><strong>Step 4:</strong> <mathjax>S + S + S = 120</mathjax></p><p><mathjax>3S = 120</mathjax></p><p><mathjax>S = 40</mathjax></p><p><strong>Step 5:</strong> The sum of an AP is: <mathjax>\sum_{i=1}^{30} a_i = \frac{30}{2}(a_1 + a_{30}) = 15 \times S = 15 \times 40 = 600</mathjax></p><p>∴ <mathjax>\sum_{i=1}^{30} a_i = 600</mathjax></p>
Correct Answer: 600