Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>If \(a_1, a_2, \ldots, a_n\) are in A.P. with common difference \(d \neq 0\), then the sum of the series \(d[\sec a_1 \sec a_2 + \sec a_2 \sec a_3 + \cdots + \sec a_{n-1} \sec a_n]\) is</p>
<p>\(\cosec a_n - \cosec a\)</p>
<p>\(\cot a_n - \cot a\)</p>
<p>\(\sec a_n - \sec a_1\)</p>
<p>\(\tan a_n - \tan a_1\)</p>
Step-by-Step Solution
Key Concept: Recognize that sec(a)sec(b) terms can be decomposed using the telescoping identity: d·sec(a)sec(b) = tan(b) - tan(a) when b - a = d. This converts the product series into a telescoping sum.
<p><strong>Step 1:</strong> Use the trigonometric identity for consecutive terms in A.P. with common difference d:</p><p>For terms in A.P.: a_{i+1} - a_i = d</p><p>We use: tan(A) - tan(B) = sin(A-B)/(cos A cos B) = d·sec(A)sec(B) when A - B = d</p><p><strong>Step 2:</strong> Rewrite each term:</p><p>d·sec(a_i)sec(a_{i+1}) = tan(a_{i+1}) - tan(a_i)</p><p><strong>Step 3:</strong> Sum the telescoping series:</p><p>d[sec a_1 sec a_2 + sec a_2 sec a_3 + ... + sec a_{n-1} sec a_n]</p><p>= [tan(a_2) - tan(a_1)] + [tan(a_3) - tan(a_2)] + ... + [tan(a_n) - tan(a_{n-1})]</p><p><strong>Step 4:</strong> Most terms cancel (telescoping):</p><p>= tan(a_n) - tan(a_1)</p><p>∴ Answer: C (tan(a_n) - tan(a_1))</p>
Correct Answer: C