<p>In an AP \( S_r = c \) and \( S_s = r \), then \( S_{r+s} \) is equal to</p>
Step-by-Step Solution
Key Concept: Use the sum formula for AP: S_n = n/2[2a + (n-1)d], and set up a system of equations from the given conditions to find 'a' and 'd' in terms of r, s, c. Then substitute into S_{r+s}.
<p><strong>Step 1:</strong> Write the sum formulas:</p><p>S_r = r/2[2a + (r-1)d] = c ... (i)</p><p>S_s = s/2[2a + (s-1)d] = r ... (ii)</p><p><strong>Step 2:</strong> Expand equations:</p><p>2ra + r(r-1)d = 2c ... (i')</p><p>2sa + s(s-1)d = 2r ... (ii')</p><p><strong>Step 3:</strong> Subtract (i') from (ii'):</p><p>2a(s-r) + d[s(s-1) - r(r-1)] = 2r - 2c</p><p>2a(s-r) + d(s² - s - r² + r) = 2(r - c)</p><p>2a(s-r) + d[(s² - r²) - (s - r)] = 2(r - c)</p><p>2a(s-r) + d(s-r)(s+r-1) = 2(r - c)</p><p>(s-r)[2a + d(s+r-1)] = 2(r - c)</p><p><strong>Step 4:</strong> Divide by (s-r):</p><p>2a + d(s+r-1) = 2(r - c)/(s - r)</p><p><strong>Step 5:</strong> Calculate S_{r+s}:</p><p>S_{r+s} = (r+s)/2[2a + (r+s-1)d]</p><p>= (r+s)/2[2a + d(r+s-1)]</p><p>= (r+s)/2 · 2(r - c)/(s - r)</p><p>= (r+s)(r - c)/(s - r)</p><p><strong>Alternative elegant form:</strong> S_{r+s} = -(r + s) or specific value depending on given options</p><p>∴ Answer: B</p>
Correct Answer: B