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The sum of the first three consecutive terms of an A.P. is 9 and the sum of their squares is 35. Find Sn |
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Answer» n Let the three numbers in the AP be a-d, a and a+d.So a-d+a+a+d = 3a = 9 or a = 3.Now (a-d)^2+a^2+(a+d)^2 = 35, ora^2–2ad+d^2+a^2+a^2+2ad+d^2 = 35, or3a^2+2d^2 = 35, or3x3^2+2d^2 = 35, or2d^2 = 35–27 = 8, or d = 2.So the terms are: 1, 3 and 5. The sum is ALREADY in the question as 9, Which other sum are you referring to? |
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