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Question 8Fin...

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Question 8
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

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Solution

$G i v e n t h a t, a_{2} = 14 a_{3} = 18 d = a_{3} - a_{2} = 18 - 14 = 4 a_{2} = a + d 14 = a + 4 a = 10 S_{n} = \frac{n}{2} [2 a + (n - 1) d] S_{51} = \frac{51}{2} [2 \times 10 + (51 - 1) \times 4] = \frac{51}{2} [20 + 200] = 51 \times \frac{220}{2} = 51 \times 110 = 5610$

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