The sum of the terms of an arithmetic progression of 50 terms is 2500. The sum of the least 10 terms of the progression is 100. Find the largest term of the progression.
Explanation:
Let the first term be ‘a’ and common difference be ‘d’.
S50 = 50/2 × [2a + 49d] = 2500
⇒ 2a + 49d = 100 …(1)
Also, S10 = 10/2 × [2a + 9d] = 100
⇒ 2a + 9d = 20 …(2)
Solving (1) and (2) we get,
a = 1 and d = 2
∴ T50 = a + 49d = 99
Hence, 99.
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