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Question:

The sum of the 2nd and the 7th terms of an AP is 30. If its 15th term is 1 less than twice its 8th term, find the AP.

Solution:

Given:a2+a7=30Let the first term be a and common difference bed.So,a+d+a+6d=302a+7d=30— (1)15thterm is 1 less than twice the8thterm,So,a+14d=2(a+7d)–1a+14d=2a+14d–1⇒a=1Substitute the value ofain equation1,2×1+7d=30⇒d=4Therefore, A.P. are1,5,9,13,.