An important and exam oriented question from arithmetic progression chapter as it was already asked in previous year paper of 2014 in which we have been asked to find the AP, if the sum of the 2nd and the 7th terms of an AP is 30. If it’s 15th term is 1 less than twice its 8th.
Book RS Aggarwal, class 10, chapter 5A, question no 35.
Given: a2+a7=30
Let the first term be a and common difference be d.
So, a+d+a+6d=30
2a+7d=30 — (1)
15th term is 1 less than twice the 8th term,
So, a+14d=2(a+7d)–1
a+14d=2a+14d–1
⇒a=1
Substitute the value of a in equation 1,
2×1+7d=30
⇒d=4
Therefore, A.P. are 1,5,9,13,.....