Q8 of 40 Page 1

The value of p for which (2p + 1), 10 and (5p + 5) are three consecutive terms of an AP is

A . –1


B . –2


C . 41


D . 2


OR


The number of terms of an AP 5, 9, 18, … 185 is


A . 31 B . 51


C . 41 D . 40


D


9 – 2p = 5p – 5


9 + 5 = 5p + 2p


⇒ 14 = 7p


⇒ p = 2


OR


an = a + (n - 1)d


185 = 5 + (n - 1) × 4


185 – 5 = 4n – 4


180 = 4n – 4


184 = 4n



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