Q. 74.5( 17 Votes )

# Find the 31^{st} term of an AP whose 11^{th}term is 38 and the 16^{th} term is 73

Answer :

Given

11^{th}term = 38

⇒ a + (11-1)d = 38 → __1__

16^{th} term = 73

⇒ a + (16-1)d = 73 → __2__

By subtracting the both equations we will get ‘d’

(a +10d) – (a+15d) = 38 -73

-5d = -35

d = 7

By substituting “d” in equation __1__

a +10d = 38

a + 10×7 = 38

a = -32

a_{31} = a + (31-1) d

= -32 + 30×7

= 178

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