Q. 84.7( 3 Votes )

The 6th and 17th terms of an A.P. are 19 and 41 respectively. Find the 40th term.

Answer :

Given,


6th term of an A.P is 19 and 17th terms of an A.P. is 41


a6 = 19 and a17 = 41


We know, an = a + (n – 1)d where a is first term or a1 and d is common difference and n is any natural number


When n = 6:


a6 = a + (6 – 1)d


a6 = a + 5d


Similarly, When n = 17:


a17 = a + (17 – 1)d


a17 = a + 16d


According to question:


a6 = 19 and a17 = 41


a + 5d = 19 ………………(i)


And a + 16d = 41…………..(ii)


Subtracting equation (i) from (ii):


a + 16d – (a + 5d) = 41 – 19


a + 16d – a – 5d = 22


11d = 22



d = 2


Put the value of d in equation (i):


a + 5(2) = 19


a + 10 = 19


a = 19 – 10


a = 9


As, an = a + (n – 1)d


a40 = a + (40 – 1)d


a40 = a + 39d


Now put the value of a = 9 and d = 2 in a40


a40 = 9 + 39(2)


a40 = 9 + 78


a40 = 87


Hence, 40th term of the given A.P. is 87


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