# The 6th</sup

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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