Question

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

Answer 1

Given:

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

So, a₆ = 19 and a₁₇ = 41

We know, a_n = 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

= a + 5d

Similarly, When n = 17:

a₁₇ = a + (17 – 1)d

= a + 16d

According to question:

a₆ = 19 and a₁₇ = 41

a + 5d = 19 ……………… (i)

And a + 16d = 41………….. (ii)

Let us subtract equation (i) from (ii) we get,

a + 16d – (a + 5d) = 41 – 19

a + 16d – a – 5d = 22

11d = 22

d = 22/11

= 2
Put the value of d in equation (i):

a + 5(2) = 19

a + 10 = 19

a = 19 – 10

= 9

As, a_n = a + (n – 1)d

a₄₀ = a + (40 – 1)d

= a + 39d

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

a₄₀ = 9 + 39(2)

= 9 + 78

= 87

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