The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the

1.	The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.
Answer» a₆ = 19 = a + (n – 1) d = 19 = a + 5d …(i) A₁₇ = 41 = a + (n – 1) d = 41 = a + 16d …(ii) Subtracting (i) from (ii), we get 22 = 11d d = 2 Now substituting the value of d in (i): 19 = a + 10 = a = 9 A₄₀ = a + (40 – 1) 2 = 9 + 78 = 87

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

Answer»

a₆ = 19 = a + (n – 1) d

= 19 = a + 5d …(i)

A₁₇ = 41 = a + (n – 1) d

= 41 = a + 16d …(ii)

Subtracting (i) from (ii), we get

22 = 11d

d = 2

Now substituting the value of d in (i):

19 = a + 10

= a = 9

A₄₀ = a + (40 – 1) 2

= 9 + 78

= 87