Question

The 6th and 17th term of ap is 19 and 41 respectively find 40th term

Answer 1

Given, the 6th and 17th term of an A.P. are 19 and 41 respectively.

⇒ a6 = a + (6 -1)d = 19

⇒ a + 5d = 19 ... (1)

Also, a17 = a + (17 -1)d = 41

⇒ a + 16d = 41 ... (2)

On subtracting (1) from (2), we get

a + 16d - (a + 5d) = 41 - 19

⇒ 11d = 22

⇒ d = 2

On putting d = 2 in (1), we get

a + 5(2) = 19

⇒ a = 9

Now, 40th term of the a.p. is:

a40 = a + (40 -1)d = a + 39d

= 9 + 39(2) = 9 + 78 = 87 

Answer 2

Answer:

Step-by-step explanation:

Given that 6th term is 19 and the 17th term is 41 respectively, have to find the 40 th term of the A.P

We know that a+(n-1)*d

Therefore, a+(6-1)*d=19

=> a+5d=19 ---- (1)

a+(17-1)*d=41

=>a+16d=41 ----(2)

On Subtracting (1) and (2) we get

a+16d-a+5d= 41-19

=> 11d = 22

=> d = 22/11

= 2

Putting d=2, On equation -- (1) we get

a+(6-1)*2=19

=> a = 9

Therefore, To find 40 th term

a40 = 9+(40-1)*2

= 87

Hence, The 40 th term of this A.P is 87