1.

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 = \(\frac{22}{11}\)

⇒ 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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