Question -
Answer -
Given:
6th termof an A.P is 19 and 17th terms of an A.P. is 41
So, a6 =19 and a17 = 41
We know, an =a + (n – 1) d [where a is first term or a1 and d is commondifference 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 subtractequation (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 inequation (i):
a + 5(2) = 19
a + 10 = 19
a = 19 – 10
= 9
As, an =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, 40th termof the given A.P. is 87.