Question -
Answer -
Given:
10th termof an A.P is 41, and 18th terms of an A.P. is 73
So, a10 =41 and a18 = 73
We know, an =a + (n – 1) d [where a is first term or a1 and d is the commondifference and n is any natural number]
When n = 10:
a10 =a + (10 – 1)d
= a + 9d
When n = 18:
a18 =a + (18 – 1)d
= a + 17d
According to question:
a10 =41 and a18 = 73
a + 9d = 41 ………………(i)
And a + 17d =73…………..(ii)
Let us subtractequation (i) from (ii) we get,
a + 17d – (a + 9d) =73 – 41
a + 17d – a – 9d = 32
8d = 32
d = 32/8
d = 4
Put the value of d inequation (i) we get,
a + 9(4) = 41
a + 36 = 41
a = 41 – 36
a = 5
we know, an =a + (n – 1)d
a26 =a + (26 – 1)d
= a + 25d
Now put the value of a= 5 and d = 4 in a26
a26 =5 + 25(4)
= 5 + 100
= 105
Hence, 26th termof the given A.P. is 105.