Given :
10th term of an A.P is 41, and
18th terms of an A.P. is 73
⇒ a10 = 41 and a18 = 73
We know,
an = a + (n – 1)d
Where a is first term or a1 and d is the common difference and n is any natural number
When n = 10 :
∴ a10 = a + (10 – 1)d
⇒ a10 = a + 9d
Similarly,
When n = 18 :
∴ a18 = a + (18 – 1)d
⇒ a18 = a + 17d
According to question :
a10 = 41 and
a18 = 73
⇒ a + 9d = 41 ………………(i)
And a + 17d = 73…………..(ii)
Subtracting equation (i) from (ii) :
a + 17d – (a + 9d) = 73 – 41
⇒ a + 17d – a – 9d = 32
⇒ 8d = 32
⇒ d = \(\frac{32}{8}\)
⇒ d = 4
Put the value of d in equation (i) :
a + 9(4) = 41
⇒ a + 36 = 41
⇒ a = 41 – 36
⇒ a = 5
As,
an = a + (n – 1)d
a26 = a + (26 – 1)d
⇒ a26 = a + 25d
Now,
Put the value of a = 5 and
d = 4 in a26
⇒ a26 = 5 + 25(4)
⇒ a26 = 5 + 100
⇒ a26 = 105
Hence,
26th term of the given A.P. is 105.