Q. 11

The 10th and 18th term of an A.P. are 41 and 73 respectively, find 26th term.

Answer :

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 = 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


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