1.

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 = \(\frac{32}{8}\) 

⇒ 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.


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