Question

Find the 41 term of an ap whose 13th term is 79 and 26 the 10 is 157

Answer 1

Let the first term of the A.P be a and the common difference be d.
Now,
∵ aⁿ = a + ( n -1 )d

∴ a₁₃ = a +  12d = 79
∴ a + 12 d = 79 ₋₋₋₋₋₋₋₋₋₋⇒ 1

and,
a₂₆ = a + 25d = 157
∴ a + 25d = 157---------⇒ 2 

Now, solving both equations 1 and 2 by elimination method

  a + 12d = 79
  a + 25d = 157
(-) (-)          (-)
______________
       - 13d = - 78
d = - 78 / -13
∴ d = 6

Now, substituting the value d = 6 in equation 1 we get

a + 12d = 79
a + 12(6) = 79
a = 79 - 72
∴ a = 7

Now,
a₄₁ = a + 40d
= 7 + 40*6
= 7 + 240

∴ a₄₁ = 247

∴ The 41st term of the given A.P is 247.

Hope this helps you !!
# Dhruvsh