Question

an ap consist of 5 term of which 3rd term is 12 and the last term is 106 find the 29 term

Answer 1

As it contains 5 terms,

3rd term = 12

a+2d = 12       -----------------1equation 
_______________________

last term = 5th term 

5th term =  106 

a+4d = 106  -----------2equation.
___________________________

subtract  both equation,

a+ 4d = 106 
a+ 2d = 12 
-   -       -

2d = 94

 d= 94/2 

d= 47 

putting the value of d in 1equation,

a+ 2(47) = 12

a + 94 =12

a= 12 -94
 
a= -82

Now,

29th term = a+28d

29 th term = -82 + [28*47]

29 th  term  =  1316 - 82


29th term = 1234




i hope this will help you


-by ABHAY