If seventh and eleventh terms of an arithmetic progression are 31 and 47 respectively. then fifteenth termis

Let a and d are first term and common difference of an A.P.

$$n^{th}$$ term will be  $$a_{n}$$ = a + (n-1)d then we have 

$$a_{7}$$ = a + (7-1)d     

$$a_{7}$$ = a + 6d             equation 1

$$a_{11}$$ = a + (11-1)d    

$$a_{11}$$ = a + 10d           equation 2 

now eq2 - eq1 

we get  4d = 16  then  d = 4    put the value in eq 1 

we get  a  = 31 

so the 15^(th) term will be 

$$a_{15}$$ = 31 + (15-1)4 

$$a_{15}$$ = 63  answer