Question 936170
formula for n th term ( tn) = a+(n-1)d
where a= first term
      d= common difference
      n = no of terms
so 3rd term  (t3)= a+(3-1)d
                  = a+2d
similarly
          7 the term   t(7)  = a+(7-1)d
                               =a+6d

6 th term  (t6)=  a+5d
  4 th term (t4) = a+3d
  sum of the 3rd and 7th term is 24
so 
            (a+2d) +(a+6d) = 24
                 2a+8d  =24...................eq(1)

difference between the 6th and 4th is 4
 (a+5d)-(a+3d) = 4
     2d  = 4
 divide with 2 on both sides
  2d/d  = 4/4
 d= 2
put d=2 in eq(1) 
         2a+8*2=24
            2a+16=24
          2a  =24-16
   2a= 8
   divide with 2 on both sides
  {{{ 2a/2  = 8/2}}}
          a= 4
so , first term   (a) = 4
sum of n terms in Arithmetic series  =  {{{ (n/2)(2a+(n-1)d)}}}
 sum of first 14 terms  = {{{ (14/2)(2*4 +(14-1)2)}}}
             ={{{7(8+13*2)}}}
           ={{{7(8+26)}}}
     ={{{7(34)}}}
       ={{{238}}}
Result : first term  = 4 , sum  = 238