19th term of AP = a+ (n-1)d
= a+ (19-1)d
52 = a+ 18d --------------1
38th term of AP = a+(n-1)d
= a+ (38-1)d
148 = a+ 37d -------------2
adding eq1 and eq 2
a + 37d = 148
a + 18d = 52
+ + +
= 2a + 55d = 200 ----------eq3
s56 = n/2 + [ 2a + (n-1)d]
= 56/2 [ 2a + (56-1)d ]
= 28 [ 2a + 55d )
= 28( 200) ---------------from eq3
= 56000
therefore sum of first 56 terms of the given AP is 5600
