Question

Find four consecutive terms in an A.P. whose sum is -54 and the sum of 1 st and 3rd term is -30

Answer 1

let the first term be a and commen difference be d
then sum of first four terms will be 4/2*(2a+(4-1)d)=-54
the first term is a and the third term is a + 2d sum of which is -30
2a + 2d = -30
a + d = -15
a= -15-d
put in another equation
-54 = 2(2( -15-d)+3d)
-54 =2( -30 -2d + 3d)
-54=-60 + 2d
2d = 6
d =3
a = -15-3
a = -18
so the four consecutive term of ap is
-18, -15, -12 , -9