Question

find four consecutive terms in an ap whose sum is -54 and the sum of the 1st and the 3rd term is -30​

Answer 1

okay so let the 4 terms be (a-2d),(a-d),(a+d),(a+2d)

lets apply the condition

(a-2d)+(a-d)+(a+d)+(a+2d)=-54

4a=-54

a=-54÷4=-13.5

(a-2d)+(a+d)=-30

2a-d=-30

lets substitute value of a

2×(-13.5)-d=-30

-27-d=-30

-d=-30+27=-3

d=3

hence the terms are

(a-2d)=-13.5-2×3=-19.5

(a-d)=-13.5-3=-16.5

(a+d)=-13.5+3=-10.5

(a+2d)=-13.5+2×3=-7.5

hope this anwer helps

Answer 2

Answer:

Let the four consecutive terms is an A.P. be a-d,a,a+d,a+2d

from the first condition,

(a-d)+a+(a+d)+(a+2d) = -54

a-d+a+a+d+a+2d = -54

4a+2d = -54

2a+d = -27...( Dividing both the sides by 2)...(1)

from the second condition,

(a-d)+(a+d) = -30

a-d+a+d = -30

a = -15

putting a = -15 in equation (1)

2(-15)+d = -27

-30+d = -27

d = -27+30

d = 3