For solving these type of questions we must always take the numbers as (a-3d),(a-d),(a+d),(a+3d)
So,
Let the investment made by Ram in these years be (a-3d),(a-d),(a+d),(a+3d)
Then,
(a-3d) + (a-d) + (a+d) +(a+3d) = 2000
=) 4a = 2000
=) a = 500
Now, again A/Q
(a-3d)^{2} +(a-d)^{2} +(a+d)^{2} +(a+3d)^{2} = 1200000
=) 4(a^{2} + 5 d^{2}) = 1200000
=) (a^{2} + 5 d^{2}) = 300000
=) 250000 + 5d^{2} = 300000
=) 5d^{2} = 50000
=) d^{2} = 10000
=) d = ±100
∴ d = 100 ( As the investment increases every year so d = -100 not possible)
Hence the investments made are 200,400,600,800 = (a) Ans.
Note - If there are 3 number in series then take the numbers as (a-d),a,(a+d)
If there are 5 number in series then take the numbers as (a-2d),(a-d), a,(a+d),(a+2d)
…….. so on
While If there are 4 number in the series then take the numbers as (a-3d),(a-d), ,(a+d),(a+3d)
If there are 6 number in the series then take the numbers as (a-5d),(a-3d),(a-d), ,(a+d),(a+3d), (a+5d)
……………. so on
Correct Answer: 200,400,600,800