# Arithmetic and Geometric Progressions

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Airhtmetic Progressions

## Investments

Ram invests a total sum of 2000 rupees on government bonds in 4 years. If these investments are in A.P and the sum of squares of the investments is 1200000. Find the investment made by ram in each year respectively. It is also known that he always invest more than the previous year.
1. 200,400,600,800
2. 875,625,375,125
3. 125,375,625,875
4. 50,350,650,950

Choice A. 200,400,600,800

## Detailed Solution

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(a2 + 5 d2) = 1200000

=) (a2 + 5 d2) = 300000

=) 250000 + 5d2 = 300000

=) 5d2 = 50000

=) d2 = 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

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## More questions from Progressions

With some simple but very powerful ideas, one can cut down on a lot of working when it comes to progressions. For example, anchoring a progression around its middle term can be very useful. Reinforce these ideas with these questions.