# Interesting Questions On Progressions – 2IIM

Question

Consider a set of numbers from 1 to 751. How many Arithmetic Progressions can be formed from this set of numbers such that the first term is 1 and the last term is 751 and the AP has at least 3 terms.

Solution

Numbers from 1 to 751 such that the first term is one and the last term is 751. From the basic definition of AP,

a = 1, a + (n-1)d = 751.

Subtracting both, we get -> (n-1)d = 750.

Now, if we find the number of ways 750 can be written as a product of two numbers, we have our answer. (After excluding cases that have n<3)

750 can be written as -> 1 x 750, 2 x 375, 3 x 250, 5 x 150 … and so on.

1 x 750 is a case and so is 750 x 1, this is the same with all other ways of writing the product. Effectively, if we count the number of factors in the number, then we should have the answer.

750 = 53 x 21 x 31. => Number of factors = 4 x 2 x 2 = 16.

Now we need to exclude the case where n will be lesser than 3, this means that (n-1) should be greater than or equal to 2.

One case gets eliminated -> 1 x 750.

So totally there are 16 – 1 = 15 cases.

Question

Let Ai be a sequence such that An/An-1 =p for all n. Let Bi be a sequence such that  Bn/Bn-1 be q for all n>1

Further, let A1 = ¼ = p, B1 = ¾ = q.

Find the smallest n such that An + Bn < 0.1

Solution

The above sequences are nothing but Geometric Progressions. The first one has first term = common ratio = ¼. The second one has first term = common ratio = ¾

So, we are effectively solving for (¼)n + (¾)n  <  0.01

Now, we can skip to trial and error. (¾)n will be far higher than (¼)n , so we can focus on that.

Let us take n = 4; (¾)n =  81/256. This is way higher than 0.01. This is closer to 1/3. If we square this, that will be close to 0.01.

So, we are looking at (¾)8 = 6561/65536 . This is just more than 0.1. We can be reasonably confident that n = 9 will work. The (¼)9 component gives us a minuscule number and can be ignored.

So, n =9 works best

———–