# Squares in series

Easier ones this time:

(1) Find three numbers which are in an Arithmetic Progression and the conditions being:

• all numbers are distinct

• each number should be a perfect square

• as multiple sets are possible, find the least one (means where the range - that is, the difference between the largest number and the smallest number - is the lowest)

(2) Why it is impossible to find the next term in the series: 1, 9, ___ (Not because, the 3rd term is NOT provided....there is a special reason for it !!) ?

• Is (2) supposed to follow the same rules as (1)? – Styx Jul 18 '17 at 12:11
• Yes, (i) and (ii) are applicable but not the question # 1 itself is applicable @Styx. – Mea Culpa Nay Jul 18 '17 at 12:29

(1) I believe the shortest is:

$1,25,49$ - the common difference in this arithmetic progression being $24$.

Fun fact: it's impossible to find four perfect squares in arithmetic progression.

(2) Because in order to form an arithmetic progression, the next term would have to be 17, which isn't square? Or because your question is too broad and there are multiple possible ways of continuing the sequence 1, 9, ___ ?

• Hint for #2: >! The numbers '1' and '9' are odd numbers and also they are perferct squares... – Mea Culpa Nay Jul 19 '17 at 9:43

(1)

1, 25, 49 - the difference between smallest and largest is 48