Four distinct positive integers a, b, c, and d form an arithmetic sequence. Also, the integers a, b,and d form a geometric sequence. If the sum of the.four integers is a perfect cube, what is thesmallest possible value of a? Explain how you know this is the smallest possible value of a.

If we have an arithmetic sequence, then it is of the form n, n+m , n+2m, n+3m, .... Here n is the first term of the sequence and m is the common difference. Now, we are given that a,b,c,d are in arithmetic sequence. Therefore, a=nb=n+mc=n+2md=n+3m We can replace n=a in b,c,d and get: a=nb=a+mc=a+2md=a+3m ....(1)Now, a,b,d form a geometric sequence. A geometric sequence is of the form s, sr, sr2, ... Here s is the first term and r is the common ratio. Now, we see that product of first and third term is equal to square of the second term. That is s×sr2=sr2 Therefore, since a,b,d form a geometric sequence, therefore, ad=b2 Using the values from (1): ad=b2⇒aa+3m=a+m2⇒a2+3am=a2+m2+2am⇒0=m2-am⇒m2-am=0⇒mm-a=0⇒m=0 or m=a Since, m is the common difference therefore m≠0 Hence, we get m=aNow, from (1) we get: a=nb=a+mc=a+2md=a+3m Since, m=a we get: a=nb=a+a=2ac=a+2a=3ad=a+3a=4a Now, we are given that the sum of the four integers is a perfect cube. Let the sum of the four integers be z3 Therefore, a+b+c+d=z3⇒a+2a+3a+4a=z3⇒10a=z3 Now, for 10a to be a perfect cube we must have a=102=100 Then 10a=10×100=1000=103 Hence, the smallest possible value of a is 100 Answer: The smallest possible value of a is 100