Question

write any two arthemetic sequences such that the sum of any number of terms from the first term became a perfect square.​

Answer 1

Answer:

hello

Step-by-step explanation:

There is really only one such sequence.

Clearly the first term must be a perfect square which we will call a2 . With common difference d , the sum of n terms must be

n[a2+n−12d]

which we shall write as

n(nd+2a2−d)=2k2

for some integer k . Since there is a k for every integer n , there must be a k for every prime p . Thus

p∣∣pd+2a2−d

for every prime p .

In other words, every prime must divide 2a2−d and so d=2a2 . Then your sequence looks like

a2,3a2,5a2,7a2,…

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Answer 2

Answer:

There is really only one such sequence.

There is really only one such sequence.Clearly the first term must be a perfect square which we will call a2. With common difference d , the sum of n terms must be

There is really only one such sequence.Clearly the first term must be a perfect square which we will call a2. With common difference d , the sum of n terms must ben[a2+n-12d]

There is really only one such sequence.Clearly the first term must be a perfect square which we will call a2. With common difference d , the sum of n terms must ben[a2+n-12d]which we shall write as

There is really only one such sequence.Clearly the first term must be a perfect square which we will call a2. With common difference d , the sum of n terms must ben[a2+n-12d]which we shall write asn(nd+2a2-d)=262

There is really only one such sequence.Clearly the first term must be a perfect square which we will call a2. With common difference d , the sum of n terms must ben[a2+n-12d]which we shall write asn(nd+2a2-d)=262for some integer k. Since there is a k for

There is really only one such sequence.Clearly the first term must be a perfect square which we will call a2. With common difference d , the sum of n terms must ben[a2+n-12d]which we shall write asn(nd+2a2-d)=262for some integer k. Since there is a k forevery integer n, there must be a k for

There is really only one such sequence.Clearly the first term must be a perfect square which we will call a2. With common difference d , the sum of n terms must ben[a2+n-12d]which we shall write asn(nd+2a2-d)=262for some integer k. Since there is a k forevery integer n, there must be a k forevery prime p. Thus

There is really only one such sequence.Clearly the first term must be a perfect square which we will call a2. With common difference d , the sum of n terms must ben[a2+n-12d]which we shall write asn(nd+2a2-d)=262for some integer k. Since there is a k forevery integer n, there must be a k forevery prime p. Thusplpd+2a2-d

for every prime p.

for every prime p.In other words, every prime must divide 2a2-d and so d=2a2. Then your sequence

for every prime p.In other words, every prime must divide 2a2-d and so d=2a2. Then your sequencelooks like

for every prime p.In other words, every prime must divide 2a2-d and so d=2a2. Then your sequencelooks likea2,3a2,5a2,7a2,..

Hope it's helpful for you