Question

Prove that the arithmetic sequence 7,11,15 does not contain perfect squares

Answer 1

Step-by-step explanation:

Given :-

7,11,15,...

To find:-

Prove that the arithmetic sequence 7,11,15 does not contain perfect squares.

Solution:-

Given that :

7, 11, 15,... are in the Arithmetic Progression

First term (a) = 7

Common difference (d) = 11-7 = 4

We know that

nth term of an AP = an = a+(n-1)d

=> an = 7+(n-1)(4)

=> an = 7+4n-4

=> an = 4n+3

Let an be the perfect square

an is in the form of bq+r

Where b = 4 and r = 3 for some integer m

We know that

Euclid's Division Lemma

a = bq+r, 0≤r<b

So the possible values of r = 0,1,2,3

if r = 0 then a = 4m+0

a = 4m

=> a² = 16m² = 4(4m²) = 4n ≠ 4n+3

Where , n = 4m²

and

If r = 1 then a = 4m+1

=> a² = (4m+1)²

=> a² = 16m²+8m+1

=> a² = 4(4m²+2m)+1

=> a² = 4n+1 ≠4n+3

Where , n = 4m²+2m+1

If r = 2 then a = 4m+2

=> a² = (4m+2)²

=> a² = 16m²+16m+4

=> a² = 4(4m²+4m+4)

=> a² = 4n ≠ 4n+3

Where n = 4m²+4m+4

If r = 3 then a = 4m+3

=> a² = (4m+3)²

=> a² = 16m²+24m+9

=> a² = 16m² +24m +8+1

=>a² = 4(4m²+6m+2)+1

=> a² = 4n+1 ≠ 4n+3

Where, n = 4m²+6m+2

So , above conditions we did not get a perfect square

So there is no perfect square term in the AP

Hence ,Proved.

Answer:-

The arithmetic sequence 7,11,15 does not contain perfect squares.

Used formulae:-

• nth term of an AP = an = a+(n-1)d
• a = First term
• d = Common difference
• n = Number of terms

Euclids Division Lemma:-

• For any two Positive integers a and b there exist two positive integers q and r satisfying a = bq+r, 0r<b.
• (a+b)² = a²+2ab+b²