Question

1. Consider the arithmetic sequence 10 ,17 ,24,........
a) what is the Algebraic form
b) prove that there is no perfect square in
this Sequence​

Answer 1

Step-by-step explanation:

The arithmetic sequence 10, 17, 24.

Here, First term a=10

The common difference d=17-10=7

The nth term of the sequence is

an= a+(n-1)d

an=10+(n-1)7

an=10+7n-7

an=7n+3

Let x be a natural number and its square is the nth term,

x²=7n+3

x²-3=7n

n=x²-3/7

Now, for all integers from 0 to 30, n does not come out to be an integer.

Therefore, the arithmetic sequence 10, 17, 24… contains no perfect squares.

Answer 2

Given :  arithmetic sequence 10,17,24

To Find : Prove that the arithmetic sequence 10,17,24  contains no perfect squares

Solution:

10 , 17  , 24

aₙ = a + (n-1)d

a = 10

d = 7

=> aₙ = 10 + (n-1)7

=> aₙ = 7n +  3  

Algebraic form 7n + 3

Any number Can be of form  

7k , 7k + 1 , 7k + 2  , 7k + 3  , 7k+4  , 7k + 5 , 7k  +  6

(7k)² == 49k² = 7.7k² = 7n ≠7n +  3  

(7k+1)² == 49k²  + 14k + 1 = 7.(7k² + 2k) + 1= 7n + 1  ≠7n +  3  

(7k+2)² == 49k²  + 28k + 4 = 7.(7k² + 4k) + 4= 7n + 4  ≠7n +  3  

(7k+3)² == 49k²  + 42k + 9 = 7.(7k² + 6k+1 ) + 2= 7n + 2  ≠7n +  3  

(7k+4)² == 49k²  + 56k + 17 = 7.(7k² + 8k+2) + 2= 7n + 2  ≠7n +  3  

(7k+5)² == 49k²  + 70k + 25 = 7.(7k² + 10k+3) + 4= 7n + 4  ≠7n +  3  

(7k+6)² == 49k²  + 84k + 36 = 7.(7k² + 12k+5) + 1= 7n + 1  ≠7n +  3  

Hence Proved that arithmetic sequence 10,17,24  contains no perfect squares

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