the n th term of arithmetic sequence is 8n-4 ,Prove that the sum of any number of terms from the first is always a perfect square.
Answers
Step-by-step explanation:
Given :-
The nth term of arithmetic sequence is 8n-4
To find :-
Prove that the sum of any number of terms from the first is always perfect square ?
Solution :-
Given that
The nth term of arithmetic sequence = 8n-4
an = 8n-4 ----------(1)
Put n = 1 in (1) then
=> a1 = 8(1)-4
=> a1 = 8-4
=> a1 = 4
First term = 4
Put n = 2 in (1) then
a2 = 8(2)-4
=> a2 = 16-4
=> a2 = 12
Common difference (d) = a2-a1
=> d = 12-4
=> d = 8
We know that
Sum of first n terms of an AP is
Sn = (n/2)[2a+(n-1)d]
Put n = 2 then
S2 = 4+12 = 16 = 4² = Perfect square
Put n= 3 then
S3 = (3/2)[2(4)+(2)8)]
=> S3 = (3/2)[8+16]
=> S3 = (3/2)(24)
=> S3 = (3×12)
=> S3 = 36 =6² = Perfect square
Sum of the first n terms
=> Sn = (n/2)[2(4)+(n-1)(8)]
=> Sn = (n/2)[8+8n-8]
=> Sn = (n/2)(8n)
=> Sn = (n×8n)/2
=> Sn = n×4n
=> Sn = 4n²
=> Sn = (2n)²
=> It is a perfect square.
Hence, Proved.
Answer :-
The sum of any number of terms from the first is always perfect square.
Used formulae:-
Sum of first n terms of an AP is
Sn = (n/2)[2a+(n-1)d]
Where ,
a = First term
d = Common difference
n = number of terms