Question

Consider the arithmetic sequence 4,12,20.a, Prove that the sum of consecutive terms of this sequence (starting from the first term) is always a perfect square ​

Answer 1

$\large{\text{\underline{Solution:-}}$

Let the sequence be $\{a_{n}\}$.

Since it is an arithmetic sequence of the first term of 4 and a common difference of 8,

$\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{n}=\dfrac{n\{2a+(n-1)d\}}{2}$

this is a case where $a=1,d=8$, we get

$\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{n}=\dfrac{n\{2\times4+8(n-1)\}}{2}$

$\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{n}=\dfrac{n\times8n}{2}\implies\therefore\displaystyle\sum^{n}_{k=1}a_{n}=(2n)^{2}$

Hence, the sum of the first $n$ terms of the arithmetic sequence sum to a perfect square $(2n)^{2}$.

Answer 2

Given :-

The Arithmetic sequence 4,12,20....

To Prove :-

Prove that the sum of consecutive terms of this sequence (starting from the first term) is always a perfect square ​

Solution :-

Here

First term = a = 4

Common difference = a₂ - a₁ = 12 - 4 = 8

We know that

Sₙ = n/2[2a + (n - 1)d]

Sₙ = n/2[2(4) + (n - 1)8]

Sₙ = n/2[8 + 8n - 8]

Sₙ = n/2[8n]

Sₙ = 8n × n/2

Sₙ = 8n²/2

Sₙ = 4n²

Sₙ = (2 × n)²

Sₙ = (2n)²

Hence, Proved