The initial trio of terms in an arithmetic progression are $v_1$, $5v_1-8$ and $3v_1+8$.

Question

SLPaper 1

The initial trio of terms in an arithmetic progression are $v_1$ , $5v_1-8$ and $3v_1+8$ .

Demonstrate that $v_1=4$ .

[2]

Verified

Solution

EITHER

uses $v_2 - v_1 = v_3 - v_2$ (M1)

$5v_1 - 8 - v_1 = 3v_1 + 8 - 5v_1 + 8$ $6v_1 = 24$ A1

uses $v_2 = \frac{{v_1 + v_3}}{2}$ (M1)

$5v_1 - 8 = \frac{{v_1 + (3v_1 + 8)}}{2}$ $3v_1 = 12$ A1

THEN

so $v_1 = 4$ AG

[2 marks]

Establish that the total of the first $n$ terms of this arithmetic progression is a perfect square.

[4]

Verified

Solution

$d = 8$ (A1)

uses $S_n = \frac{n}{2}\left(2v_1 + (n-1)d\right)$ M1

$S_n = \frac{n}{2}\left(8 + 8(n-1)\right)$ A1

$= 4n^2$

$= \left(\frac{2n}{2}\right)^2$ A1

Note: The final A1 can be awarded for clearly explaining that $4n^2$ is a perfect square.

so total of the first $n$ terms is a perfect square AG

[4 marks]

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