Question

The first four terms of an arithmetic sequence are: a, a + d, a + 2d, a + 3d. The first

Answer 1

The first eight terms of the sequence are 27, 27, 39, 87, 117.378, 147.755, 178.132 and 208.509.

Determination of a given set of successive values of a sequence

By (1) we have that $a = 27 - A$, and we simplify the system of equations as follows:

$(27-A)+d + A\cdot r = 27$

$d + (r-1)\cdot A = 0$ (2b)

$(27-A) +2\cdot d + A\cdot r^{2} = 39$

$2\cdot d + (r^{2}-1)\cdot A = 12$ (3b)

$(27-A) +3\cdot d + A\cdot r^{3} = 87$

$3\cdot d + (r^{3}-1)\cdot A = 60$ (4b)

By (2b), we simplify the system of equations once again:

$2\cdot (1-r)\cdot A+(r^{2}-1)\cdot A = 12$

$[2\cdot (1-r)+(r^{2}-1)]\cdot A = 12$ (3c)

$3\cdot (1-r)\cdot A + (r^{3}-1)\cdot A = 60$

$[3\cdot (1-r) + (r^{3}-1)]\cdot A = 60$ (4c)

And by equalising (3c) and (4c) we have an expression in terms of $r$:

$\frac{12}{[2\cdot (1-r)+(r^{2}-1)]} = \frac{60}{[3\cdot (1-r)+(r^{3}-1)]}$

$12\cdot [3\cdot (1-r)+(r^{3}-1)] = 60\cdot [2\cdot (1-r) + (r^{2}-1)]$

$36\cdot (1-r) +12\cdot (r^{3}-1) = 120\cdot (1-r)+60\cdot (r^{2}-1)$

$84\cdot (1-r) +60\cdot (r^{2}-1)-12\cdot (r^{3}-1) = 0$

$84-84\cdot r +60\cdot r^{2}-60-12\cdot r^{3}-12 = 0$

$-12\cdot r^{3}+60\cdot r^{2}-84\cdot r +2 = 0$ (5)

The roots of this third order polynomial are: $r_{1} \approx 0.0242$, $r_{2} = 2.488 + i\,0.831$ and $r_{3} \approx 2.488-i\,0.831$. Since $r$ must be a real number, then $r \approx 0.0242$.

By (4c) we have the value of $A$:

$A = \frac{60}{3\cdot (1-0.0242)+(0.0242^{3}-1)}$

$A \approx 31.130$

By (2b) we find the value of $d$:

$d = (1-0.0242)\cdot 31.130$

$d = 30.377$

And by (1) we find the value of $a$:

$a = 27-A$

$a = 27-31.130$

$a = -4.13$

The first eight terms are calculated below:

$n_{1} = 27$

$n_{2} = 27$

$n_{3} = 39$

$n_{4} = 87$

$n_{5} = [-4.13 + 4\cdot (30.377)]+(31.130)\cdot (0.0242)^{4} = 117.378$

$n_{6} = [-4.13+5\cdot (30.377)+(31.130)\cdot (0.0242)^{5}] = 147.755$

$n_{7} = [-4.13+6\cdot (30.377)\cdot (31.130)\cdot (0.0242)^{6}] = 178.132$

$n_{8} = [-4.13+7\cdot (30.377)\cdot (31.130)\cdot (0.0242)^{7}] = 208.509$

The first eight terms of the sequence are 27, 27, 39, 87, 117.378, 147.755, 178.132 and 208.509. $\blacksquare$

Remark

The statement present typing mistakes and is poorly formatted. Correct form is shown below:

The first four terms of an arithmetic sequence are: $a$, $a + d$, $a + 2\cdot d$, $a + 3d$. The first four terms of another sequence are: $A$, $A\cdot r$, $A\cdot r^{2}$, $A\cdot r^{3}$. The eight terms satisfy:

$a + A = 27$ (1)

$(a+d)+A\cdot r = 27$ (2)

$(a + 2\cdot d) + A\cdot r^{2} = 39$ (3)

$(a + 3\cdot d) + A\cdot r^{3} = 87$ (4)

By using the substitution $a = 27-A$, or otherwise, find the all eight terms.

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