Progression arithmetic/geometric

Question

I tried to equate the system but could not find the solution. Could anyone help?

The sequence $(a_1, a_2, a_3, \ldots)$ is an arithmetic progression with common difference $3$, and the sequence $(b_1, b_2, b_3, \ldots)$ is an increasing geometric progression. Knowing that $a_2 = b_3$, $a_{10} = b_5$ and $a_{42} = b_7$, the value of $b_4-a_4$ is:

Given Options

• a. 2

• b. 0

• c. 1

• d. -1

Answer 1

Using the property $b_n^2=b_{n-k}b_{n+k}$ for geometric series, we have

$$(a_0+30)^2=a_{10}^2=b_5^2=b_3b_7=a_2a_{42}=(a_0+6)(a_0+126)$$

so

$$a_0^2+60a_0+900=a_0^2+132a_0+756$$

which simplifies to $144=72a_0$. Thus $a_0=2$, hence $a_n=3n+2$. Now

$$b_4^2=b_3b_5=a_2a_{10}=8\cdot32=16\cdot16$$

so $b_4=16$, and thus $b_4-a_4=16-14=2$.

Answer 2

We have $$ \left\{ \begin{array}{rlr} a_2=&b_3&(1)\\ a_{10}=&b_{5}&(2)\\ a_{42}=&b_{7}&(3)\\ \end{array} \right. \Longleftrightarrow \left\{ \begin{array}{rl} a_1+1\cdot 3=&b_1\cdot s^{2}&(1)\\ a_1+9\cdot 3=&b_1\cdot s^{4}&(2)\\ a_1+41\cdot 3=&b_1\cdot s^{6}&(3)\\ \end{array} \right. $$ $$ \left\{ \begin{array}{rl} a_1+1\cdot 3=&b_1\cdot s^{2}&(1)\\ a_1+9\cdot 3=&b_1\cdot s^{4}&(2)\\ a_1+41\cdot 3=&b_1\cdot s^{6}&(3)\\ \end{array} \right. \Longleftrightarrow \left\{ \begin{array}{rl} 24=&b_1\cdot (s^{4}-s^{2})&(2)-(1)\\ 96=&b_1\cdot (s^{6}-s^{4})&(3)-(2)\\ \end{array} \right. $$

Note that $96=b_1\cdot (s^{6}-s^{4})=[b_1\cdot (s^4-s^2)]\cdot s^2=24\cdot s^2$ implies $$ s^2=\frac{96}{24}=4 \\ s^4=16 \\ s^6=64 $$ Then $b_1=2$ and $a_1=5$.

Answer 3

Hint: What is $a_2$ in terms of $a_1$? How about $a_3$? How about $a_n$? Now, what is $b_2$ in terms of $b_1$? How about $b_n$? Once you have that, you have three equations and three unknowns ($a_1$, $b_1$, and the geometric progression ratio). Solve them.