Question

Let $a_1, a_2, a_3,...$ be a G.P. of increasing positive terms. If $a_1a_5 = 28$ and $a_2 + a_4 = 29$, then $a_6$ is equal to:

• A. 628
• B. 812
• C. 526
• D. 784

Answer 1

Answer: (D)

784

Answer 2

Let the GP be:
  $a_1 = a,\; a_2 = ar,\; a_3 = ar^2,\; \ldots$

Given:

1. $a_1 \cdot a_5 = a \cdot (ar^4) = a^2 r^4 = 28$                 (1)
2. $a_2 + a_4 = ar + ar^3 = ar(1+r^2) = 29$                (2)

Notice that $a_6 = a r^5$. Write

$$a r^5 = (a r) r^4.$$

Thus, if we find $a r$ and $r^4$, we can get $a_6$.

Step 1: Express $a$ in terms of $a r$.
Let $A = a r$.
From (1):

$$a^2 r^4 = \left(\frac{A}{r}\right)^2 r^4 = A^2 r^2 = 28 \quad \Longrightarrow \quad r^2 = \frac{28}{A^2}.$$

Step 2: Substitute in (2):

$$A(1 + r^2) = 29 \quad \Longrightarrow \quad A\left(1 + \frac{28}{A^2}\right) = 29.$$

This becomes:

$$A + \frac{28}{A} = 29.$$

Multiplying both sides by $A$:

$$A^2 - 29A + 28 = 0.$$

The quadratic factors as:

$$(A-1)(A-28)=0.$$

Since the GP is increasing ($r>1$) the common ratio must be $r>1$; taking $A=a r=1$ leads to a very large $r$ (as seen below) while the other possibility gives a decreasing sequence. Check using:

• If $A = 1$, then from $r^2=\frac{28}{1}$ we get $r = \sqrt{28}=2\sqrt7$ which is $>1$ and yields $a= \frac{1}{2\sqrt7}$.
• (The alternative $A=28$ gives $r^2=28/28^2=\frac1{28}<1$, a decreasing sequence.)

So, choose $A=1$.

Step 3: Now,

$$r^2=28 \quad \Longrightarrow \quad r^4=(r^2)^2 = (28)^2 =784.$$

Then,

$$a_6 = (a r) \, r^4 = 1 \cdot 784 = 784.$$