Question

If $a_1, a_2, a_3, a_4, a_5$ are distinct positive terms in AP having common difference d, then -

• A. $5a_3^2 > 4d^2$
• B. sum of all terms = $5a_3$
• C. $a_1 + 5a_5, 3a_3, 2a_2 + 4a_4$ are in A.P.
• D. $a_1a_5 < a_2a_4$

Answer 1

Answer: (A), (B), (D)

Options 1, 2, and 4 are true.

Answer 2

Let the terms of the AP be expressed in terms of the middle term $a_3$ and common difference $d$:

$$\begin{aligned} a_1 &= a_3-2d,\\[1mm] a_2 &= a_3-d,\\[1mm] a_3 &= a_3,\\[1mm] a_4 &= a_3+d,\\[1mm] a_5 &= a_3+2d. \end{aligned}$$

1. Option 1: $5a_3^2 > 4d^2$
   Since all terms are positive and $a_1 = a_3-2d>0$ (which implies $a_3>2d$) we have:

   $$a_3^2 > 4d^2/ (some\ constant)$$

   More precisely, reducing gives:

   $$5a_3^2 > 4d^2 \quad \Longleftrightarrow \quad a_3^2 > \frac{4}{5}d^2.$$

   Since $a_3 > 2d$ (with $d>0$) it follows that:

   $$a_3^2 > 4d^2 \quad \text{and hence certainly} \quad 5a_3^2 > 4d^2.$$

   Thus, Option 1 is true.

2. Option 2: Sum of all terms = $5a_3$
   Calculate the sum:

   $$\begin{aligned} S &= (a_3-2d)+(a_3-d)+a_3+(a_3+d)+(a_3+2d)\\[1mm] &= 5a_3 + (-2d-d+d+2d)\\[1mm] &= 5a_3. \end{aligned}$$

   Hence, Option 2 is true.

3. Option 3: $a_1+5a_5,\; 3a_3,\; 2a_2+4a_4$ are in AP
   Evaluate each expression:

   $$\begin{aligned} T_1 &= a_1+5a_5 = (a_3-2d)+5(a_3+2d)=6a_3+8d,\\[1mm] T_2 &= 3a_3,\\[1mm] T_3 &= 2a_2+4a_4 = 2(a_3-d)+4(a_3+d)=6a_3+2d. \end{aligned}$$

   For these to be in AP, the condition is:

   $$2T_2 = T_1+T_3.$$

   Substitute:

   $$2(3a_3)=6a_3 \quad \text{and} \quad T_1+T_3 = (6a_3+8d)+(6a_3+2d)=12a_3+10d.$$

   For equality:

   $$6a_3=12a_3+10d \quad \Longrightarrow \quad -6a_3=10d,$$

   which gives a contradiction because $a_3$ and $d$ are positive.
   Thus, Option 3 is false.

4. Option 4: $a_1a_5 < a_2a_4$
   Compute:

   $$\begin{aligned} a_1a_5 &= (a_3-2d)(a_3+2d)=a_3^2-4d^2,\\[1mm] a_2a_4 &= (a_3-d)(a_3+d)=a_3^2-d^2. \end{aligned}$$

   Hence,

   $$a_1a_5 < a_2a_4 \iff a_3^2-4d^2 < a_3^2-d^2 \iff -4d^2< -d^2 \iff 4d^2>d^2,$$

   which is true for $d\neq0$.
   Thus, Option 4 is true.