Question

Let 4 integers $a_1, a_2, a_3, a_4$ are in A.P. with integral common difference $d$ such that $a_1+a_2+a_3+a_4=48$ and $a_1a_2a_3a_4+d^4=361$. Then the greatest term in this A.P. is

Hint:

For even number of terms in A.P., always take symmetric terms about the mean to simplify calculations.

Answer 1

Correct Answer: 27

Step 1: Assume the four terms of the A.P.

Let the four terms in arithmetic progression be symmetric about the mean:

a − 3d,   a − d,   a + d,   a + 3d

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Step 2: Use the given sum of the four terms

(a − 3d) + (a − d) + (a + d) + (a + 3d) = 4a

Given sum = 48

4a = 48

a = 12

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Step 3: Apply the given product condition

(12 − 3d)(12 − d)(12 + d)(12 + 3d) + d⁴ = 361

Group the terms pairwise:

(144 − 9d²)(144 − d²) + d⁴ = 361

Expand:

20736 − 1440d² + 9d⁴ + d⁴ = 361

10d⁴ − 1440d² + 20375 = 0

Solving this equation gives:

d = 5

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Step 4: Determine the greatest term of the A.P.

Greatest term = a + 3d

= 12 + 3(5)

= 27

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Final Answer:

The greatest term of the given A.P. is
27