Let $$a_{1},a_{2},a_{3},a_{4}$$ be an A.P. of four terms such that each term of the A.P. and its common difference $$l$$ are integers. If $$a_{1} +a_{2}+a_{3}+a_{4}= 48$$ and $$a_{1} a_{2}a_{3}a_{4} + l^{4} = 361,$$ then the largest term of the A.P. is equal to

Here is the step by step solution using your requested method.

Let the four terms of the arithmetic progression be $$a - 3d$$, $$a - d$$, $$a + d$$, and $$a + 3d$$.

Observe that the common difference is the second term minus the first term, which is $$2d$$. The problem states the common difference is $$l$$, which means $$l = 2d$$.

The sum of all four terms is given as 48.

$$(a - 3d) + (a - d) + (a + d) + (a + 3d) = 48,$$ $$4a = 48,$$ $$a = 12.$$

Next, use the second condition given in the problem. The product of the four terms plus the common difference to the power of four equals 361.

$$(a - 3d)(a - d)(a + d)(a + 3d) + l^4 = 361$$

Group the terms using the difference of squares formula.

$$(a^2 - 9d^2)(a^2 - d^2) + (2d)^4 = 361$$

Substitute the value of $a$ which is 12.

$$(144 - 9d^2)(144 - d^2) + 16d^4 = 361$$

Expand the brackets.

$$20736 - 144d^2 - 1296d^2 + 9d^4 + 16d^4 = 361$$

Combine the similar terms.

$$25d^4 - 1440d^2 + 20736 = 361 $$ which gives $$ 25d^4 - 1440d^2 + 20375 = 0$$

Divide the entire equation by 5.

$$5d^4 - 288d^2 + 4075 = 0$$

Apply the quadratic formula to solve for $$d^2$$.

$$d^2 = \frac{288 \pm \sqrt{288^2 - 4 \cdot 5 \cdot 4075}}{2 \cdot 5}$$ $$d^2 = \frac{288 \pm \sqrt{82944 - 81500}}{10}$$ $$d^2 = \frac{288 \pm \sqrt{1444}}{10}$$ $$d^2 = \frac{288 \pm 38}{10}$$

This provides two possible values for $$d^2$$.

• First value: $$d^2 = \frac{326}{10} = 32.6$$
• Second value: $$d^2 = \frac{250}{10} = 25$$

The problem specifies that the common difference $$l$$ and all terms are integers. If $$d^2$$ is 32.6, then $$d$$ and subsequently $$l$$ will not be integers. Therefore, we select the second case where $$d^2 = 25$$.

This means $$d = 5$$ or $$d = -5$$.

Let us find the terms using $$d = 5$$ and $$a = 12$$.

• First term: $$12 - 3(5) = -3$$
• Second term: $$12 - 5 = 7$$
• Third term: $$12 + 5 = 17$$
• Fourth term: $$12 + 3(5) = 27$$

The terms of the progression are -3, 7, 17, and 27. The largest term among these is 27.