JEE Mathematics 2023 Question with Solution

Given: $a_1 = 8$ and $a_1, a_2, a_3, \dots, a_n$ is an A.P. The sum of the first four terms is $50$ and the sum of the last four terms is $170$.

Find: The product of the middle two terms.

For the first four terms,

$$8 + (8 + d) + (8 + 2d) + (8 + 3d) = 50$$ $$32 + 6d = 50$$ $$6d = 18$$ $$d = 3$$

Now for the last four terms,

$$(8 + (n-4)d) + (8 + (n-3)d) + (8 + (n-2)d) + (8 + (n-1)d) = 170$$

Substituting $d = 3$,

$$4(8) + (4n - 10)3 = 170$$ $$32 + 12n - 30 = 170$$ $$12n = 168$$ $$n = 14$$

Since there are $14$ terms, the middle two terms are $a_7$ and $a_8$.

$$a_7 = 8 + 6d = 8 + 18 = 26$$ $$a_8 = 8 + 7d = 8 + 21 = 29$$

Therefore,

$$a_7 \cdot a_8 = 26 \times 29 = 754$$

So, the product of the middle two terms is $754$.

Given: The first four terms sum to $50$ and the last four terms sum to $170$.

Find: The product of the middle two terms.

From the first four terms,

$$32 + 6d = 50$$

so

$$d = 3$$

For the last four terms,

$$32 + (4n - 10)d = 170$$

With $d = 3$,

$$32 + (4n - 10)3 = 170$$

which gives

$$n = 14$$

Hence the middle terms are the $7^{\text{th}}$ and $8^{\text{th}}$ terms:

$$a_7 = 8 + 6 \cdot 3 = 26, \quad a_8 = 8 + 7 \cdot 3 = 29$$

Therefore, the required product is $26 \times 29 = 754$.