Question

The product of two concecutive terms of an arithemetic sequence 5,8,11....is 598. Find the position of the two terms?

Answer 1

Given :-

◉ An AP which has infinite terms.

5,8,11,...

From the AP, we have

• Common difference, d = 3
• First term, a = 5

◉ Product of two consecutive terms of the given AP is 598.

To Find :-

◉ The two terms

Solution :-

Let the nth term of the AP be one of the two numbers then the other one would be n - 1 term.

Now, It is given that the product of nth and n - 1 term is 598, so we have

⇒ aₙ × aₙ₋₁ = 598

⇒ (a + (n - 1)d)(a + (n - 1 - 1)d) = 598

⇒ (5 + (n - 1)3)(5 + (n - 2)3) = 598

⇒ (5 + 3n - 3)(5 + 3n - 6) = 598

⇒ 25 + 15n - 30 + 15n + 9n² - 18n - 15 - 9n + 18= 598

⇒ 9n² - 3n = 600

⇒ 3n² - n = 200

⇒ 3n² - n - 200 = 0

Comparing the given quadratic with the standard form of a quadratic equation i.e., ax² + bx + c = 0, we get

• a = 3 , b = -1 , c = -200

Using the quadratic formula,

⇒ n = { -b ± √(b² - 4ac) } / 2a

⇒ n = (1 ± √2401) / 6

⇒ n = (1 ± 49) / 6

Case 1

Taking positive sign,

⇒ n = (1 + 49) / 6

⇒ n = 50 / 6

⇒ n = 25/3

But, position of a term can't be in decimal hence we neglect this value.

Case 2

Taking negative sign,

⇒ n = (1 - 49) / 6

⇒ n = -48/6

⇒ n = -8

Now, since the position of both the terms are negative which means negative of the terms.

Also, When even numbers of negative number are multiplied then the result will be a positive number.

So, The terms are 8th term and 7th term.

Let us verify,

⇒ -a₈ × -a₇ = 598

⇒ a₈ × a₇ = 598

⇒ (a + 7d)(a + 6d) = 598

⇒ (5 + 7×3)(5 + 6×3) = 598

⇒ (5 + 21)(5 + 18) = 598

⇒ 26 × 23 = 598

⇒ 598 = 598

Hence, Verified.