Question

product of two consecutive terms of an arithmetic 5 8 11 etc is 598 find the position of terms​

Answer 1

Finding the positing of the term

Answer: Required positions are 7th and 8th.

Explanation:

Given A P is 5 , 8 , 1 1

first term = a = 5    

common difference d = 11 - 8 = 8 - 5 = 3

let positing of the first required term = n

so position of next required consecutive term = n + 1

nth term $a_{n}$ = a + (n - 1 ) d = 5 + (n - 1) 3 = 3n + 2

(n+1)th term $a_{(n+1)}$ = a + (( n + 1) - 1 ) d = 5 + (n ) 3 = 3n + 5

given that product of two consecutive term is 598

⇒ (3n + 2) ( 3n +  5 ) = 598

⇒ 9n² + 15n + 6n + 10 = 598

⇒ 9n² + 21n + 10 - 598 = 0

⇒ 9n² + 21n  - 588 = 0

⇒ 3 ( 3n² + 7n - 196) = 0

⇒ 3n² + 7n - 196 = 0

using quadratic formula to solve above quadratic equation

n = (-b ± √(b²-4ac)/2a

in above case a = 3 , b = 7 and c = -196

⇒n = (-7 ± √(7²-4× 3 ×(-196) )/6

⇒n =(-7 ± 49 )/6

⇒ n = (-7 + 49 )/6  or n = (-7 - 49 )/6

⇒ n = 42/6 or n =  -56/6

as n cannot be negative , considering only positive term

that means n = 42/6 = 7

position of one term is n = 7

and position of consecutive term is n + 1 = 7 + 1 = 8

we can verify by finding 7th and 8th term and multiplying them

$a_{7}$ = a + (7 -1) d = 5 + 6 x 3 = 23

$a_{8}$ = $a_{7}$ + d = 23 + 3 =  26

product = 23 x 26 = 598

Hence required position are 7th and 8th.