Question

product of 2 consecutive turms of the as 5,8,11 is 598 .find the position of the terms multiplied . dey ethenkilum malayalikalundo​

Answer 1

Given:- product of 2 consecutive terms of ap is 598. AP = 5,8,11.

To Find:- the positions of the terms multiplied.

Solution :-

Let assume that the position of the terms be n+1 and n+2.

So, According to our given scenario,

(a+ nd)[a + (n+ 1)d] = 598

⇒9n² + 39n - 558 = 0

⇒3n² + 13n - 186 = 0.......[divided by 3]

⇒3n² + 31n - 18n - 186 = 0

⇒n( 3n + 31 ) - 6( 3n + 31 ) = 0

⇒(n - 6)(3n + 31) = 0

Therefore, n = 6 or n = -31/3

and n = -31/3 is neglected here!

Hence, the positions of the terms multiplied are 7 and 8.

Answer 2

$\huge{\boxed{\rm{\red{Question}}}}$

Product of 2 consecutive turms of the as 5,8,11 is 598 .find the position of the terms multiplied .

$\huge\underline\mathfrak\red{Answer}$

${\bigstar}\large{\boxed{\sf{\pink{Given \: that}}}}$

• product product of two consecutive terms of AP is 598 .
• AP = 5 , 8 , 11

${\bigstar}\large{\boxed{\sf{\pink{To \: find}}}}$

• The position of the term multiplied.

${\bigstar}\large{\boxed{\sf{\pink{Solution}}}}$

• let assume that position of the terms will be n + 1 and n + 2.

So according to the given problem (a+nd)[a+(n+1)+d] = 598

➡ 9n² + 39n + 558 = 0

➡ 3n² + 13n + 186 = 0

➡ 3n² + 31n - 18n - 186 = 0

➡ n(3n + 31) - 6(3n +31) = 0

➡ (n-6) (3n+31) = 0

• Therefore , n = 6 or n = -31/3 nd n = -31/3 is neglected !

Hence, positions of the terms multiplied are 7 and 8.