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

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

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

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

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

Product of two consecutive terms of AP is 598 .

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

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

The position of the term multiplied.

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.

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Answer 2

Answer:

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

your answer is in attachment....^_^

Step-by-step explanation:

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