Question

Is 667 a term of an AP 11 , 18 , 25 , ...?

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Answer 1

$\huge\mathfrak{Heya\: Mate}$

$\huge\bold{Solution}$

Given:-

AP is 11 , 18 , 25....

Here , a = 11 , d = 18 - 11 = 9 and a(n) = 667

:- a + (n - 1)d = a(n)

11 + (n - 1)7 = 667

=> 11 + 7n - 7 = 667

=> 7n + 4 = 667

=> 7n = 663

=> n = 663/7

which is not a whole number.

Hence ,it is not term of an AP

Hope it helps you Out :)

Answer 2

let Tn =667 ;a=11;d=18-11=7

so by formula

Tn =a+(n+1)d

667=11+(n-1)×7

667=11+7n-7

667=4+7n

667-4=7n

663=7n

n=663/7

n=94.71

so we know that n can neither be in decimal form

so 667 is not the term of this a.p.

hope it will really help

by

Ankita ❤️❤️