Question

6. Check whether, –150 is a term of the AP : 11, 8, 5, 2 . . .​

Answer 1

$\boxed{\fcolorbox{red}{blue}{Here,a=11,d=8−11=−3}}$

Let −150 be the nth term of the given A.P.

We know that,an

=a+(n−1)d

−150=11+(n−1)×−3

−150=11−3n+3

$\boxed{\fcolorbox{red}{lime}{(−150)+(−14)=−3n}}$

$\boxed{\fcolorbox{red}{aqua}{−3n=−164}}$

$\boxed{\fcolorbox{red}{lime}{3n=164}}$

n3= $\boxed{\fcolorbox{red}{pink}{164}}$

$\boxed{\fcolorbox{red}{blue}{So,it is clear that n is not an integer.}}$

Therefore,−150 is not a term of the given A.P.

$<marquee behaviour-move> <font color="lime"> <h1>@sɴᴇʜᴀsʜɪsʜ</ ht> </marquee>$

Answer 2

For the given, A.P. 11, 8, 5, 2, …

First term, a = 11

Common difference, d = a2 - a1 = 8 - 11 = -3

Let -150 be the nth term of this A.P.

As we know, for an A.P.,

an = a + (n - 1) d

-150 = 11 + (n - 1)(-3)

-150 = 11 - 3n + 3

-164 = -3n

n = 164/3

Clearly, n is not an integer but a fraction.

Therefore:

-150 is not a term of this A.P.

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