Question

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Answer 1

Answer:

ANSWER

The given sequence is an AP in which first term a=122 and common difference d=−6. We have to find value of n for which t

n

is first negative number. Then,

t

n

<0

⇒a+(n−1)d<0

⇒122+(n−1)×−6<0

⇒128−6n<0

⇒6n>128⇒n>21

6

2

Since, 22 is the natural number just greater than 21

6

2

. So, n=22

Thus, t

22

term of the given sequence is first negative term and given by

t

22

= a+(n-1)d = 122+(22-1)(-6) = -4