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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
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