Which term of the AP: 121, 117, 113, ..... is its first negative term?
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[Hint : Find n for an< 0]
Solution and Explanation
Given A.P. is \(121, 117, 113 ……\)
\(a = 121\) and \(d = 117 − 121 = −4\)
\(a_n = a + (n − 1) d\)
\(a_n = 121 + (n − 1) (−4)\)
\(a_n= 121 − 4n + 4\)
\(a_n= 125 − 4n \)
We have to find the first negative term of this A.P.
Therefore, \(an < 0\)
\(125 - 4n < 0\)
\(125 < 4n\)
\(n > \frac {125}{4}\)
\(n > 31.25\)
Therefore, 32nd term will be the first negative term of this A.P.
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