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ÂŤ2 A - q Al -2 L [H ) AL M. ,,d.lfirat ./WIJAL Ao | |
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Answer» Given:first term(a)= 121common difference (d)= 117- 121 = -4∵ n th term of an AP an= a + (n – 1)d⇒121+(n-1)×(-4)⇒121-4n+4⇒12+4-4n⇒125 -4nan= 125 -4n For first negative term , an <0⇒ 125-4n<0⇒125<4n⇒4n>125⇒n>125/4⇒n> 31 1/4 least integral value of n= 32 Hence, 32nd term of the given AP is the first negative term. |
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