A) yes
B) no
C) - 148 is not a term of this A.P.
D) - 149 is a term of this AP.
Correct Answer: B
Solution :
(b): Here, \[{{a}_{1}}=11,{{a}_{2}}=8,{{a}_{3}}=5,{{a}_{4}}=2\] \[{{a}_{2}}-{{a}_{1}}=8-11=-3;{{a}_{3}}-{{a}_{2}}=5-8=-3\]; \[{{a}_{4}}-{{a}_{3}}=2-5=-3\] The successive difference of terms is constant. So, numbers are in AP. Common difference d =? 3. Let ? 150 be the nth term of the given AP. We know that, the nth term of an AP is \[{{a}_{n}}={{a}_{1}}+(n-1)d\] where n = natural number. \[\Rightarrow -150=11+(n-1)(-3)\] \[\Rightarrow -3(n-1)=-150-11=-161\] \[\Rightarrow n-1=\frac{161}{3}\] \[\Rightarrow n=\frac{161}{3}+1=\frac{164}{3}\,\,\,\therefore n\notin N\] So, our assumption was wrong and ? 150 is not a term of the given AP.
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