A) 4th
B) 5th
C) 6th
D) 7th
Correct Answer: B
Solution :
(b): Let a be the first term and d the common difference of an AP. Also, let kth term of AP be zero. \[\therefore {{a}_{k}}=a+(k-1)d\,\,\therefore {{a}_{3}}=a+2d=4\] \[[\because {{a}_{3}}=4(given)]......(i)\] And \[[\because {{a}_{9}}=-8(given)]......(ii)\] On subtracting Eq. (i) from Eq. (ii), we get \[6d=-12\Rightarrow d=\frac{-12}{6}=-2\] \[\therefore \] From Eq. (i), \[a+2\times (-2)=4\] \[\Rightarrow a-4=4\,\,\,\,\Rightarrow a=4+4=8\] kth term: \[{{a}_{k}}=0\Rightarrow a+(k-1)d=0\,\,\] \[\Rightarrow 8+(k-1)(-2)=8\,\,\,\Rightarrow (k-1)(-2)=-8\] \[\Rightarrow k-1=\frac{-8}{-2}=4\]\[\Rightarrow k=4+1=5\] Hence, 5th term of this AP is zero.You need to login to perform this action.
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