A) 16
B) 8
C) 4
D) 2
Correct Answer: B
Solution :
Let a, b, c, d be four numbers of the sequence. Now, according to the question \[{{b}^{2}}=ac\] and c - b = 6 and a - c = 6 Also, given\[\] \[\therefore \]\[{{b}^{2}}=ac\Rightarrow {{b}^{2}}=a\left[ \frac{a+b}{2} \right]\]\[(\because 2c=a+b)\] \[\Rightarrow \]\[{{a}^{2}}-2{{b}^{2}}+ab=0\] Now,\[c-b=6\]and\[a-c=6,\]gives\[a-b=12\] \[\Rightarrow \]\[b=a-12\] \[\therefore \]\[{{a}^{2}}-2{{b}^{2}}+ab=0\] \[\Rightarrow \]\[{{a}^{2}}-2{{(a-12)}^{2}}+a(a-12)=0\] \[\Rightarrow \]\[{{a}^{2}}-2{{a}^{2}}-288+48a+{{a}^{2}}-12a=0\] \[\Rightarrow \]\[36a=288\]\[\Rightarrow \]\[a=8\] Hence, last term is d= a = 8.You need to login to perform this action.
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