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x^-1, x^2, x^3, x^4, x^5, x^6, x^7

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by sanju09 » Tue Jan 12, 2010 3:07 am
In the sequence x^-1, x^2, x^3, x^4, x^5, x^6, x^7, the average of first three numbers is 13 and that of first four numbers is 30. What is the last number of the sequence?
(A) 128
(B) 343
(C) 729
(D) 2187
(E) 8384
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by rohan_vus » Tue Jan 12, 2010 3:48 am
Should be D

Sum of 1st 3 in seq = 39
Sum of 2st 4 in seq = 120..
So the 4th number in seq = 120 -39 = 81 ==> x^4 = 81..so x = 3 or -3 ..But ans choices doesnt talk abt any -ve values so take x = 3 ( +ve one)

X^7 = x^4 *x^3 = 81*27..

Thus last unit didgit should be 7.(We dont need even to carry out all multiplications) ..only naswer choice D gives that and hence the ans
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by sanju09 » Tue Jan 12, 2010 3:57 am
Should be D

Sum of 1st 3 in seq = 39
Sum of 2st 4 in seq = 120..
So the 4th number in seq = 120 -39 = 81 ==> x^4 = 81..so x = 3 or -3 ..But ans choices doesnt talk abt any -ve values so take x = 3 ( +ve one)

X^7 = x^4 *x^3 = 81*27..

Thus last unit didgit should be 7.(We dont need even to carry out all multiplications) ..only naswer choice D gives that and hence the ans


rohan's booze-filled elucidation says that all :D
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by rohan_vus » Tue Jan 12, 2010 4:05 am
Yes indeed :mrgreen:
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