@bodhi_vriksha said:For k = 2.4 and r = 4, k/r = 0.6 =3/5For k = 10/3 and r = 2, k/r = 5/3 Team BV - Vineet
missed to visualize..thanks sir 
k^+47 = 16k-16 or,
k^2-16k+63 = 0
16+_/ 256 - 252 / 2 = 16+2/2 , 16-2/2 => 9+7 = 16 (a)??
@bodhi_vriksha said:Ok ...here goes the next one...The sum of all the possible values of integral k such that (k^2-2k) ^ (k^2+47) = (k^2-2k) ^ (16k-16) is(a) 16 (b) 17 (c) 18 (d) 19 (e) none of theseTeam BV - Vineet
@bodhi_vriksha said:Ok ...here goes the next one...The sum of all the possible values of integral k such that (k^2-2k) ^ (k^2+47) = (k^2-2k) ^ (16k-16) is(a) 16 (b) 17 (c) 18 (d) 19 (e) none of theseTeam BV - Vineet
19 ?
@bodhi_vriksha said:Ok ...here goes the next one...The sum of all the possible values of integral k such that (k^2-2k) ^ (k^2+47) = (k^2-2k) ^ (16k-16) is(a) 16 (b) 17 (c) 18 (d) 19 (e) none of theseTeam BV - Vineet
K takes value 7, 9, 1, 2
answer: d) 19
answer: d) 19
@bodhi_vriksha said:Ok ...here goes the next one...The sum of all the possible values of integral k such that (k^2-2k) ^ (k^2+47) = (k^2-2k) ^ (16k-16) is(a) 16 (b) 17 (c) 18 (d) 19 (e) none of theseTeam BV - Vineet
K = 0, 1, 2, 7 and 9
Sum = 19.. ?
@bodhi_vriksha said:Ok ...here goes the next one...The sum of all the possible values of integral k such that (k^2-2k) ^ (k^2+47) = (k^2-2k) ^ (16k-16) is(a) 16 (b) 17 (c) 18 (d) 19 (e) none of theseTeam BV - Vineet
possible values are -> 0+9+7+1+2=19
@bodhi_vriksha said:Ok ...here goes the next one...The sum of all the possible values of integral k such that (k^2-2k) ^ (k^2+47) = (k^2-2k) ^ (16k-16) is(a) 16 (b) 17 (c) 18 (d) 19 (e) none of theseTeam BV - Vineet
roots = 9 ,7 + 1+2 (satisfying the condition manually) = 19
Posted very late : 19 should be the ans
@bodhi_vriksha said:Ok ...here goes the next one...The sum of all the possible values of integral k such that (k^2-2k) ^ (k^2+47) = (k^2-2k) ^ (16k-16) is(a) 16 (b) 17 (c) 18 (d) 19 (e) none of theseTeam BV - Vineet
k = 9,7,2,1
9+7+2+1=19
my take 19
19 is correct indeed and the values of k are 9,7,2, and 1
Team BV - Vineet
@albiesriram said:
a2 -1 + a2 + a4 - 1 + a4 +..... = 137
2 ( a2 + a4 + a6 ... + a98 ) = 137 + 49 = 186
( a2 + a4 + ... + a98 ) = 93 ?
@bodhi_vriksha said:Ok ...here goes the next one...The sum of all the possible values of integral k such that (k^2-2k) ^ (k^2+47) = (k^2-2k) ^ (16k-16) is(a) 16 (b) 17 (c) 18 (d) 19 (e) none of theseTeam BV - Vineet
Equating the powers k^2 + 47 = 16k – 16, we get
k = 9 or 7.
Now equating the base k^2 – 2k to 0,1 and –1.
k^2 – 2k = 0⇒ k = 2.
k^2 – 2k = 1⇒ no integer solution possible.
k^2 – 2k = –1⇒ k = 1.
We get the values of k as 1, 2, 7 and 9.
So the sum of all the possible values of k is 19.
k = 9 or 7.
Now equating the base k^2 – 2k to 0,1 and –1.
k^2 – 2k = 0⇒ k = 2.
k^2 – 2k = 1⇒ no integer solution possible.
k^2 – 2k = –1⇒ k = 1.
We get the values of k as 1, 2, 7 and 9.
So the sum of all the possible values of k is 19.
@albiesriram said:
a1, a2, a2-a1, -a1, -a2, -(a2-a1) ...repeating...
Sum of 6 = 0
Hence, 2a2 - a1 = 1985
a2 - a1 = 1492
a2 = 493 and a1 = -999
Sum till 2001 terms = 2a2 = 986 ?