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05 Jul 2026, 09:43
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T4: -1
T7: 12
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
63% (02:46) correct
37%
(03:18)
wrong
based on 671
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Not Attempted Yet
For positive values of \(n\), \(T_n\) is defined such that \(T_{n+2}-2T_n^2-T_{n+1}=0\), also given is \(T_5=2\). Select the values of \(T_4\) & \(T_7\) in the table such that they are jointly compatible with these conditions.
Make only two selections, one in each column.
Make only two selections, one in each column.
| T4 | T7 | Value |
| -8 | ||
| -2 | ||
| -1 | ||
| 2 | ||
| 12 |
ShowHide Answer
Official Answer
T4: -1
T7: 12
05 Jul 2026, 09:58
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For positive values of \(n\), \(T_n\) is defined such that \(T_{n+2}-2T_n^2-T_{n+1}=0\), also given is \(T_5=2\). Select the values of \(T_4\) & \(T_7\) in the table such that they are jointly compatible with these conditions.
Make only two selections, one in each column.
Use \(n=4\):
\(T_6=2T_4^2+T_5\)
Since \(T_5=2\):
\(T_6=2T_4^2+2\)
Use \(n=5\):
\(T_7=2T_5^2+T_6\)
\(T_7=2*2^2+(2T_4^2+2)\)
\(T_7=2T_4^2+10\)
So, the selected pair must satisfy \(T_7=2T_4^2+10\).
Only the pair \(T_4=-1\) and \(T_7=12\) satisfies this equation:
\(T_7=2(-1)^2+10=12\)
So the correct selections are:
\(T_4=-1\)
\(T_7=12\)
GMAT-Club-Forum-7cwlmyy6.png [ 8.41 KiB | Viewed 2597 times ]
Make only two selections, one in each column.
Use \(n=4\):
\(T_6=2T_4^2+T_5\)
Since \(T_5=2\):
\(T_6=2T_4^2+2\)
Use \(n=5\):
\(T_7=2T_5^2+T_6\)
\(T_7=2*2^2+(2T_4^2+2)\)
\(T_7=2T_4^2+10\)
So, the selected pair must satisfy \(T_7=2T_4^2+10\).
Only the pair \(T_4=-1\) and \(T_7=12\) satisfies this equation:
\(T_7=2(-1)^2+10=12\)
So the correct selections are:
\(T_4=-1\)
\(T_7=12\)
Attachment:
GMAT-Club-Forum-7cwlmyy6.png [ 8.41 KiB | Viewed 2597 times ]
General Discussion
14 Jul 2026, 09:38
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Cneltonmp
For positive values of \(n\), \(T_n\) is defined such that \(T_{n+2}-2T_n^2-T_{n+1}=0\), also given is \(T_5=2\). Select the values of \(T_4\) & \(T_7\) in the table such that they are jointly compatible with these conditions.
\(T_{n+2}-2T_n^2-T_{n+1}=0\)
n = 5
\(T_7-2T_5^2-T_6 = T_7 - 2*2^2 - T_6=0\)
\(T_7 = 8 + T_6\) (1)
n = 4
\(T_6-2T_4^2-T_5= T_6-2T_4^2-2 = 0\)
\(T_6 = 2T_4^2 + 2 \) (2)
Putting value of T_6 in (1)
\(T_7 = 8 + T_6 = 8 + 2 + 2T_4^2 = 10 + 2T_4^2\)
\(T_4 = -1 ; T_7 = 10 + 2(-1)^2 = 12\)
For positive values of \(n\), \(T_n\) is defined such that \(T_{n+2}-2T_n^2-T_{n+1}=0\), also given is \(T_5=2\). Select the values of \(T_4\) & \(T_7\) in the table such that they are jointly compatible with these conditions.
\(T_{n+2}-2T_n^2-T_{n+1}=0\)
n = 5
\(T_7-2T_5^2-T_6 = T_7 - 2*2^2 - T_6=0\)
\(T_7 = 8 + T_6\) (1)
n = 4
\(T_6-2T_4^2-T_5= T_6-2T_4^2-2 = 0\)
\(T_6 = 2T_4^2 + 2 \) (2)
Putting value of T_6 in (1)
\(T_7 = 8 + T_6 = 8 + 2 + 2T_4^2 = 10 + 2T_4^2\)
\(T_4 = -1 ; T_7 = 10 + 2(-1)^2 = 12\)
| \(T_4\) | \(T_7\) |
| -1 | 12 |
