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Re: Algebra: PS Diagnostic Test
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07 Oct 2022, 16:52
39. The infinite sequence of integers \(a_1\), \(a_2\), …, \(a_n\), … is such that \(a_1 = -17\) and \(a_{n-1} - 7 < a_n < a_{n-1} - 2\) for \(n > 1\). If \(a_x=-53\), then how many different values can x take?
A. 1
B. 4
C. 5
D. 6
E. 7
Solution:
a1=-17 then as per the inequality of the question stem
-24 < a2 < -19 => a2 = {-23,-22,-21,-20} Because sequence consists of integers only
Now, if you consider a3 will have 4 different inequality ranges based on the value of a2 which will be:
-30 < a3 < -25, -29 < a3 < -24, -28 < a3 < -23, -27 < a3 < -22
Combining the entire inequality will yield: -30 < a3 < -22 => a3 = {-29,-28,-27,-26,-25,-24,-23} All inclusive
Using this logic, we can see that the inequality range of any term of the sequence can be determined by taking adding -6 to the highest term and -3 to the lowest term
And the inclusive integers of the term can be determined by adding 1 to the lowest term of the inequality and subtracting 1 from the highest term of the inequality
So, based on this logic, we can determine:
a4 = {-35 to -26} All inclusive
a5 = {-41 to -29} All inclusive
a6 = {-47 to -32} All inclusive
a7 = {-53 to -35} All inclusive : x=7 can = -53 (1)
a8 = {-59 to -38} All inclusive : x=8 can = -53 (2)
a9 = {-65 to -41} All inclusive : x=9 can = -53 (3)
a10 = {-71 to -44} All inclusive : x=10 can = -53 (4)
a11 = {-77 to -47} All inclusive : x=11 can = -53 (5)
a12 = {-83 to -50} All inclusive : x=12 can = -53 (6)
a13 = {-89 to -53} All inclusive : x=13 can = -53 (7)
Now since -53 is the last term of a13, we know that following terms will be less than -53 in their entire range so no use going on from here
We have 7 possible terms where the term can = -53
Answer - E