A 4 A_4 tells you A 1 A_1

Algebra Level 1

2 2 4 4 1 1 3 3

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6 solutions

Ramji Varadarajan
Mar 15, 2014

From given condition, a=_, b=3+2 a ,c=3+2 b ,d=3+2*c such that 4th number (d) is 37

For d=3+2 c, substitute d=37 we get , 37=3+2 c , 37-3=2*c, 34/2=c , c=17

For c=3+2 b, substitute c=17 we get , 17=3+2 b , 17-3=2*b, 14/2=b , b=7

For b=3+2 a, substitute b=7 we get , 7=3+2 a , 7-3=2*a, 4/2=a , a=2

Hence, the Final Solution (a=2).

Hence, the First number is 2.

Justin Ruaya
Dec 5, 2016

NOTEPAD MASTER RACE

Let first be a, 2nd is 2 * a + 3, 3rd is 2 * (2 * a + 3) + 3, 4th is 2 * ( 2 * (2 * a + 3) + 3) + 3, So, 2 * ( 2 * (2 * a + 3) + 3) + 3 = 37, which gives a = 2

4th = 37, 3rd = (37-3)/2=17, 2nd = (17-3)/2=7, 1st = (7-3)/2=2

Junno Martinez
Nov 30, 2016

Easier to work backwards!

SOLUTION: If every number after the first is (3 + 2 * (n) ), where n is the previous number in the sequence, and 37 is the fourth number in that sequence, then we can write the equation for the 4th number, 37; 3 + 2 * (n) = 37 => n = 17, which is the third number. Then 3 + 2 * (n) = 17 => n = 7, which is the second number. Then 3 + 2 * (n) = 7 => n = 2, which is the first number.

Done!

Prajwal Kavad
Apr 8, 2014

Let first term be a, 2nd is 2 * a + 3, 3rd is 2 * (2 * a + 3) + 3, 4th is 2 * ( 2 * (2 * a + 3) + 3) + 3, So, 2 * ( 2 * (2 * a + 3) + 3) + 3 = 37, which gives a = 2

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