THE ARITHMETIC PROGRESSION 2,5,8,11 AND 3,10,17,24... HAVE SOME COMMAN VALUES.WHAT IS THE LARGEST VALUE LESS THEN 500 THAT THEY HAVE IN COMMAN?
To find the largest value less than 500 that two arithmetic progressions have in common, we need to determine the general formula for both progressions and then find their intersection.
First, let's determine the general formula for each arithmetic progression.
For the progression 2, 5, 8, 11, the common difference is 3. We can use the formula:
an = a1 + (n - 1)d
where:
an = nth term
a1 = first term
d = common difference
Substituting the given values into the formula, we can find the nth term for the first progression:
a1 = 2
d = 3
an = 2 + (n - 1)3
an = 3n - 1
Now, let's determine the general formula for the second arithmetic progression.
For the progression 3, 10, 17, 24, the common difference is 7. Using the same formula, we have:
a1 = 3
d = 7
an = 3 + (n - 1)7
an = 7n - 4
To find the common values between the two progressions, we need to solve the equation:
3n - 1 = 7n - 4
Simplifying, we get:
4n - 3n = 4 - 1
n = 3
Now, substituting n = 3 into either equation, we can find the common value:
an = 3n - 1
a3 = 3(3) - 1
a3 = 8
Therefore, the common value less than 500 between the two progressions is 8.
the nth term in the first:
an= 2+3n
the mth term in the second:
yn= 3+7m
if an=ym
an-ym=0
3n-7m=1
3n=7m+1
this means above that 7m +1 is divisible by 3 3,6,9,12,15 (see that m=2 works, and n=5).
Lets look at the divisibility rule at the high end.
7m+1 divisible by three; 501, 498, 495, 492, .... Now, which of those minus one are divisible by 7?
489, 486, 483, ...
Another method is multiplication by 3. A number of the form 10x + y has the same remainder when divided by 7 as 3x + y. So get the leftmost digit of the original number, multiply by 3, add the next digit, get the remainder by 7, and continue from the beginning: multiply by 3, add the next digit, etc. For example, the number 371: 3×3 + 7 = 16 remainder 2, and 2×3 + 1 = 7. This method can be used to find the remainder of division by 7
So here subtract one, then test.
489, 486, 483, ...
3*4+8=20 r 6, 3*6+8=24 not divisible
485: 3*4+8=2r6, 3*6+5=23 not divisible
482: 3*4+*=2r6, 6*3+2=20 not divisible
479: in my head, 2r5, 15+9=24 not div
476: 2r5, 15+6=21 divisible...
so 7y+1=477, or 7y=476
3n-7y=1 or 3n=477
solve for n, m and you have it.