An arithmetic progression of consecutive odd positive integers has 12

Question

An arithmetic progression of consecutive odd positive integers has 12 terms. If the sum of the squares of the first and last terms is 4850, what is the product of the first and last terms?

A. 1248
B. 1763
C. 2183
D. 2265
E. 2340

KrishnakumarKA1 · Answer

let the numbers be a, a+2, a+4, .........a+22

according to the question a^2 +(a +22)^2 = 4850

or 2 a^2 + 44a +484 = 4850
or 2 a^2 +44a - 4366 = 0
or a^2 + 22a - 2183 = 0

solving we get a = 37 or -59
since the progression consists of positive integer, a =37

a(a+22) =37*59 = 2183

Option C

jedit · Answer

I managed to get to this equation, but how do you quickly solve this within a minute or two. Could you share the quick method you used. The numbers are far off, but I still couldn't use brain math to work out multiples of two odd numbers 22 apart. Thank you.

KrishnakumarKA1 · Answer

Calculate D = b^2 - 4ac = 484 + 4*2183 = 9216 since we know a is integer, \sqrt{9216} will be an integer also we can predict the range of 9216 ie 8100(sq of 90)<9216<10000(sq of 100) so 9216 will be the square of number between 90-100 In this range for unit digit to be 6 for square, the number should have unit digit as 4 or 6 therefore the number should be either 94 or 96 now there are two ways 1# either calculate the square for these two number 2# calculate the square for 95 = 9025 ( use shortcut for calculating square for number ending with 5) since D is greater than square of 95 \sqrt{d} will be 96. Now simply use the conventional formula I hope that clears your doubt

gracie · Answer

another approach:
let x and y=first and last terms
we know that the range (y-x) of the progression=22
xy= /2
xy=(4850-484)/2
xy=2183

khrystal · Answer

let the numbers be a, a+2, a+4, .........a+22
according to the question a^2 +(a +22)^2 = 4850
or 2 a^2 + 44a +484 = 4850
or 2 a^2 +44a - 4366 = 0
or a^2 + 22a = 2183 ...... lets call this as equation (1)
there is no need to solve that equation further because question asked for the product of first and last term , which would be 
 a*(a+22)
or a^2 + 22a , which is equal to 2183 from the equation(1)

Therefore the answer will be C

ScottTargetTestPrep · Answer

We can let x = the first (or smallest) odd integer, so x + 22 = the last (or largest) odd integer. Note that we need to find the value of x(x + 22) = x^2 + 22x. We can create the following equation and manipulate it to obtain the answer quickly:

x^2 + (x + 22)^2 = 4850

x^2 + x^2 + 44x + 484 = 4850

2x^2 + 44x = 4366

Divide both sides of the above equation by 2. The result is the answer to the question posed.

x^2 + 22x = 2183

Answer: C

IanStewart · Answer

If x is the median of the list, then x-11 and x+11 are the smallest and largest values. We know (x-11)^2 + (x + 11)^2 = 4850, so 2x^2 + 242 = 4850, and x^2 = 2425 - 121 = 2304. Since the question asks us to find (x-11)(x+11) = x^2 - 121, the answer is 2304 - 121 = 2183.

Using this approach, it's also easy to find the exact list if you need it: since x^2 = 2304, and since x is an integer, then the only plausible value of x is 48 (we know x isn't far below 50, since 50^2 = 2500, and x^2 must have a units digit of 4). So the smallest and largest values in the list are x - 11 = 37 and x + 11 = 59.

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An arithmetic progression of consecutive odd positive integers has 12

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An arithmetic progression of consecutive odd positive integers has 12

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