The first term, the mean, the last term, and the sum of an arithmetic

Question

The first term, the mean, the last term, and the sum of an arithmetic progression of nine terms are all perfect squares. What is the sum of these four squares?

A. 289
B. 300
C. 376
D. 415
E. 487

poojachaudhary0305 · Accepted Answer

Let the sequence be a-4d, a-3d, a-2d, a-d, a, a+d, a+2d, a+3d, a+4d

First term + Last term + mean + sum of terms = a-4d + a+4d + a + 9a = 12a

The only term divisible by 12 is 300

gracie · Answer

hi,

let x=first term
y=last term
sum of four squares=x+y+(x+y)/2+ 9➡
6(x+y)
300 is only multiple of 6
x=1, y=49, mean=25, sum=225
1+49+25+225=300
B

I hope this helps.

jfranciscocuencag · Answer

gracie

Hello!

Could you please provide an answer?

Regards!

JimitMehta · Answer

Hi Grace,

Can you please explain why did you multiply 9 with the mean?

Regards,
JM

gracie · Answer

hi, JM

because there are 9 terms in the sequence.
9 terms*mean (25)=sum of total sequence (225)

I hope this helps,
gracie

Jayvnandu1991 · Answer

I solved it this way.

take the first number as 6, 2nd number as 10, then the mean becomes 8   
the addition of the squares of all three gives 200. subtracting it from the answer choices given, Only 300 yields a perfect square i.e. 100 
Ans. B

jfranciscocuencag · Answer

Hello  gracie !

Could you please help me with the following?

How could it be 49 the last term if its an arithmetic progression of 9 terms?

Shouldn't be:

nth = a + d(n-1)

1(a),4,9,16,25(mean),36,49,64,81... 9 terms in total.

So

a = 1
nth = 81
Mean = 25
Sum of terms = 285

Mmmm now that I am writing it I guess is wrong cuz it should be a geometric progression, isn't it?

foryearss · Answer

guys , let the arithmetic progression of nine terms be x
so what is the sum of an arithmetic progression of nine terms ? 8x or 9x ?

Kinshook · Answer

Given: The first term, the mean, the last term, and the sum of an arithmetic progression of nine terms are all perfect squares. 
Asked: What is the sum of these four squares?

Let a be first term, d be common difference and n be number of terms of an arithmetic progression

First term a is a perfect square
Mean = (l+a)/2 is a perfect square
Last term l = a + (n-1)d; is a perfect square
Sum of 9 terms = (9/2) (l+a); is a perfect square

a + (l+a)/2 + l + 9(l+a)/2 = 6a + 6l = 6(a+l)
300 is the only multiple of 6
a + l = 50;

Let a be 1 and l be 49; (n-1)d = 48; 
Mean = 25
Sum of 9 terms = 9*25 = 225
6(a+l) = 6*50 = 300

IMO B

Archit3110 · Answer

sum of squares = 1/6 * ( n) *( n+1)*(2n+1)
n=4
sum of squares = 1/6 *4*5*9
sum of squares = 30
so answer option has to be multiple of 30
which is only option B 300

Fdambro294 · Answer

When you use the Sum of Perfect Squares formula, aren’t you taking the Sum of the 1st N perfect Squares?
 
Thus, the equation took the sum of:

(1)^2 + (2)^2 + (3)^2 + (4)^2 = 30

Does that logically get us to the right answer? Just want to check to see if I’m missing something

Thank you much for any help!

Posted from my mobile device

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The first term, the mean, the last term, and the sum of an arithmetic

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The first term, the mean, the last term, and the sum of an arithmetic

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