There are 88 numbers a 1- a 2- a 3- - a 88 and each of them is either equal to -3 or -1 - Given that a 12- a 22- a 882-280- then the value of a14-a24-a884-4-500 icorrect answer -3- wrong answer 0

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Forming simultaneous linear equations

There are 88 ...

Question

There are $88$ numbers $a_{1}, a_{2}, a_{3}, \dots, a_{88}$ and each of them is either equal to $- 3$ or $- 1.$ Given that $a_{1}^{2} + a_{2}^{2} + \dots + a_{88}^{2} = 280,$ then the value of $\frac{a_{1}^{4} + a_{2}^{4} + \dots + a_{88}^{4}}{4} - 500$ is
(correct answer + 3, wrong answer 0)

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a_{1}, a_{2}, a_{3}, . . . . . . . ., a_{25}

a_{3} + a_{5} + a_{12} + a_{14} + a_{21} + a_{23} = 60

a_{2} + a_{7} + a_{19} + a_{24} =

a_{1}, a_{2}, a_{3}, a_{4}

a_{1}^{2} + a_{2}^{2} + a_{3}^{2} + a_{4}^{2} = 1

(a_{1} - a_{2})^{2} + (a_{2} - a_{2})^{2} + (a_{3} - a_{4})^{2} + (a_{4} - a_{1})^{2}

a_{1}, a_{2}, a_{3}, a_{4}

a_{1} + a_{2} + a_{3} + a_{4} = 0

a_{1}^{2} + a_{2}^{2} + a_{3}^{2} + a_{4}^{2} = 1

(a_{1} - a_{2})^{2} + (a_{2} - a_{3})^{2} + (a_{3} - a_{4})^{2} + (a_{4} - a_{1})^{2}

a_{1}, a_{2}, a_{3}, . . . .

a_{1}, a_{5}, a_{10} + a_{15} + a_{20} + a_{24} = 225

a_{1} + a_{2} + a_{3} + . . . + a_{24}

a_{1}, a_{2}, a_{3}, . . . ., a_{11}

a_{1} = 15, 27 - 2 a_{2} > 0

a_{k} = 2 a_{k - 1} - a_{k - 2}

k = 3, 4, . . . . . . ., 11

\frac{a_{1}^{2} + a_{2}^{2} + . . . + a_{11}^{2}}{11} = 90

\frac{a_{1} + a_{2} + . . . + a_{11}}{11}

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