Answer:
The four such consecutive integers are 8, 9 , 10 and 11.
Step-by-step explanation:
Let first integer = k
Then (k+1), ((k+1)+1), and (((k+1)+1)+1) are the next consecutive integers are.
Or, k , (k+1), (k+2) and (k+3) are four consecutive integers.
Now according to the question,
3(k +(k+3)) = -5(k + (k+1))-114
or simplifying this , we get
3(2k+3) = -5(2k+1) - 114
or, 6k + 9 = -10k - 5 - 114
⇒ 6k + 10k = -119 - 9
or, 16k = -128
or, k = 128/16 =8
Hence, the first integer = k = 8
The next consecutive integers are (k+1), (k+2) and (k+3) = 9, 10 and 11
Hence, the four such consecutive integers are 8, 9 , 10 and 11.
Four Consecutive integers satisfying the given conditions are -8,-7,-6,-5.
let us take the consecutive integers as x, x+1, x+2, x+3
An integer is a number that is not a fraction, but a whole number.
It is given that:
3 times the sum of the first and the fourth was 114 less than the product of -5 and the sum of the first two consecutive integers.
3 times the sum of first and fourth integers:
3(x+x+3) = 6x+9
Product of -5 and the sum of the first two consecutive integers :
(-5)* (x+x+1) = -10x-5
According to the question,
(-10x-5)-(6x+9) =114
-16x-14 = 114
-16x = 128
x=-8
so integers are -8, -8+1, -8+2, -8+3 i.e. -8,-7,-6,-5
Therefore, consecutive integers are -8,-7,-6,-5.
To get more about integers visit:
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