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Random4422
28.12.2019 •
Mathematics
Plz it's due
i'll give brainliest to first correct answer and all my points
the sum of a sequence of consecutive integers is 342. the largest integer in the sequence is 3 times greater than the smallest integer in the sequence. what is the smallest integer and how many integers are in the sequence?
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Ответ:
smallest integer=9,
total number of integers are in the sequence = 19.
Step-by-step explanation:
let a be the first term and the number of terms be n. as they are consecutive,the common difference will be 1. these will form Arithmetic Progression series.the sum of n terms in AP series is :[ where, l: last term]
or, sum of n terms in AP series is :[where, d: common difference]
using the first formula,
342 =![\frac{n}{2} (a+3a)](/tpl/images/0435/3698/789e1.png)
342=![\frac{n}{2} (4a)](/tpl/images/0435/3698/c06d4.png)
342=2an .......equation 1.
using the second formula,
342=![\frac{n}{2} (2a+{n-1}l)](/tpl/images/0435/3698/84ba6.png)
684=n[2a+n-1]
684=2an+n^2-n
from equation 1,
342=n^2-n .......equation 1
after solving, the solution to the above equation is 19.
substituting this in equation 2,
171=a[19]
a=9
so, smallest integer= a = 9,
total number of integers are in the sequence = n = 19.
Ответ:
Smallest integer is 9
number of integers is 19
Step-by-step explanation:
Step-by-step explanation:
Sum on consecutive integers is
n(a+l)/2
where a is the first/smallest integer, l is the last/largest integer and n is the number of integers
n(a+l)/2 = 342 (1)
l = 3a (2)
l = a + (n-1) (3)
Using (2) and (3)
3a = a+(n-1)
2a = n-1
n = 2a+1 (4)
Using (1) (2) and (4)
(2a+1)(a+3a)/2 = 342
(2a+1)(4a) = 684
4(2a²+a) = 684
2a²+a= 684÷4
2a²+a = 171
2a²+a-171 = 0
2a²+19a-18a-171 = 0
a(2a+19)-9(2a+19) = 0
(2a+19)(a-9) = 0
a = 9 or a = -19/2
Since a is an integer, it has to be 9
a = 9
n = 2a+1 = 2(9)+1
n = 19
Ответ:
dcfadvav
Step-by-step explanation: