Consecutive integers

Let $x$ be the 1st of the 5 consecutive number.

$x + (x + 1) + (x + 2) + (x + 3) + (x + 4) = 250$

$5x + 10 = 240$

$5x = 230$

$x = 46$

But the answer is not 46, because the question had stated that the answer is second of the 5 consecutive number

So,

$46 + 1 =47$

The answer is $47$

let x = first number, x + 1 = 2nd number, x + 2 = 3rd number, x + 3 = 4th number, and x + 4 = 5th number

According to the problem, the sum of these numbers is 240, therefore $x+(x+1)+(x+2)+(x+3)+(x+4)=240$ By combining like terms and adding all constants, we can have $5x+10=240$ Adding -10 both sides of the equation $5x+10+ (-10)=240+(-10)$ $5x=230$ Divide both sides by 5 $\frac {5x}{5}=\frac {230}{5}$ $x=46$ Therefore the second number is $x+1=46+1=47$ .

let: x = 1st number x+1 = 2nd number x+2 = 3rd number x+3 = 4th number x+4 = 3rd number

sol'n: 240 = x + (x+1) + (x+2) + (x+3) + (x+4) 240 = 5x+10 5x = 240-10 5x = 230 x = 46

therefore:

x+1 = 46+1 = 47 2nd number is 47

x+(x+1)+(x+2)+(x+3)+(x+4)=240

5x+10=240

5x=230

x=46

So the second number is x+1=

46+1=47

let 1st number of consecutive series is x. so, sum of 5 consecutive integers is x+x+1+x+2+x+3+x+4=240. 5x+10=240. x=46. 2nd consecutive number of x is 47

let the five numbers are x,x+1,x+2,x+3,x+4,x+4 so their sum will be 5x+10=240 from this x=46 and the second number is 47

let x be the number x +(x+1)+(x+2)+(x+3)+(x+4)=240 5x +10=240 5x = 230 x=46 and second number is 47

Actually, it's a really simple problem. The numbers are consecutive , so, we can suppose that the first one in the sequence is "a", then:

a + (a+1) + (a+2) + (a+3) + (a+4) = 240

5a + 10 = 240

5a = 230

a = 46

But we want the second in the sequence, then we want a+1, that is 47.

let n be the 1st number, each consecutive number will be n+ 1, n + 1 +1 , and so on .Hence,

n + (n+1) + (n+2) + (n+3) + (n+4) = 240
5n + 10 = 240
5n = 230
n = 46

2nd number = n + 1 = 47

as they r consecutive numbers so,

x+(x+1)+(x+2)+(x+3)+(x+4)=240

5x+10=240

so,X=46

so,the second number is 46+1=47

We have 5 consecutive integers.Let us designate them as x,(x+1),(x+2),(x+3) and (x+4) respectively.

So : x+x+1+x+2+x+3+x+4 = 240

 or  5x+10=240



 or  5x=240-10=230


   x=230/5=46

Hence, the second number = (x+1) = 46 + 1 = 47

five consecutive numbers 40*5 = 200, balance = 240-200 = 40, sum of 1 to 10 is 55. balance = 55-40 = 15, the first 1 to 5 gives sum value of 15. neglect. remaining 6,7,8,9,10, second one is 7.add 7 with 40 ===> 40+7 =47, The second integer is 47 ,

Let the first integer be x then 2nd=x+1, 3rd=x+2, 4th=x+3, 5th=x+4, Adding all we get, x+x+1+x+2+x+3+x+4=5x+10=240(given), or, 5x = 230, x=230/5=46 Then 2nd=47.

240/5=48 So the numbers are: 46, 47, 48, 49 and 50. Solution: 47

240/5=48 So, 48 should be the median of those 5 consecutive integers and so its 3rd one

2nd one would be 47 .

Let $x$ be the first of the five integers. It then implies that the second, third, fourth, and fifth consecutive integers are $(x+1), (x+2), (x+3),$ and $(x+4)$ , respectively. Then:

$x + (x + 1) + (x + 2) + (x +3) +(x + 4) = 240$

$5x + 10 = 240$

$5x = 230$

$x = 46$

Since the second of the five numbers, which is $(x+1)$ , is being asked, then $x + 1 = 46 + 1 = \boxed{47}$ .

PS: You can change the arbitrary values of the integers, letting $x$ be the second of the integers instead, for example. However, I did this way of solving it for the problem to be done easier (like not having to do subtraction).

let the 5 consecutive numbers be (a-2d , a-d , a , a+d , a+2d) then a-2d+a-d+a+a+d+a+2d=240 a=48 then second consecutive no. is (a-d) = (48 -1)= [47] 47 correct ans........

the avg. of 5 consecutive integers is 48.
five consecutive integers are 46,47,48,49,50.
so the 2nd integer is 47

46+47+48+49+50=240

let the first integer be x. so, five consecutive integers will be- x, x+1, x+3, x+4, x+5. then, the sum of all these integers is 250. i.e. x+x+1+x+2+x+3+x+4+x+5= 240. ie. 5x+10=240. then 5x=230. then, x=46 therefore second ineger=x +1=47

by selectin 5 consecutive terms of A.P. as : a-2d , a-d , a , a+d, a+2d ... then on adding ... 5a=240 this gives. a=48.. then 2 term from beginning=1 term preceeding our selected term... and terms are consecutive... d=1 , so ans = a-d = 48-1=47.

Let the numbers be (n-2), (n-1), n, (n+1) and (n+2) .

Then 5n = 240

Hence n = 48

But this is not the answer, since n is the third number. The answer is *n-1 = 48 - 1 = 47 *

take 5 integers as x-2,x-1,x,x+1,x+2; add up and equate to given 240 we get x and follows the answer

46+(47)+48+49+50=240

let x in the first no,then x + (x+1) + (x+2) + (x+3) + (x+4) = 240 => 5x +10 = 240 => 5x = 230 => x = 46 then the 2nd no wil be 47.

Let the first number be x so, According to the question x + x +1 + x + 2 + x + 3 + x + 4 =240 or, 5x + 10 = 240 or, 5x = 240 -10 or, 5x = 230 so, x = 230/5 so, x = 46 so, the second number is x + 1 = 46 + 1 = 47

let the first integer be x. now the second, third, fourth and fifth integers will be (x+1),(x+2),(x+3) and (x+4). When we add them we get 5x+10 According to question 5x+10=240. -------> x+2=48 -----> x=46 hence the second no. is x+1=47

N + (N+1) + (N+2) + (N+3) + (N+4) = 240 ,N+1 = 47

5 consicutive no are x,x+1,x+2,x+3,x+4 atq.. 5x+10=240 5(x+2)=240 x+2=48 x=46 second no is x+1 ans=47

x+(x+1)+(x+2)+(x+3)+(x+4)=240 5x+10=240 5x=230 x=46 since asked 2nd, so x+1=47

It is a multiple choice question, so that makes the question easier. You can just try each number and see which one works. The correct answer is 47, because 46+47+48+49+50=240.

Let the $5$ consecutive integers be $n, n+1, n+2, n+3, n+4$ .

The sum of the $5$ consecutive integers = $5n+10$ .

Then

$5n+10 = 240$

$5n = 230$

$n = 46$ .

Hence the second of the $5$ consecutive integers is $46 + 1 = \boxed{47}$ .

x+(x+1)+(x+2)+(x+3)+(x+4)=240

5x+10=240

5x=230

x=230/5

x=46

So, the second number is (x+1) which is simplified as (46+1)=47

take x,x+1,x+2,x+3,x+4 as the five consecutive integers which equal 240 then after adding them up we get it as 5x + 10=240 which further equals x as 46. Substitute the value of x in the second consecutive integer x+1 as 46+1=47.

Let x be the middle integer. Then the first integer is (x-2), (x-1), x, (x+1) and (x+2) Sum= 1st+2nd+3rd+4th+5th integer 240=(x-2)+(x-2)+x+(x+1)+(x+2) 240=5x 48=x

Then substitute the value of x (48-2)=46 (48-1)=47 48 (48+1)=49 (48+2)=50

therefore, the second integer is 47

There are in all 5 integers. Sum of 5 integers equal to 240 Let the first consecutive number be = X Let the second consecutive number be = X + 1 Let the third consecutive number be = X + 2 Let the fourth consecutive number be = X + 3 Let the fifth consecutive number be = X + 4 Therefore equation :- (X)+(X+1)+(X+2)+(X+3)+(X+4) = 240 (5X)+(10) = 240 5X = 240-10 5X = 230 X = 230/5 X = 47

(x) + (x+1) + (x+2) + (x+3) + (x+4) = 240 5x+ ((1+4)+(2+3)) = 240 5x + 10 = 240 5x = 230 x = 46

Second cons = x+1 Which equals 46+1 Thus, $\boxed{47}$

240=5/2(2a+(5-1)1). then we get a=46. the second is 47

(2a + (N-1)d )(N/2) = 240 d = 1 N = 5 find a and 2nd term is a + 1

Assume numbers be n, n+1, n+2, n+3, n+4. As per condition given sum gives 240. i.e., n+n+1+n+2+n+3+n+4=240. We get n=46. Like this the second number is n+1 is 47.

assume those 5 consecutive numbers are n,n+1,n+2,n+3 and n+4 sum of these 5 numbers is 240......so n+n+1+n+2+n+3+n+4=240 by solving this we will get n=46 5 numbers are 46,47,48,49,50

let x, x+1, x+2, x+3 and x+4 be the 5 consecutive integers.. solving for x: 5x+10=240 5x=230 x=46 therefore the second will be 47

Let x as first number then 5 consecutive numbers will be a , a+1 , a+2 , a+3 , a+4 sum of all these = 240 so 5a + 10 = 240 5a = 230 a=46 so second no = a+1 = 46+1 = 47