Question

The sum of 5 consecutive integers is -250. What is the 4th integer

Answer 1

Answer:

-49

Step-by-step explanation:

$S_{5}$=-250

n/2{2a+(5-1)d}=-250

5/2{2a+4d}= -250

{2a+4d}=-250×2/5

2a+4(1)=-100          [since all the integers are consecutive therefore the

                               difference is 1]                              

2a=-100-4

a=-104/2

a=-52

$a_{4}$=a+(n-1)d

$a_{4}$=-52+3×1

⇒-52+3

⇒-49

Answer 2

$\huge\bold{\underline{To \: Find}} \: -$

The fourth term when sum of five consecutive integers is -250.

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$\large\sf{\underline{Given}}:$

• Sum of these 5 integers = -250

$\huge\bf{\underline{\underline{Solution}}} \: –$

➡ Let the first integer be 'x'

So, the rest four consecutive integers would be (x+1), (x+2), (x+3), (x+4)

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Acc. to the question –

$\sf{(x)+(x+1)+(x+2)+(x+3)+(x+4) = - 250} \\ \sf{\implies5x + 10 = - 250} \\ \sf{\implies5x = - 250 - 10} \\ \sf{ \implies5x = - 260} \\ \sf{\implies x = {\frac{-260}{5}}} = (- 52) \\{\boxed{\boxed{ \green{ \bf x = - 52}}}}$

→ Now, as x=(-52), other integers are

1. x = -52
2. (x+1) = -51
3. (x+2) = -50
4. (x+3) = -49
5. (x+4) = -48

∴ The 4th consecutive integer is -48.

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