Question

If the sum of four consecutive integers is 266, find the integers.

Answer 1

Hint:Let the first integer be n so the other consecutive integers will be (n + 1), (n + 2) and (n + 3). Then make the equation, n + (n + 1) + (n + 2) + (n + 3) which is equal to 266. Hence find the value of n by simplifying the linear equation.

Complete step-by-step answer:
In the question we are stated that sum of four consecutive integers is 266 and we have to find these integers.
Now as we are given about four consecutive integers so let the first integer be n so the next 3 integers would (n + 1), (n + 2), (n + 3).
Hence the four integers are n, n +1, n + 2, n + 3. We are given that their sum is equal to 266. So, we can write,
n + (n + 1) + (n + 2) + (n + 3) = 266
or, 4n + 6 = 266
Now subtracting 6 from both the sides we get,
4n + 6 – 6 = 266 – 6
Or, 4n = 260
So, the value of n is \[\dfrac{260}{4}\] or 65.
Hence the value of n is 65. So, the value of (n + 1), (n + 2) and (n + 3) will be (65 + 1), (65 + 2) and (65 + 3) or 66, 67, 68.
Hence the integers are 65, 66, 67, 68.

Note: After supposing four integers as n, n + 1, n + 2, n + 3 we can find the sum using formula of arithmetic progression which is \[\dfrac{n}{2}\left( a+l \right)\], where n is number of terms which is 4, a is \[{{1}^{st}}\] term which is n and l is last term which is n + 3 and then equate it with 266.