Consider four consecutive integers in an incr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the four consecutive integers be $a-3d, a-d, a+d, a+3d$. However,since they are consecutive integers,the common difference must be $1$. Let th

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Let the four consecutive integers be $a-3d, a-d, a+d, a+3d$. However,since they are consecutive integers,the common difference must be $1$. Let the integers be $x, x+1, x+2, x+3$.We are given that one of these integers is the sum of the squares of the other three.Case $1$: $x+3 = x^2 + (x+1)^2 + (x+2)^2$$x+3 = x^2 + x^2 + 2x + 1 + x^2 + 4x + 4$$3x^2 + 5x + 2 = 0$$(3x+2)(x+1) = 0$$x = -1$ or $x = -2/3$ (not an integer).If $x = -1$,the integers are $-1, 0, 1, 2$.Check: $2 = (-1)^2 + 0^2 + 1^2 = 1 + 0 + 1 = 2$. This holds true.The sum of these integers is $(-1) + 0 + 1 + 2 = 2$.

Consider four consecutive integers in an increasing arithmetic progression. One of these integers is equal to the sum of the squares of the other three. What is the sum of all the numbers?

Similar Questions

If $a, b, c$ are in Arithmetic Progression,then $(a + 2b - c)(2b + c - a)(a + 2b + c) = \dots$

The maximum value of the sum of the Arithmetic Progression $50, 48, 46, 44, \dots$ is:

If $S_1, S_2, S_3, \dots, S_m$ are the sums of $n$ terms of $m$ $A.P.s$ whose first terms are $1, 2, 3, \dots, m$ and common differences are $1, 3, 5, \dots, 2m - 1$ respectively,then $S_1 + S_2 + S_3 + \dots + S_m = $

If $a, b, c, d, e, f$ are in an arithmetic progression,then $e - c = \dots$

The ratio of the sum of $m$ and $n$ terms of an $A.P.$ is $m^2:n^2$. Then the ratio of the $m^{th}$ and $n^{th}$ term will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module