Solve the equation: 1+4+7+10+ldots+x=287 | vedclass

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

Question

(B) The given series $1, 4, 7, 10, \ldots, x$ is an Arithmetic Progression $(AP)$ where the first term $a = 1$ and the common difference $d = 4 - 1 =

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) The given series $1, 4, 7, 10, \ldots, x$ is an Arithmetic Progression $(AP)$ where the first term $a = 1$ and the common difference $d = 4 - 1 = 3$.Let the number of terms be $n$. The $n^{th}$ term is given by $a_n = a + (n - 1)d$.Thus,$x = 1 + (n - 1) 	imes 3 = 3n - 2$.The sum of $n$ terms of an $AP$ is given by $S_n = \frac{n}{2}(a + l)$,where $l$ is the last term $x$.Given $S_n = 287$,we have $287 = \frac{n}{2}(1 + x)$.Substituting $x = 3n - 2$ into the equation:$287 = \frac{n}{2}(1 + 3n - 2)$$574 = n(3n - 1)$$3n^2 - n - 574 = 0$.Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$n = \frac{1 \pm \sqrt{(-1)^2 - 4(3)(-574)}}{2(3)} = \frac{1 \pm \sqrt{1 + 6888}}{6} = \frac{1 \pm \sqrt{6889}}{6} = \frac{1 \pm 83}{6}$.Since $n$ must be a positive integer,$n = \frac{84}{6} = 14$.Now,find $x$: $x = 3n - 2 = 3(14) - 2 = 42 - 2 = 40$.

Solve the equation: $1+4+7+10+\ldots+x=287$

Similar Questions

The $n^{th}$ term of an $A.P.$ is given by $T_{n} = 5 - 6n$. Find the sum of the first $n$ terms of the $A.P.$

Find the number of terms in the finite $A.P.$ $6, 5.5, 5, \ldots, -12$.

The $10^{\text{th}}$ term of the $AP: 5, 8, 11, 14, \ldots$ is

The $A.P.$ with the first term $-3$ and the common difference $4$ is $ \ldots \ldots \ldots \ldots . $

For an $A.P.$,the $12^{th}$ term is $4$ and the $20^{th}$ term is $-20$. Find the $n^{th}$ term of the $A.P.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module