The solution of the equation (x + 1) + (x + 4 | vedclass

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

Question

(A) The given equation is $(x + 1) + (x + 4) + (x + 7) + \dots + (x + 28) = 155$.This is an arithmetic progression $(A.P.)$ where the first term $a =

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) The given equation is $(x + 1) + (x + 4) + (x + 7) + \dots + (x + 28) = 155$.This is an arithmetic progression $(A.P.)$ where the first term $a = (x + 1)$ and the common difference $d = 3$.Let $n$ be the number of terms. The last term $l = (x + 28)$.The formula for the $n^{th}$ term is $l = a + (n - 1)d$.Substituting the values: $(x + 28) = (x + 1) + (n - 1)3$.$27 = (n - 1)3 \Rightarrow n - 1 = 9 \Rightarrow n = 10$.The sum of an $A.P.$ is given by $S_n = \frac{n}{2}(a + l)$.Substituting the values: $155 = \frac{10}{2}[(x + 1) + (x + 28)]$.$155 = 5(2x + 29)$.$31 = 2x + 29$.$2x = 2 \Rightarrow x = 1$.

The solution of the equation $(x + 1) + (x + 4) + (x + 7) + \dots + (x + 28) = 155$ is

Similar Questions

$\prod\limits_{n = 1}^{10} {\left( {\frac{{\left( {6\sum\limits_{i = 0}^n i } \right) + 1}}{{\left( {6\sum\limits_{j = 0}^n {(j - 1)} } \right) + 1}}} \right)} $ is equal to:

If $a, b, c$ are in $H.P.$,then

Five numbers are in $A.P.$,whose sum is $25$ and product is $2520$. If one of these five numbers is $-\frac{1}{2}$,then the greatest number amongst them is

Which term of the geometric sequence $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots$ is $\frac{1}{19683}$?

The sum of $1^3 + 2^3 + 3^3 + 4^3 + ..... + 15^3$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module