The greatest positive integer k, for which 49 | vedclass

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

Question

(C) The given sum is a geometric series: $S = 1 + 49 + 49^2 + \ldots + 49^{125}$.Using the formula for the sum of a geometric progression,$S = \frac{4

Vedclass · Accepted Answer

$63$

Vedclass · Answer

(C) The given sum is a geometric series: $S = 1 + 49 + 49^2 + \ldots + 49^{125}$.Using the formula for the sum of a geometric progression,$S = \frac{49^{126}-1}{49-1} = \frac{49^{126}-1}{48}$.We can write $49^{126}-1$ as $(49^{63})^2 - 1^2 = (49^{63}-1)(49^{63}+1)$.Thus,$S = \frac{(49^{63}-1)(49^{63}+1)}{48}$.For $49^k+1$ to be a factor of $S$,we observe that $49^{63}+1$ is a factor of $S$ because $48$ divides $(49^{63}-1)$ (since $49 \equiv 1 \pmod{48}$,so $49^{63} \equiv 1^{63} \equiv 1 \pmod{48}$,implying $49^{63}-1$ is a multiple of $48$).Therefore,the greatest positive integer $k$ is $63$.

The greatest positive integer $k,$ for which $49^k+1$ is a factor of the sum $49^{125}+49^{124}+\ldots+49^{2}+49+1$ is

Similar Questions

If $a_1, a_2, \dots, a_{50}$ are in a geometric progression,then $\frac{a_1 - a_3 + a_5 - \dots + a_{49}}{a_2 - a_4 + a_6 - \dots + a_{50}} = \dots$

The sum of a $G.P.$ with common ratio $r = 3$ is $364$,and the last term is $243$. Find the number of terms $n$.

Suppose the sides of a triangle form a geometric progression with common ratio $r$. Then,$r$ lies in the interval

If $x, G_1, G_2, y$ are the consecutive terms of a $G.P.$,then the value of $G_1 G_2$ is

If the $2^{\text{nd}}$ and $5^{\text{th}}$ terms of a $G$.$P$. are $24$ and $3$ respectively,then the sum of the $1^{\text{st}}$ six terms is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module