A plastic toy consists of 14 circular discs o | vedclass

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

Question

(B) The radii of the $14$ discs form an arithmetic progression: $1, 2, 3, \dots, 14$.The circumference of a disc with radius $r$ is given by $C = 2\pi

Vedclass · Accepted Answer

$660$

Vedclass · Answer

(B) The radii of the $14$ discs form an arithmetic progression: $1, 2, 3, \dots, 14$.The circumference of a disc with radius $r$ is given by $C = 2\pi r$.The sum of the circumferences of all $14$ discs is $S = 2\pi(1) + 2\pi(2) + 2\pi(3) + \dots + 2\pi(14)$.$S = 2\pi(1 + 2 + 3 + \dots + 14)$.Using the sum formula for the first $n$ natural numbers, $S_n = \frac{n(n+1)}{2}$, where $n = 14$:$S = 2\pi 	imes \frac{14(14+1)}{2} = 2\pi 	imes \frac{14 	imes 15}{2} = 2\pi 	imes 105 = 210\pi$.Taking $\pi \approx \frac{22}{7}$:$S = 210 	imes \frac{22}{7} = 30 	imes 22 = 660\, cm$.

$A$ plastic toy consists of $14$ circular discs of different radii. The radius of the smallest disc is $1\, cm$, the radius of the next bigger disc is $2\, cm$, and so on, such that the radius of the biggest disc is $14\, cm$. Find the sum of the circumferences of all the discs in $cm$.

Similar Questions

For each of the following $A.P.s$,find the $n^{th}$ term: $10.5, 11.7, 12.9, 14.1, \dots$

The sum of three numbers in $A.P.$ is $12$ and the sum of their cubes is $288$. Find those numbers.

Find the number of terms of the finite $A.P.$ $-1, -\frac{5}{6}, -\frac{2}{3}, \ldots, \frac{10}{3}$.

Justify whether it is true to say that the following are the $n^{\text{th}}$ terms of an $AP.$
$(i)$ $2n-3$
$(ii)$ $3n^{2}+5$
$(iii)$ $1+n+n^{2}$

The sum of the $5^{\text{th}}$ and the $7^{\text{th}}$ terms of an $AP$ is $52$ and the $10^{\text{th}}$ term is $46$. Find the $AP$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module