Question
Find the sum of the first 1000 positive integers.
✓
Answer
According to question we are given that a spiral is made up of successive semi-circles, with centres alternately at $A$ and $B$, starting with centre at $A$, of radii $0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm$, ....as shown in Fig.Let $I _1, I _2, I _3, I _4, \ldots I _{13}$ be the lengths (circumferences) of semi-circles of radii $r_1=0.5 cm, r_2=1.0 cm, r_3=1.5 cm, r_4=2.0 cm, r_5=2.5 cm, \ldots$ respectively.
Now, Semi-perimeter of circle = $\pi\cdot r $
Therefore,
$l _ { 1 } = \pi r _ { 1 } = \pi \times 0.5 = \frac { \pi } { 2 } \mathrm { cm }$
$l _ { 2 } = \pi r _ { 2 } = \pi \times 1 = 2 \left( \frac { \pi } { 2 } \right) \mathrm { cm }$
$l _ { 3 } = \pi r _ { 3 } = \pi \times \frac { 3 } { 2 } = 3 \left( \frac { \pi } { 2 } \right) \mathrm { cm }$
$l _ { 4 } = \pi r _ { 4 } = \pi \times 2 = 4 \left( \frac { \pi } { 2 } \right) \mathrm { cm }$
and
$l _ { 13 } = \pi r _ { 13 } = \pi \times \frac { 13 } { 2 } \mathrm { cm } = 13 \left( \frac { \pi } { 2 } \right) \mathrm { cm }$
Therefore total length of the spiral $= I _1+ I _2+ I _3+\ldots+ I _{13}$
$\bf= \left\{ \frac { \pi } { 2 } + 2 \left( \frac { \pi } { 2 } \right) + 3 \left( \frac { \pi } { 2 } \right) + \dots + 13 \left( \frac { \pi } { 2 } \right) \right\} $
$\bf= \frac { \pi } { 2 } ( 1 + 2 + 3 + \cdots + 13 ) $
$\bf= \frac { \pi } { 2 } \times \frac { 13 } { 2 } ( 1 + 13 ) \quad \left[ \text { Using } S _ { n } = \frac { n } { 2 } ( a + l ) \right]$
$\bf= \frac { \pi } { 2 } \times \frac { 13 } { 2 } \times 14$ = $\bf\frac { 1 } { 2 } \times \frac { 22 } { 7 } \times 13 \times 7$ = $\bf {143 cm}$
which is required length of the spiral made up of thirteen consecutive semi-circles.
Now, Semi-perimeter of circle = $\pi\cdot r $
Therefore,
$l _ { 1 } = \pi r _ { 1 } = \pi \times 0.5 = \frac { \pi } { 2 } \mathrm { cm }$
$l _ { 2 } = \pi r _ { 2 } = \pi \times 1 = 2 \left( \frac { \pi } { 2 } \right) \mathrm { cm }$
$l _ { 3 } = \pi r _ { 3 } = \pi \times \frac { 3 } { 2 } = 3 \left( \frac { \pi } { 2 } \right) \mathrm { cm }$
$l _ { 4 } = \pi r _ { 4 } = \pi \times 2 = 4 \left( \frac { \pi } { 2 } \right) \mathrm { cm }$
and
$l _ { 13 } = \pi r _ { 13 } = \pi \times \frac { 13 } { 2 } \mathrm { cm } = 13 \left( \frac { \pi } { 2 } \right) \mathrm { cm }$
Therefore total length of the spiral $= I _1+ I _2+ I _3+\ldots+ I _{13}$
$\bf= \left\{ \frac { \pi } { 2 } + 2 \left( \frac { \pi } { 2 } \right) + 3 \left( \frac { \pi } { 2 } \right) + \dots + 13 \left( \frac { \pi } { 2 } \right) \right\} $
$\bf= \frac { \pi } { 2 } ( 1 + 2 + 3 + \cdots + 13 ) $
$\bf= \frac { \pi } { 2 } \times \frac { 13 } { 2 } ( 1 + 13 ) \quad \left[ \text { Using } S _ { n } = \frac { n } { 2 } ( a + l ) \right]$
$\bf= \frac { \pi } { 2 } \times \frac { 13 } { 2 } \times 14$ = $\bf\frac { 1 } { 2 } \times \frac { 22 } { 7 } \times 13 \times 7$ = $\bf {143 cm}$
which is required length of the spiral made up of thirteen consecutive semi-circles.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Evaluate: $\tan 40^{\circ} \times \tan 50^{\circ}$
→Sum of two numbers is 105 and their difference is 45. Find the numbers.
A piece of wire 22 cm long is bent into the form of an arc of a circle subtending an angle of $60^{\circ}$ at its centre. Find the radius of the circle. (Use $\pi=22/7$)
→In an A.P. if the sum of third and seventh term is zero, find its 5th term.
Find the nature of the roots of the quadratic equation $2 x^2-3 x+5=0$. If the real roots exist. Find it.
→In Fig. 4, O is the centre of a circle. The area of sector OAPB is $\frac{5}{18}$ of the area of the circle. Find x.
→Find the roots of the quadratic equation $x^2-3 x-10=0$ by factorization.
→Prove the given identity, where the angles involved are acute angles for which the expressions are defined.
$\left( \frac { 1 + \tan ^ { 2 } A } { 1 + \cot ^ { 2 } A } \right) = \left( \frac { 1 - \tan A } { 1 - \cot A } \right) ^ { 2 }$ $= tan^2 A$
→$\left( \frac { 1 + \tan ^ { 2 } A } { 1 + \cot ^ { 2 } A } \right) = \left( \frac { 1 - \tan A } { 1 - \cot A } \right) ^ { 2 }$ $= tan^2 A$
The side of a square is equal to the side of an equilateral triangle. The ratio of their areas is:
(sec A + tan A) (1 – sin A)