Class 2. Nuda
(1) Body is large thimble-shaped or conical.
(2) Tentacles are absent.
(3) Mouth is wide and pharynx is large.
(4) The meridional vessels are produced into a complex system of anastomosing branches.
Example : Beroe
(4) Anus and madreporite aboral.
(5) Pedicellariae present.
(6) Free-living, slow-creeping, predaceous and scavengerous.
Examples : Astropecten, Luidia, Goniaster, Oreaster (= Pentaceros), Asterina, Solaster, Pteraster, Echinaster, Asterias, Heliaster, etc.
Class 2. Ophiuroidea
(1) Brittle-stars and allies.
(2) Body star-like with arms sharply marked off from the central disc.
(3) Pedicellariae absent.
(4) Stomach sac-like; no anus.
(5) Ambulacral grooves absent or covered by ossicles; tube feet without suckers.
(6) Madreporite oral.
Examples : Ophiura, Ophiothrix, Ophioderma, Ophiopholis, Gorgonocephalus, Asteronyx.
Class 3. Echinoidea
(1) Body not divided into arms; globular (sea urchins), or flattened disc-like (sea-cakes).
(2) Mouth at lower pole, covered by 5 strong and sharp teeth, forming a biting and chewing apparatus called “Aristotle's Lantern”.
(3) Tube-feet slender with suckers.
(4) Skin ossicles fused to form a rigid globular, disc like, or heart-shaped shell or test with movable spines.
(5) 3–jawed pedicellariae present in skin.
(6) Gut long, slenderical and coiled. Anus present.
(7) Larval forms pluteus and Echinopluteus.
Examples : Echinus, Clypeaster, Echinarachinus, Echinocardium, etc.
(3) Mouth surrounded by many hollow retractile tentacles.
(4) Tube feet usually present; sucker-like.
(5) Skin more... 
(3) The shell is composed of a longitudinal series of 8 plates.
(4) The foot is flat and ventral.
(5) The radula is well developed.
(6) Respiration by 8 to 60 pairs of gills.
(7) Unisexual; only one gonad; trochophore larval stage.
Example : Chiton, Cryptochiton, etc.
Introduction
The ordered collection of objects is called sequence. The sequence having specified patterns is called progression. In this chapter besides discussing about the arithmetic progression, we will also discuss about the geometric progression and arithmetic-geometric progression. The various numbers occurring in the sequence is called term of the sequence. A sequence having finite number of terms is called finite sequence, where as the sequence having infinite number of terms is called infinite sequence.
The real sequence is that sequence whose range is a subset of the real number.
A series is defined as the expression denoting the sum of the terms of the sequence. The sum is obtained after adding the terms of the sequence. If \[{{a}_{1}},{{a}_{2}},{{a}_{3}},---,{{a}_{n}}\] is a sequence having n terms, then the sum of the series is given by,
\[\sum\limits_{n=1}^{m}{{{a}_{n}}={{a}_{1}}+{{a}_{2}}+{{a}_{3}}+----{{a}_{n}}}\] 
Arithmetic Progression (A.P.)
A sequence is said to be in A.P, if the difference between the consecutive terms is a constant. The difference between the consecutive terms of an AP is called common difference and any general term is called nth term of the sequence.
If \[{{a}_{1}},{{a}_{2}},{{a}_{3}},---,{{a}_{n}}\] be the nth terms of the sequence in A.P., then nth terms of the sequence is given by \[{{a}_{n}}=a+\left( n-1 \right)d\],
Where 'a' is the first term of the sequence
'd' is the common difference
'n' is the number of terms of the sequence.
Sum of N-Terms of the A.P.
If \[{{a}_{1}},{{a}_{2}},{{a}_{3}},---,{{a}_{n}}\] be the n terms of the sequence in A.P., then the sum of n-terms of the sequence is given \[{{S}_{n}}=\frac{n}{2}\left[ 2a+(n-1)d \right]\]
Arithmetic Mean
If 'a' and 'b' are any two terms of A.P., then the arithmetic mean is given by \[AM=\frac{a+b}{2}\]
Properties of AP
(a) If a constant is added or subtracted from each term of the AP sequence, then the resulting sequence is also in AP.
(b) If each term of an AP sequence is multiplied or divided by a constant, then the resulting sequence is also in AP.
For a sequence in A.P., the sum of n terms of the sequence is \[2n+3{{n}^{2}}\], find the 20th term of the sequence:
(a) 121
(b) 112
(c) 119
(d) 124
(e) None of these
Answer: (c)
Explanation
We have, \[{{S}_{n}}=2n+3{{n}^{2}}\]
\[{{S}_{1}}=5,{{S}_{2}}=16\]
Thus common difference is \[d={{S}_{2}}-2a=16-10=6\]
Hence the 20th term of the sequence is \[=6\times 320-1=119\]
Find the number of odd integers starting with 15 and will give the sum as 975.
(a) 25
(b) 15
(c) 20
(d) 30
(e) None of these
Answer: (a)
Explanation
Here we have a = 15 and common difference = 2
Since we know that, \[{{S}_{n}}=\frac{n}{2}\left[ 2a+(n-1)d \right]\]
But \[{{S}_{n}}=975\]
\[\therefore \,\,\,\frac{n}{2}\left[ 2a+(n-1)d \right]=975\]
\[\Rightarrow \,\,\frac{n}{2}\left[ 2\times 15+(n-1)\times 2 \right]=975\]
\[\Rightarrow \,\,{{n}^{2}}+14n-=975=0\]
\[\Rightarrow \,\,\,\,\,\,n=25\,\,\text{or}\,\,n=-39\]
Since number of terms cannot be negative, hence n = - 39 is rejected. Therefore n = 25.
If the 15th term of an AP is 45 and 20th term is 60, and then find the 30th term of the AP.
(a) 70
(b) 90
(c) 110
(d) 120
(e) None of these
Answer: (b)
Explanation
We have,
\[{{A}_{15}}=A+14d=45\]
\[{{a}_{20}}=a+19d=60\]
On solving the above equation we get,
a = 3 and d = 3
Therefore, \[{{a}_{30}}=a+29d\]
\[\Rightarrow \,\,{{a}_{30}}=3+29\times 3=90\]
If the number of different cards of different colours Thomas has are in AP. If he has cards of seven different colours in the order of VIBGYOR such that third more... 
Geometric Progression (G.P.)
A sequence is said to be in G.P. if the ratio between the consecutive terms is constant. The sequence \[{{a}_{1}},{{a}_{2}},{{a}_{3}},---,{{a}_{n}}\] is said to be in G.P. if the ratio of the consecutive term is a constant.
If 'r' is the common ratio, then the nth term of the sequence is given by \[{{a}_{n}}=a\,\,{{r}^{n-1}}\]
The sum of n terms of the G.P. sequence is given by
\[{{S}_{n}}=\frac{a({{r}^{n}}-1)}{r-1}If\,\,r\,>\,1\,and\,{{S}_{n}}=\frac{a(1-{{r}^{n}})}{1-r}if\,r\,>1\]
Sum to infinity is a G.P. series is given by \[{{S}_{\propto }}=\frac{a}{1-r}.\]
Geometric Mean (G.M.)
If 'a' and 'b' are any two terms of G.P., then the geometric mean is given by \[GM=\sqrt{ab}\]
Properties of GP
(a) If each term of GP is multiplied or divided by a constant, then the resulting sequence is also in GP.
(b) If each term of the GP is raised to the same power then the resulting sequence is also in GP.
(c) The reciprocal of each term of GP also results in GP.
The sum of the series of the sequence given by \[\frac{2}{3},-1,\frac{3}{2},.....\]. to 5 terms is given by:
(a) 0
(b) \[\frac{55}{24}\]
(c) \[\frac{65}{34}\]
(d) \[\frac{65}{24}\]
(e) None of these
Answer: (b)
Explanation
Since the above sequence is in GP, and first term is \[a=\frac{2}{3}\] and common ratio \[=r=-\frac{3}{2}\]
The sum of n terms of GP is given by \[{{S}_{n}}=\frac{a(1-{{r}^{n}})}{1-r}\]
\[\Rightarrow \,\,{{S}_{n}}\frac{\frac{2}{3}\left\{ 1-{{\left( -\frac{3}{2} \right)}^{5}} \right\}}{1-\left( -\frac{3}{2} \right)}\]
\[\Rightarrow \,\,{{S}_{n}}\frac{4}{15}\left\{ 1+\frac{243}{32} \right\}\]
\[\Rightarrow \,\,{{S}_{n}}=\frac{55}{24}\]
Robert and James were playing a game and Robert asks James to find the three numbers which are in GP such that their sum is 19 and their product is 216. The three numbers are.
(a) (4, 6, 9)
(b) (3, 6, 12)
(c) (5, 10, 20)
(d) (7, 14, 28)
(e) None of these
Answer: (a)
Explanation
Let 'a' be the first term of the GP and V be the common ratio. Then the three terms which are in GP can be written\[as\frac{a}{r},a,ar.\].
According to question,
\[\frac{a}{r}\times a\times ar=21\]
\[\Rightarrow \,\,{{a}^{3}}=216\]
\[\Rightarrow \,\,a=6\]
Also, \[\frac{a}{r}+a+ar=19\]
\[\Rightarrow \,\,\frac{6}{r}+6++6r=19\]
\[\Rightarrow \,\,6{{r}^{2}}-13r+6=0\]
\[\Rightarrow \,\,\frac{3}{2}or\frac{2}{3}\]
There is a sequence of numbers such that they are in GP and sum to infinite of the number of terms of the sequence is 15 and the sum of their square is 45. The first term of the sequence is given by:
(a) 3
(b) 5
(c) 9
(d) 13
(e) None of these
Answer: (b)
Explanation
The sum of infinite number of terms in GP is given by \[{{S}_{\propto }}=\frac{a}{1-r}\]
\[\Rightarrow \,\,15=\frac{a}{1-{{r}^{2}}}----(1)\]
Also the sum of square of the terms of the GP is given by,
\[{{S}_{\propto }}=\frac{{{a}^{2}}}{1-{{r}^{2}}}\]
\[\Rightarrow \,\,45=\frac{{{a}^{2}}}{1-{{r}^{2}}}----(2)\]
Solving (1) and (2) we get,
a = 5
more... 
Harmonic Progression (H.P.)
The sequence is said to be in H.P. If the reciprocal of its terms gives the A.P. It has got wide application in the field of geometry and theory of sound. The questions are generally solved by inverting the terms and using the property of arithmetic progression.
Three numbers a, b, c are said to be in HP if,
\[\frac{a}{c}=\frac{a-b}{a-c}\]
Harmonic Mean (HM)
If 'a' and 'b' be any two terms, then their harmonic mean is given by \[HM=\frac{2ab}{a+b}\].
Relation between AM, GM, and HM
Since we know that,
\[AM=\frac{a+b}{2},\,\,GM=\sqrt{ab}\,and\,HM=\frac{2ab}{a+b}\]
Then,
\[AM\times HM=\frac{a+b}{2}\times \frac{2ab}{a\times b}=ab={{G}^{2}}\]
\[AM\times HM=G{{M}^{2}}\]
Form the above relation we can say that AM > GM and GM is intermediate value between AM and HM, therefore GM > HM. Hence we can say that AM > GM > HM.
Also the relation between A and G is given by,
\[AM-GM=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{2}}\]
The harmonic mean between two numbers is \[\frac{48}{5}\] and geometric mean is 12. The two numbers are:
(a) (3 & 20)
(b) (2 & 12)
(c) (6 & 24)
(d) (5 & 32)
(e) None of these
Answer: (c)
Explanation
Let the two number be ‘a’ and ‘b’.
Then, \[HM=\frac{2ab}{a+b}\,and\,GM=\sqrt{ab}\]
Putting the value of HM and GM in the above relation we get,
\[\frac{48}{5}=\frac{2ab}{a+b}\,and\,12=\sqrt{ab}\]
On solving these two equations we get, A = 6 & b = 24
If \[{{p}^{th}}\] term of HP is equal to \[{{q}^{th}}\] and the p, then (p + q) term of the series is.
(a) \[\frac{pq}{p+q}\]
(b) \[\frac{p-q}{p+q}\]
(c) \[\frac{p-q}{pq}\]
(d) \[\frac{p+q}{pq}\]
(e) None of these
Answer: (a)
Explanation
Let 'a' and 'd' be the first term and common difference of an AP,
Then the \[{{p}^{th}}\,and\,{{q}^{th}}\] term of the AP is \[{{a}_{p}}(p-1)d\,and\,{{a}_{p}}=a+(q-1)d\]
For HP series the corresponding terms are,
\[\Rightarrow \,\,\frac{1\,}{pq}=a+(p-1)d\,and\,\frac{1}{p}=a+(q-1)d\]
On solving the above equation we get,
\[a=\frac{1}{pq}and\,\,d=\frac{1}{pq}\]
Therefore, \[{{(p+q)}^{th}}\,term\,=\frac{p+q}{pq}\]
Hence \[{{(p+q)}^{th}}\,\] of the HP is given by \[\frac{pq}{p+q}\]
For any two numbers the ratio of HM : GM is 12 :13, and then the ratio of the two numbers is given by:
(a) 3 : 8
(b) 2 : 5
(c) 4 : 9
(d) 5 : 7
(e) None of these
Answer: (c)
Explanation
Let the two number be 'a' and 'b'. Then,
\[HM=\frac{2ab}{a+b\,}and\,GM=\sqrt{ab}\]
\[\Rightarrow \,\,\,\,\frac{HM}{GM}=\frac{\frac{2ab}{a+b}}{\sqrt{ab}}\]
\[\Rightarrow \,\,\,\frac{12}{13}=\frac{2\sqrt{ab}}{a+b}\]
\[\Rightarrow \,\,\,\frac{13}{12}=\frac{a+b}{2\sqrt{ab}}\]
By componendo and dividendo, we get
\[\Rightarrow \,\,\,\frac{25}{1}=\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}\]
\[\Rightarrow \,\,\,\frac{25}{1}=\frac{{{\left( \sqrt{a}+\sqrt{b} \right)}^{2}}}{{{\left( \sqrt{a}-\sqrt{b} \right)}^{2}}}\]
\[\Rightarrow \,\,\frac{5}{1}=\frac{\left( \sqrt{a}+\sqrt{b} \right)}{\left( \sqrt{a}-\sqrt{b} \right)}\]
\[\Rightarrow \,\,\frac{a}{b}=\frac{9}{4}\]
The number of bricks arranged in a complete pyramid on a square base of side 10 units is given by.
(a) 290
(b) 385
(c) 425
(d) 525
(e) None of these
more... | 3.11 | ||
| + | 0.313 | |
| 36.723 | \[\leftarrow \] (answer should be reported to one decimal place) |
| 0.241 | ||
| + | 0.09 | \[\leftarrow \] (has 2 decunak okaces) |
| 3.4731 | \[\leftarrow \] (answer should be reported to 2 decimal places) |
| 2.831 | \[\leftarrow \] (has 3 decimal places) | |
| - | 24.5492 | |
| 38.2818 | \[\leftarrow \] (answer should be reported to 3 decimal places after rounding off) |
| 142.06 | ||
| x | 0.23 | \[\leftarrow \] (two significant figures) |
| 32.6738 | \[\leftarrow \] (answer should have two significant figures) |
| 51.028 | ||
| x | 1.31 | \[\leftarrow \] (three significant figures) |
| 66.84668 |
You need to login to perform this action.
You will be redirected in
3 sec
