Sequences and Series Test 36

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Sequences and Series Test 36

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Weekly Quiz Competition

Question 1
1 / -0

The number of permutations of letters of the word "PARALLAL" atken four at a time must be,

A
$$216$$

B
$$244$$

C
$$286$$

D
$$1680$$

Solution

Permutation of word PARALLAL taken four at a time
Four letter word might be $$LLL$$_
So total of $$16$$ words [As $$4$$ways, $$4$$spots]
For $$AALL$$, $$6$$ different arrangements
$$AA$$_ _, can be $${ 4 }_{ { C }_{ 2 } }$$ ways $$=6$$ ways
Arranged in $$12$$ ways is total of $$72$$ words
And extra words can be formed in $$120$$ ways
$$16+6+144+120=286$$ ways
Question 2
1 / -0

$$1+1.1!+2.2!+3.3!+...+n.n!$$ is equal to

A
$$n!$$

B
$$(n-1)!$$

C
$$(n+1)!$$

D
$$n$$

Solution

given,

$$1+1.1!+2.2!+3.3!+......+n.n!$$

$$=1+(2-1).1!+(3-1)!.2!+.......+(n+1-1).n!$$

$$=1+2.1!-1.1!+3.2!-1.2!+...+(n+1)n!-1n!$$

$$=1+2!-1!+3!-2!+4!-3!+......(n+1)!-n!$$

$$=1+(n+1)!-1$$

$$=(n+1)!$$
Question 3
1 / -0

$$\sum _{ k=1 }^{ 10 }{ k.k! } =$$

A
10!

B
11!

C
10!+1

D
11!-1

Solution

Given,

$$\sum _{k=1}^{10}kk!$$

$$a_k=kk!$$

$$a_1=1\cdot \:1!=1$$

$$a_2=2\cdot \:2!=4$$

Similarly,

$$a_3=18$$

$$a_4=96$$

$$a_5=600$$

$$a_6=4320$$

$$a_7=35280$$

$$a_8=322560$$

$$a_9=3265920$$

$$a_{10}=36288000$$

$$=1+4+18+96+600+4320+35280+322560+3265920+36288000$$

$$=39916799$$

$$=11!-1$$
Question 4
1 / -0

The number of words which can be made out of the letters of the word $$'MOBILE'$$ when consonants always occupy odd places is _______.

A
$$20$$

B
$$36$$

C
$$30$$

D
$$720$$

Solution
Question 5
1 / -0

The number of was dividing $$52$$ cards amongst four players equally, are

A
$$\dfrac {52!}{(13!)^{4}}$$

B
$$\dfrac {52!}{(13!)^{4}4!}$$

C
$$\dfrac {52!}{(12!)^{4}4!}$$

D
None of these
Question 6
1 / -0

13, 16, 22, 33, 51 ?

A
89

B
78

C
102

D
69

E
none of these

Solution

REF.Image
$$ 13,16,22,33,51, ?$$
Question 7
1 / -0

655, 439, 314, 250, 223 ?

A
215

B
210

C
195

D
190

E
none of these

Solution

Solution $$\rightarrow $$ Given terms are 655,439,314,250,223,?
As $$a_{1}=655$$
$$a_{2}= 439$$
$$a_{3}=314$$
$$a_{4}= 250$$
$$a_{5}= 223$$
$$a_{6}=?$$
So, $$a_{1}-a_{2}=655- 439= 216=6^{3}$$
$$a_{3}-a_{2}= 439- 314= 125=5^{3}$$
$$a_{3}-a_{4}= -250+314= 64=4^{3}$$
$$a_{4}-a_{5}=-223+250= 27=3^{3}$$
So, we can observe difference b/w terms are cubes
of 6,5,4,3 so, is similar way
$$a_{5}-a_{6}=2^{3}$$
$$223-a_{6}= 8$$
$$a_{6} = 215$$
So, we find next term is 215
Question 8
1 / -0

The area of the triangle vertices $$(1,0),(7,0)$$ and $$(4,4)$$ is ___ square units.

A
$$8$$

B
$$10$$

C
$$12$$

D
$$14$$

Solution
Question 9
1 / -0

Find the next term
$$32,49,83,151,287,559,?$$

A
$$1118$$

B
$$979$$

C
$$1103$$

D
$$1120$$

Solution
Question 10
1 / -0

$$4,6,12,30,90,315,?$$

A
$$945$$

B
$$1102$$

C
$$1260$$

D
$$1417.5$$

E
$$None\ of\ these$$

Solution

$$=315\times 4$$
$$=1260.$$
Hence, the answer is $$1260.$$