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Colleges

Triton International College

Location: Subidhanagar, Tinkune

Phone No: 123456790

Courses: BCA, BBA, BIM, BBS, BIT

Triton International College

Location: Subidhanagar, Tinkune

Phone No: 123456790

Courses: BCA, BBA, BIM, BBS, BIT

Sequence and Series

8. The sum to infinity of the G.P : 1/4, -1/16 ...... is [2075]

5

-5

-1/5

1/5

9. The value of a so that 8a + 4, 6a - 2, 3a + 7 will form an A.P is [2075]

15

2

9

10

Solutions
2b = a + c
2(6a - 2) = 8a + 4 + 3a + 7 12a - 4 = 11a + 11
12a - 11a = 11 + 4
a = 15

9. Sum of infinity of 1 - 1/2 + 1/4 - 1/8 is [2074]

0

2/3

1/3

1

Solutions
First you must find r, the common ratio.
You find this with the following formula: r = t2/t1 ⇒ -(1/2) / 1 ⇒ -1/2
⇒s∞ = a/1-r
⇒s∞ = 1/ 1 - (-1/2)
⇒s∞ = 2/3

8. If k + 2, 4k - 6, 3k - 2 and n are in AP then K is [2074]

-6

0

3

-3

Solutions
According to AP Property AP = first term+third term is equal to twice of second term that is
(k+2)+(3k−2)=2(4k−6)
⇒4k=8k−12
⇒4k−8k=−12
⇒−4k=−12
⇒4k=12
⇒k=12/4 =3
Hence k=3.

14. (1 + w + w 2)3 + ( 1 + w - w2)3 =

16

16w

-16

16w2

6. The fourth, seventh and tenth terms of G.P are 1,m,n respectively then [2073]

lm = m2

l2 = m2 + n2

l2 = mn

n2 = lm

7. The nth term of the series : 1.3 + 3.5 + 5.7 + ... is [2073]

(n+1)(n+2)

(n+1)(n+3)

4n2-1

n(n-1)

8.If 2p, p + 8, 3p + 1 are in A.P. Then p is [2072]

5

3

7

-5

Solutions
$2p + 3p + 1 = 2(p + 8)$
$5p + 1 = 2p + 16$
$3p = 15 $
$p = 5$

9. A,G and H are arithmetic mean, geometric mean and harmonic means respectively then [2072]

A2-GH

h2 - AG

G - AH

G2 - AH

7. If a2, b2, c2 are in H.P. then [2071]

2a2x2 = b2(a2 + c2)

a2c2 = b2(a2 + c2)

2b2c2 = a2(b2 + c2)

c2a2 = 2b2(a2 + c2)

16. The nth term of the series: 2 + 4 + 6 + 12 + 20 + ..... is [2070]

n(n+1)

(n+1)(n+2)

n(n-1)

n

17. If |x| < 1 and y=x + x2 + x4 + ....... + ∞ , then x is equal to [2070]

y 1 y

y y 1

y + 1 y

y 1 + y

Other Questions

The value of 21/4 . 41/8 . 8 1/16 ....∞

1

2

3/2

4

Solutions

P = $$ 2^{1/4} . 2^{2/8} . 2^{3/16} .... = $$
$$ 2^{{1/4}+{2/8}+{3/16}+...∞} $$
Now $$ S= {1\over4} + {2\over8} + {3 \over 8} + ... ∞ ------(1) $$
$$ {1\over2}S= {1\over8} + {2\over16} + ... ∞ -------(2) $$
$$ {1\over2}S= {{1\over4}\over 1-{1\over2}} $$
S = 1
ps = 2

If a,b,c are in AP, Then $ \frac{a}{bc},\frac{1}{c},\frac{2}{b} $ will be in

A.P

G.P

H.P

None of these

Solutions
a, b, and c are in A.P. Hence, 2b=a+c ----------- (1)
$ \frac{a}{bc}+\frac{2}{b}=\frac{a+2c}{bc}\ne \frac{2}{c} $
$ \Rightarrow \frac{a}{bc},\frac{1}{c},\frac{2}{b} $ are not in A.P
$ \frac{bc}{a}+\frac{b}{2}=\frac{2bc+ab}{2a}\ne c $ Hence, the given numbers are not in H.P. Again,
$ \frac{a}{bc}\frac{2}{b}=\frac{2a}{{{b}^{2}}c}\ne \frac{1}{{{c}^{2}}} $ Therefore, the given numbers are not in G.P.

The 15th terms of the series 2 1 2 + 1 7 13 + 1 1 9 + 20 23 + . . . is

10/39

10/21

10/23

None of these

Solutions
Reciprocals of the terms of the series are 2/5, 13/20, 9/10, 23/20,...or 8/20, 13/20, 18/20, 23/30,..its nth term is $ \frac{8+(n-1)5}{20}=\frac{5n+3}{20} $ Therefore, the 15th term is $\frac{20}{78}=\frac{10}{39} $

Let $$ S=\frac{4}{19}+\frac{44}{{{19}^{2}}}+\frac{444}{{{19}^{3}}}+... $$ up to ∞. Then S is equal to

40/9

38/81

36/171

none of these

Solutions
$$ S=\frac{4}{19}+\frac{44}{{{19}^{2}}}+\frac{444}{{{19}^{3}}}+... ------(1) $$
$$ \Rightarrow \frac{1}{19}S=\frac{4}{{{19}^{2}}}+\frac{44}{{{19}^{3}}}+... ------(2)$$
Subtracting 2 from 1 we get
$$ \frac{18}{19}S=\frac{4}{19}+\frac{40}{{{19}^{2}}}+\frac{400}{{{19}^{3}}}+... $$
$$ =\frac{\frac{4}{19}}{1-\frac{10}{19}}=\frac{4}{9} $$
$$ \Rightarrow S=\frac{38}{81} $$

21 + 22 + 23 + ... 2n =

2(2n - 1)

2(2n-1 - 1)

2(2n+1 - 1)

None of above

8th term of the series $ 2\sqrt{2}+\sqrt{2}+0+..... $ will be

$ -5\sqrt{2} $

$ 5\sqrt{2} $

$ 10\sqrt{2} $

$ -10\sqrt{2} $

Solutions
Given Series is in A.P. Here a = $ a=2\sqrt{2},\ d=-\sqrt{2} $ Hence term of the series $ =2\sqrt{2}+(8-1)(-\sqrt{2})=-5\sqrt{2}$

Find the value of k given that 4, k, k2 - 1 are consecutive terms of a geometric sequence?

$ ±{2\over\sqrt{3}} $

$ ±{\sqrt{3}} $

$ {2\over\sqrt{5}} $

$ {1\over2} $

Solutions
4/k=k/(k²-1)
4k²-4=k²
3k²=4
k=±{2/√3}

Solutions

Solutions

Sequence 2,1,4,3 .... has general term

n - (-1)n

3n + (-1)n

n + (1)n

-n + (-1)n

Solutions

Next two Term of sequence 0,3,2,5, ... are:

7,9

0,-2

4,6

4,7

Solutions

General term of sequence -5, -3, 1, 9 ... is

{2n - 7}

{2n - 6}

{2n - 5}

{2n - 1}

Solutions

Which term of sequence {n2 - n + 4}, does not give prime number

1st

13th

41st

19th

Solutions

Solutions

Solutions

Solutions

Solutions

Solutions

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Solutions
I there is