csitfamilySequence 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
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
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
(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
$5p + 1 = 2p + 16$
$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]
Other Questions
The value of 21/4 . 41/8 . 8 1/16 ....∞
1
2
3/2
4
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
$ \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 is
10/39
10/21
10/23
None of these
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
$$ \Rightarrow \frac{1}{19}S=\frac{4}{{{19}^{2}}}+\frac{44}{{{19}^{3}}}+... ------(2)$$
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} $
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} $
4k²-4=k²
3k²=4
k=±{2/√3}
Sequence 2,1,4,3 .... has general term
n - (-1)n
3n + (-1)n
n + (1)n
-n + (-1)n
Next two Term of sequence 0,3,2,5, ... are:
7,9
0,-2
4,6
4,7
General term of sequence -5, -3, 1, 9 ... is
{2n - 7}
{2n - 6}
{2n - 5}
{2n - 1}
Which term of sequence {n2 - n + 4}, does not give prime number
1st
13th
41st
19th
