Eamcet Mathematics - Sequences & Series Chapter Sample Question Paper With Answer Key
Exam Duration: 60 Mins Total Questions : 50
The arithmetic mean between two numbers is A, and S is sum of n arithmetic means between these numbers; then
- (a)
S = nA
- (b)
A = nS
- (c)
A = S
- (d)
NONE OF THESE
S=n times the mean between the two numbers = nA
If a, b, c are in G.P., then the equations \(ax^2+2bx+c = 0\) and \(dx^2+2ex+f=0\) have a common root \(\frac { d }{ a } ,\frac { e }{ b } ,\frac { f }{ c } \) are in
- (a)
A.P.
- (b)
G.P.
- (c)
H.P.
- (d)
NONE OF THESE
The sum of the first ten terms of the series \(log3+log\frac { 9 }{ 4 } +log\frac { 27 }{ 16 } +log\frac { 81 }{ 64 } +....\quad is\)
- (a)
45 log 3 - 5 log 4
- (b)
50 log 3 + 20 log 4
- (c)
55 log 3 - 45 log 4
- (d)
NONE OF THESE
\(S=log\quad 3+log\frac { { 3 }^{ 2 } }{ 4 } +log\frac { { 3 }^{ 3 } }{ { 4 }^{ 2 } } +log\frac { { 3 }^{ 4 } }{ { 4 }^{ 3 } } ...upto\quad 10\quad terms\)
=log 3+(2 log 3-log 4)+(3 log 3-2 log 4)+(4 log 3-3 log 4)+...upto 10 terms
=(log 3)(1+2+3+...+10)-(log 4)(1+2+3+...+9)
\(=(log\quad 3)\frac { 10\times \left( 10+1 \right) }{ 2 } -(log\quad 4).\frac { 9\times (9+1) }{ 2 } \)
=55 log 3-45 log 4
The sum of the series \({ cot }^{ -1 }(3)+{ cot }^{ -1 }(7)+{ cot }^{ -1 }(13)+{ cot }^{ -1 }(21)+...\) up to n terms, is
- (a)
\({ tan }^{ -1 }\left( \frac { n+2 }{ n } \right) \)
- (b)
\({ tan }^{ -1 }\left( \frac { 1 }{ n+2 } \right) \)
- (c)
\({ tan }^{ -1 }\left( \frac { n }{ n+2 } \right) \)
- (d)
NONE OF THESE
We first find the nth term tn of the series 3,7,13,21,...
Lets Sn be the sum of the first n terms of the series.
\(\frac { { S }_{ n }=3+7+13+21+...+{ t }_{ n-1 }+{ t }_{ n }\\ { S }_{ n }=\underset { - }{ 3 } \underset { - }{ + } 7\underset { - }{ +13 } \underset { - }{ + } ...\underset { - }{ + } { t }_{ n-2 }\underset { - }{ + } { t }_{ n-1 }\underset { - }{ + } { t }_{ n }] }{ 0=3+(4+6+8+...upto(n-1)terms)-{ t }_{ n } } \)
Differences are in AP,
\(\therefore \quad { t }_{ n }=3+\left( \frac { n-1 }{ 2 } \right) \left\{ 2\times 4+\left( n-2 \right) \times 2 \right\} \)
\(=3+\left( \frac { n-1 }{ 2 } \right) \left[ 2n+4 \right] \)
\({ t }_{ n }={ n }^{ 2 }+n+1\)
Let Tn be the nth term of the given series and Sn be the sum of the first n terms of the given series.
\({ T }_{ n }={ cot }^{ -1 }\left( { n }^{ 2 }+n+1 \right) ={ tan }^{ -1 }\left( \frac { 1 }{ 1+n+{ n }^{ 2 } } \right) \)
\(\left( \because \quad { cot }^{ -1 }x={ tan }^{ -1 }\left( \frac { 1 }{ x } \right) \right) \)
If a1, a2, a3, ... , an are positive real numbers whose product is a fixed number c, then the minimum value of a1 + a2 + a3 + ...+ an-1 + 2an, is
- (a)
n(2c)1/n
- (b)
(n+1)c1/n
- (c)
2nc1/n
- (d)
(n+1)(2c)1/n
We know A.M. ≥ G.M.
∴ \(\frac { { a }_{ 1 }+{ a }_{ 2 }+{ a }_{ 3 }+...+{ 2a }_{ n } }{ n } \)
≥ (a1.a2.a3...(2an))1/n
∴ a1+a2+a3+...+2an≥n(2c)1/n
(∵ a1a2a3...an=c)
∴ Minimum value of
a1+a2+a3+...aan=n(2c)1/n
Hence, the correct alternative is (a).
If \(a_1,a_2,a_3, . . . a_n\) are in A.P. where an>0 for all n then value of \(\frac { 1 }{ \sqrt { { a }_{ 1 } } +\sqrt { { a }_{ 2 } } } +\frac { 1 }{ \sqrt { { a }_{ 2 } } +\sqrt { { a }_{ 3 } } } +...+\frac { 1 }{ \sqrt { { a }_{ n-1 } } +\sqrt { { a }_{ n } } } \) is
- (a)
\(\frac { 1 }{ \sqrt { { a }_{ 1 } } +\sqrt { { a }_{ n } } } \)
- (b)
\(\frac { 1 }{ \sqrt { { a }_{ 1 } } -\sqrt { { a }_{ n } } } \)
- (c)
\(\frac { n }{ \sqrt { { a }_{ 1 } } +\sqrt { { a }_{ n } } } \)
- (d)
\(\frac { n-1 }{ \sqrt { { a }_{ 1 } } +\sqrt { { a }_{ n } } } \)
The value of \(\left( 1+\frac { { a }^{ 2 }{ x }^{ 2 } }{ 2! } +\frac { { a }^{ 4 }{ x }^{ 4 } }{ 4! } +... \right) ^2-\left( ax+\frac { { a }^{ 3 }{ x }^{ 3 } }{ 3! } +\frac { { a }^{ 5 }{ x }^{ 5 } }{ 5! } +... \right) ^2\)
- (a)
eax
- (b)
e-ax
- (c)
0
- (d)
1
Using A2-B2 = (A+B)(A-B), the given expression takes the form
\(\left( 1+ax+\frac { { a }^{ 2 }{ x }^{ 2 } }{ 2! } +\frac { { a }^{ 3 }{ x }^{ 3 } }{ 3! } +... \right) -\left( 1-ax+\frac { { a }^{ 2 }{ x }^{ 2 } }{ 2! } -\frac { { a }^{ 3 }{ x }^{ 3 } }{ 3! } ... \right) \)
=eax.e-ax=eax-ax=e0=1
The sum of the series \(1+\frac { 1+2 }{ 2! } +\frac { 1+2+3 }{ 3! } +\frac { 1+2+3+4 }{ 4! } +...\infty \) is
- (a)
e
- (b)
\(3e\over 2\)
- (c)
2e
- (d)
\(5e\over 2\)
\({ T }_{ n }=\frac { 1+2+3+...+n }{ n! } =\frac { n(n+1) }{ 2(n!) } \)
\(=\frac { 1 }{ 2 } \left[ \frac { n-1 }{ (n-1)! } +\frac { 2 }{ (n-1)! } \right] \)
\(or\quad { T }_{ n }=\frac { 1 }{ 2 } \left[ \frac { 1 }{ (n-2)! } +\frac { 2 }{ (n-1)! } \right] \)
\(\therefore \quad { T }_{ 1 }=\frac { 1 }{ 2 } \left[ \frac { 1 }{ (-1)! } +\frac { 2 }{ 0! } \right] \)
\({ T }_{ 2 }=\frac { 1 }{ 2 } \left[ \frac { 1 }{ 0! } +\frac { 2 }{ 1! } \right] \)
\({ T }_{ 3 }=\frac { 1 }{ 2 } \left[ \frac { 1 }{ 1! } +\frac { 2 }{ 2! } \right] \)
\({ T }_{ 4 }=\frac { 1 }{ 2 } \left[ \frac { 1 }{ 2! } +\frac { 2 }{ 3! } \right] \) and so on
Adding and using (-1)! as \(\infty \), we get
T1+T2+T3+T4+...
\(=\frac { 1 }{ 2 } \left[ 0+1+1+\frac { 1 }{ 2 } +...\infty \right] +\frac { 2 }{ 2 } \left[ 1+1+\frac { 1 }{ 2! } +\frac { 1 }{ 3! } +...\infty \right] \)
\(=\frac { 1 }{ 2 } e+e=\frac { 3 }{ 2 } e\)
The sum of the series \(\frac { 12 }{ 2! } +\frac { 28 }{ 3! } +\frac { 50 }{ 4! } +\frac { 78 }{ 5! } +\quad is\)
- (a)
e
- (b)
3e
- (c)
4e
- (d)
5e
Let S=12+28+50+78+...+Tn
\(\frac { S=12+28+50+...+{ T }_{ n-1 }\\ \quad \quad -\quad -\quad -\quad \quad -\quad \quad - }{ 0=12+16+22+28+...{ T }_{ n } } \)
Tn=2+(10+16+22+28+...upto n terms)
\(=2+\frac { n }{ 2 } \left[ 2\times 10+(n-1)\times 6 \right] \)
\(=2+\frac { n }{ 2 } \left[ 14+6n \right] =2+n(7+3n)\)
or Tn=3n2+7n+2
Let tn be the nth term of the given series
\({ t }_{ n }=\frac { { 3n }^{ 2 }+7n+2 }{ \left( n+1 \right) ! } =\frac { 3n\left( n+1 \right) +4n+2 }{ \left( n+1 \right) ! } \)
\(=\frac { 3n\left( n+1 \right) +4\left( n+1 \right) -2 }{ \left( n+1 \right) ! } \)
\(=\frac { 3 }{ (n-1)! } +\frac { 4 }{ n! } -\frac { 2 }{ \left( n+1 \right) ! } \)
\(\therefore \quad { \Sigma t }_{ n }=3\Sigma \frac { 1 }{ \left( n-1 \right) ! } +4\Sigma \frac { 1 }{ n! } -2\Sigma \frac { 1 }{ \left( n+1 \right) ! } \)
=3e+4e-2e=5e
If \(\frac { a+bX }{ a-bX } =\frac { b+cX }{ b-cX } =\frac { c+dX }{ c-dX } ,\left( X\neq 0 \right) \) then a, b, c and d are in
- (a)
AP
- (b)
GP
- (c)
HP
- (d)
None of these
Given, \(\frac { a+bX }{ a-bX } =\frac { b+cX }{ b-cX } =\frac { c+dX }{ c-dX } \)
On applying componendo and dividendo, we get
\(\frac { a+bX+a-bX }{ a+bX-\left( a-bX \right) } =\frac { b+cX+b-cX }{ b+cX-\left( b-cX \right) } =\frac { c+dX+c-dX }{ c+dX-\left( c-dX \right) } \)
\(\Rightarrow \quad \frac { 2a }{ 2bX } =\frac { 2b }{ 2cX } =\frac { 2c }{ 2dX } \Rightarrow \frac { a }{ bX } =\frac { b }{ cX } =\frac { c }{ dX } \)
On multiplying each term by X, we get
\(\frac { a }{ b } =\frac { b }{ c } =\frac { c }{ d } \)
Therefore, a, b, c and d are in GP.
If x, Y and z are positive integers, then value of expression ( X + Y) (Y + Z) (Z + X) is
- (a)
= 8 xyz
- (b)
> 8 xyz
- (c)
< 8 xyz
- (d)
= 4 xyz
\(AM>GM,\frac { X+Y }{ 2 } >\sqrt { XY } ,\frac { Y+Z }{ 2 } >\sqrt { YZ } and\quad \frac { Z+X }{ 2 } >\sqrt { ZX } \)
On multiplying the three inequalities, we get
\(\frac { X+Y }{ 2 } .\frac { Y+Z }{ 2 } .\frac { Z+X }{ 2 } >\sqrt { \left( XY \right) \left( YZ \right) \left( ZX \right) } \)
\(\therefore\left( \frac { X+Y }{ 2 } \right) \left( \frac { Y+Z }{ 2 } \right) \left( \frac { Z+X }{ 2 } \right) >8XYZ\)
If \({ log }_{ a }X,{ log }_{ b }X\) and \({ log }_{ c }X\) are in HP, then a,b and c are in
- (a)
AP
- (b)
HP
- (c)
GP
- (d)
None of the above
Given, \({ log }_{ a }X,{ log }_{ b }X,{ log }_{ c }X\) are in HP.
\(\Rightarrow \frac { logX }{ loga } ,\frac { logX }{ logb } ,\frac { logX }{ logc } \) are in HP.
\(\Rightarrow \frac { loga }{ logX } ,\frac { logb }{ logX } ,\frac { logc }{ logX } \) are in AP.
\(\Rightarrow{ log }_{ X }a,{ log }_{ X }b,{ log }_{ X }c\) are in AP.
So, a, b and c are in GP.
The sum upto n terms of the series \(\tan ^{ -1 }{ \frac { 1 }{ 2 } } +\tan ^{ -1 }{ \frac { 2 }{ 9 } } +\tan ^{ -1 }{ \frac { 1 }{ 8 } } +\tan ^{ -1 }{ \frac { 2 }{ 25 } } +\tan ^{ -1 }{ \frac { 1 }{ 18 } } +...\) is
- (a)
\(\tan ^{ -1 }{ \left( \frac { 1 }{ 3 } \right) } \)
- (b)
\(\frac { \pi }{ 4 } \)
- (c)
\(\tan ^{ -1 }{ \frac { 1 }{ 2 } } \)
- (d)
\(\cot ^{ -1 }{ 2 } \)
Let r th term of the series be,\({ T }_{ r }={ tan }^{ -1 }\left( \frac { \sqrt { 2 } }{ r+1 } \right) ^{ 2 }\)
∴ \({ S }_{ n }=\sum _{ r=1 }^{ n }{ { T }_{ r } } =\sum _{ r=1 }^{ n }{ tan^{ -1 } } \left( \frac { 2 }{ { r }^{ 2 }+2r+1 } \right) \)
\(=\sum _{ r=1 }^{ n }{ { tan }^{ -1 } } \frac { r+2-r }{ 1+r(r+2) } =\sum _{ r=1 }^{ n }{ { tan }^{ -1 } } (r+2)-{ tan }^{ -1 }r\)
\({ S }_{ n }={ tan }^{ -1 }(n+2)-{ tan }^{ -1 }={ tan }^{ -1 }\left( \frac { n+2-1 }{ 1+(n+2)1 } \right) \)
\(={ tan }^{ -1 }\left( \frac { n+1 }{ n+3 } \right) ={ tan }^{ -1 }\left( \frac { 1+\frac { 1 }{ n } }{ 1+\frac { 3 }{ n } } \right) \)
\(\Rightarrow { S }_{ \infty }={ tan }^{ -1 }(1)=\frac { \pi }{ 4 } \)
A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months, his saving of immediately previous month. His total saving from the start of service will be Rs. 11040 after
- (a)
19 months
- (b)
20 months
- (c)
21 months
- (d)
18 months
Let the time taken to save Rs.11040 be (n + 3) months.
For first 3 months, he saves Rs.200 each month.
∴ In (n + 3) month,
\(3\times 200+\left\{ \frac { n }{ 2 } \left[ 2(240)+(n-1)\times 40 \right] \right\} =11040\)
⇒ \(600+\frac { n }{ 2 } \left\{ 40(12+n-1) \right\} =11040\)
⇒ 600+20n(n+11)=11040
⇒ n2+11n-522=0
⇒ n2+29n-18n-522=0
⇒ n(n+29)-18(n+29)=0
⇒ (n-18)(n+29)=0
∴ n=18 [neglecting n=-29]
∴ Total time taken=(n+3)=21 months
An infinite GP has first term x and sum 5, then x belongs to
- (a)
x < - 10
- (b)
-10< x < 0
- (c)
0 < x < 10
- (d)
x > 10
\(\because a = x,S=5\)
If r be the common ratio, then \(\frac { a }{ 1-r } =5\)
\(\Rightarrow \frac { x }{ 1-r } =5\)
\(\Rightarrow \quad 1-r=\frac { x }{ 5 } \)
\(\Rightarrow \quad r=\left( 1-\frac { x }{ 5 } \right) \)
\(\because \) -1 < r < 1 (for infinite GP)
\(\Rightarrow \quad -1<1-\frac { x }{ 5 } <1\)
\(\Rightarrow \quad -2<-\frac { x }{ 5 } <0\)
\(\Rightarrow \quad 10>x>0\)
\(\therefore\) 0< x < 10
If a, b, c are in AP, then the equation (a - b) x2 + (c - a) x + (b - c) = 0 has two roots which are
- (a)
rational and equal
- (b)
rational and distinct
- (c)
irrational conjugates
- (d)
complex conjugates
\(\because\) a,b,c are in AP
\(\therefore\) b-a = c-b
or a-b = b - c ....(i)
Given equation is
(a-b) x2 + (c - a) x + (b - c) = 0 .....(ii)
at x = 1,
(a-b) + (c-a) + (b-c) = 0
Hence, x = 1 is one root of Eq. (ii), if other root is \(\beta\)
Then, \(1\times \beta =\frac { b-c }{ a-b } =1\) {from Eq. (i)}
\(\therefore\) \(\beta\) = 1
Hence, roots rational and equal.
Given that n arithmetic means are inserted between two sets of numbers a, 2b and 2a, b, where a, b \(\in \) R. Suppose further that mth mean between these two sets of numbers is same, then the ratio, a : b equals
- (a)
n - m + 1 : m
- (b)
n - m + 1 : n
- (c)
m : n -m + 1
- (d)
n : n - m + 1
a, A1,A2,......,An , 2b
2b = (n+2) th term = a + (n + 2 - 1) d
\(d=\frac { 2b-a }{ n+1 } \)
\({ A }_{ m }=a+m\left( \frac { 2b-a }{ n+1 } \right) \)
Again 2a, B1,B2,.....,Bn, b
b = (n + 2)th term = 2a + (n + 2-1) D
\(\therefore \quad D=\frac { b-2a }{ n+1 } \)
\(\therefore \quad { B }_{ m }=2a+mD=2a+m\left( \frac { b-2a }{ n+1 } \right) \)
\(\Rightarrow\) Am = Bm
\(\Rightarrow \quad a+m\left( \frac { 2b-a }{ n+1 } \right) =2a+m\left( \frac { b-2a }{ n+1 } \right) \)
\(\Rightarrow \) m(a+b) = a(n+1)
or a(n-m+1) = bm
\(\therefore\) a : b = m : n - m + 1
Let a, b, c be three positive prime numbers. The progression in which \(\sqrt { a } ,\sqrt { b } ,\sqrt { c } \) can be three terms (not necessarily consecutive) is
- (a)
AP
- (b)
GP
- (c)
HP
- (d)
none of these
If in AP
\(\sqrt { b } =\sqrt { a } +(m-1)d\)
and \(\sqrt { c } =\sqrt { a } +(n-1)d\)
\(\therefore \quad \frac { \sqrt { b } -\sqrt { a } }{ \sqrt { c } -\sqrt { a } } =\frac { m-1 }{ n-1 } =a\) ratioal number but a, b, c are prime, then \(\frac { \sqrt { b } -\sqrt { a } }{ \sqrt { c } -\sqrt { a } } \) is irrational
\(\therefore\) irrational \(\neq \) rational
So\(\sqrt { a } ,\sqrt { b } ,\sqrt { c } \) are not in AP similarly they are not in GP or HP
The sum to infinity of the series, \(1+2\left( 1-\frac { 1 }{ n } \right) +3{ \left( 1-\frac { 1 }{ n } \right) }^{ 2 }+....\) is
- (a)
n2
- (b)
n ( n+1)
- (c)
\(n{ \left( 1+\frac { 1 }{ n } \right) }^{ 2 }\)
- (d)
none of these
Let \(S=1+2\left( 1-\frac { 1 }{ n } \right) +3{ \left( 1-\frac { 1 }{ n } \right) }^{ 2 }\) ........(i)
\(\therefore \left( 1-\frac { 1 }{ n } \right) S=1\left( 1-\frac { 1 }{ n } \right) +2{ \left( 1-\frac { 1 }{ n } \right) }^{ 2 }\) ..............(ii)
(i)-(ii) gives \({S\over n}=1+\left( 1-\frac { 1 }{ n } \right) +{ \left( 1-\frac { 1 }{ n } \right) }^{ 2 }+....\infty \)
\(=\frac { 1 }{ 1-\left( 1-\frac { 1 }{ n } \right) } =n\)
\(\Rightarrow \quad S={ n }^{ 2 }\)
If an AP,a7=9 if a1a2a7 is least, the common difference is
- (a)
\(\frac { 13 }{ 20 } \)
- (b)
\(\frac { 23 }{ 20 } \)
- (c)
\(\frac { 33 }{ 20 } \)
- (d)
\(\frac { 43 }{ 20 } \)
Let d be the common difference
\(\because\) a7 = 9
\(\therefore\) a1 + 6d = 9
Let D = a1 a2 a7
=(9 - 6d) (9-5d) 9
\(=270\left\{ { \left( d-\frac { 33 }{ 20 } \right) }^{ 2 }-\frac { 9 }{ 400 } \right\} \)
For least value of D,
\(d-\frac { 33 }{ 20 } =0\)
\(\therefore\) d = 33/20
If three unequal numbers p, q, r are in HP and their squares are in AP, then the ratio p : q : r is
- (a)
\(1-\sqrt { 3 } :-2:1+\sqrt { 3 } \)
- (b)
\(1:\sqrt { 2 } :-\sqrt { 3 } \)
- (c)
\(1:-\sqrt { 2 } :\sqrt { 3 } \)
- (d)
\(1+\sqrt { 3 } :-2:1-\sqrt { 3 } \)
By hypothesis,
\(q=\frac { 2pr }{ p+r } \)
\(\Rightarrow \frac { q }{ 2 } =\frac { pr }{ p+r } =k(say)\quad \)
\(\Rightarrow \quad q=2k,pr=k(p+r)\)
Also, p2,q2 ,r2 are in AP
2q2 = p2 + r2 = (p + r)2 - 2 pr
8k2 = (p + r)2-2k (p + r)
(p+ r)2 -2(p+ r)k -.8k2 =0
\(\Rightarrow\)p + r= 4k, - 2k
\(\Rightarrow\)P + r =4k,pr = 4k2
and (p-r)2 = (p+r)2 - 4pr = 0
\(\therefore\) p = r
This is against the hypothesis
\(\therefore\) p+r=--2k,pr=-2k2
Now, (p - r)2 = 12 k2 \(\Rightarrow\) P - r = \(\pm \) 2\(\sqrt { 3 } \) k
Combine it with p + r = - 2 k to get
p = (- 1\(\pm \)\(\sqrt { 3 } \)) k and r = (-1 \(\pm \)\(\sqrt { 3 } \)) k
\(\therefore\) p:q:r::-1\(\pm \)\(\sqrt { 3 } \):2:-1\(\pm \)\(\sqrt { 3 } \)
or p: q : r:: 1\(\pm \)\(\sqrt { 3 } \):-2: 1 \(\pm \) \(\sqrt { 3 } \)
the pth term Tp of HP is q (p+q) and qth term Tq is p(p+q) when p>1, q>1, then
- (a)
Tp+q =pq
- (b)
Tpq=p+q
- (c)
Tp+q>Tpq
- (d)
Tpq>Tp+q
Tp of AP=\(\frac { 1 }{ q(p+q) } =A+(p-1)D\) .....(i)
Tq of AP=\(\frac { 1 }{ p(p+q) } =A+(q-1)D\) ......(ii)
\(\frac { 1 }{ { T }_{ p+q } } =A+(p+q-1)D\)
and \(\frac { 1 }{ { T }_{ pq } } =A+(pq-1)D\)
Now, solving Eqs. (i) and (ii), we get
A=D=\(\frac { 1 }{ pq(p+q) } \)
\(\therefore \frac { 1 }{ { T }_{ p+q } } =A+(p+q-1)D=(p+q)D=\frac { 1 }{ pq } \)
\(and\frac { 1 }{ { T }_{ pq } } =A+f(p+q-1)D=pqD=\frac { 1 }{ p+q } \quad \)
\(\Rightarrow\) Tp+q=pq and Tpq=p+q
Also, \(\because\) p > a, q > 1
\(\therefore\) pq > p +q
ie, Tp+q>Tpq
If a, b, c be three unequal positive quantities in HP, then
- (a)
a100 + c100 > 2b100
- (b)
a3 + c3 > 2b3
- (c)
a5 + c5 > 2b5
- (d)
a2 + c2 > 2b2
\(\because \quad { ({ a }^{ n/2 }-{ c }^{ n/2 }) }^{ 2 }>0\)
\(\Rightarrow \quad { a }^{ n }+{ c }^{ n }>2{ a }^{ n/2 }{ c }^{ n/2 }\) .....(i)
GM > HM
\(\sqrt { ac } >b\)
\(\Rightarrow \quad { 2(ac) }^{ n/2 }>{ 2b }^{ n }\) ......(ii)
From Eqs. (i) and (ii),
an + cn > 2bn
Putting n = 2, 3, 5, 100
If the arithmetic mean of two positive numbers a and b (a>b) is twice their geometric mean, then a : b is
- (a)
2 + \(\sqrt { 3 } \) : 2 - \(\sqrt { 3 } \)
- (b)
7 + 4\(\sqrt { 3 } \) : 1
- (c)
1 : 7 - 4 \(\sqrt { 3 } \)
- (d)
2: \(\sqrt { 3 } \)
From the given condition \(\frac { 1 }{ 2 } (a+b)=2\sqrt { ab } \Rightarrow \sqrt { \frac { a }{ b } } +\sqrt { \frac { b }{ a } } =4\)
Putting \(y=\sqrt { \frac { a }{ b } } \)
\(\Rightarrow \quad y+\frac { 1 }{ y } =4\)
\(\Rightarrow \quad y=\frac { 4\pm 2\sqrt { 3 } }{ 2 } \) .....(i)
\(\Rightarrow \quad \sqrt { \frac { a }{ b } } =2\pm \sqrt { 3 } \)
Since, the roots of Eq. (i) are \(\sqrt { \frac { a }{ b } } and\quad \sqrt { \frac { b }{ a } } \) , we have
\(\frac { \sqrt { a/b } }{ \sqrt { b/a } } =\frac { 2+\sqrt { 3 } }{ 2-\sqrt { 3 } } \Rightarrow \frac { a }{ b } =\frac { 2+\sqrt { 3 } }{ 2-\sqrt { 3 } } \Rightarrow a:b=2+\sqrt { 3 } :2-\sqrt { 3 } \)
From \(\sqrt { a/b } =2+\sqrt { 3 } ,\) we get
\(\frac { a }{ b } =(2+\sqrt { 3 } )^{ 2 }=7+4\sqrt { 3 } \Rightarrow a:b=7+4\sqrt { 3 } :1\)
\(From\quad \sqrt { \frac { b }{ a } } =2-\sqrt { 3 } ,we\quad get\)
\(\frac { b }{ a } ={ (2-\sqrt { 3 } ) }^{ 2 }=7-4\sqrt { 3 } \)
\(\Rightarrow \quad a:b=1:7-4\sqrt { 3 } \)
Suppose P is the first of n (n > 1) AM's between two positive numbers a and b; q the first of n HM's between the same two numbers.
The value of p is
- (a)
\(\frac { na+b }{ n+1 } \)
- (b)
\(\frac { na-b }{ n+1 } \)
- (c)
\(\frac { nb+a }{ n+1 } \)
- (d)
\(\frac { nb-a }{ n+1 } \)
Since, n+1 > n-1
\(\Rightarrow \quad { \left( \frac { n+1 }{ n-1 } \right) }^{ 2 }>1\)
\(\Rightarrow \quad p{ \left( \frac { n+1 }{ n-1 } \right) }^{ 2 }>p\) .....(i)
\(Now,\quad p=a+d=a+\frac { (b-a) }{ (n+1) } =\left( \frac { an+b }{ n+1 } \right) \)
Suppose P is the first of n (n > 1) AM's between two positive numbers a and b; q the first of n HM's between the same two numbers.
Final conclusion is
- (a)
q lies between p and \(\left( \frac { n+1 }{ n-1 } \right) p\)
- (b)
q lies between p and \({ \left( \frac { n+1 }{ n-1 } \right) }^{ 2 }p\)
- (c)
q does not lies between p and \(\left( \frac { n+1 }{ n-1 } \right) p\)
- (d)
q does not lies between p and \({ \left( \frac { n+1 }{ n-1 } \right) }^{ 2 }\)
Since, n+1 > n-1
\(\Rightarrow \quad { \left( \frac { n+1 }{ n-1 } \right) }^{ 2 }>1\)
\(\Rightarrow \quad p{ \left( \frac { n+1 }{ n-1 } \right) }^{ 2 }>p\) .....(i)
\(Now,\quad p=a+d=a+\frac { (b-a) }{ (n+1) } =\left( \frac { an+b }{ n+1 } \right) \)
and \(\frac { 1 }{ q } =\frac { 1 }{ a } +D=\frac { 1 }{ a } +\frac { \frac { 1 }{ b } -\frac { 1 }{ a } }{ n+1 } =\frac { (a+bn) }{ ab(n+1) } \)
\(q=\frac { ab(n+1) }{ (a+bn) } \)
Also, \(\frac { p }{ q } =p\times \frac { 1 }{ q } =\left( \frac { an+b }{ n+1 } \right) \left( \frac { a+bn }{ ab(n+1) } \right) \)
\(=\frac { ({ a }^{ 2 }n+ab{ n }^{ 2 }+ab+{ b }^{ 2 }n) }{ ab({ n+1) }^{ 2 } } \)
\(=\frac { n\left( \frac { a }{ b } +\frac { b }{ a } \right) +{ n }^{ 2 }+1 }{ { (n+1) }^{ 2 } } \)
\(\therefore \quad \frac { P }{ q } -1=\frac { n\left( \frac { a }{ b } +\frac { b }{ a } -2 \right) }{ { (n+1) }^{ 2 } } \)
\(=n\frac { { \left( \sqrt { \frac { a }{ b } } -\sqrt { \frac { b }{ a } } \right) }^{ 2 } }{ { (n+1) }^{ 2 } } \)
\(=\frac { n }{ { (n+1) }^{ 2 } } { \left( \sqrt { \frac { a }{ b } } -\sqrt { \frac { b }{ a } } \right) }^{ 2 }\)
\(\because \quad \frac { p }{ q } -1>0\Rightarrow \frac { p }{ q } >1\)
\(\Rightarrow \) p>q
\(\therefore\) From Eqs. (i) and (ii),
\(q<p<{ \left( \frac { n+1 }{ n-1 } \right) }^{ 2 }p\quad \)
Also, for p=8, n=3, then q < 8 < 32
\(\therefore\) q does not lies between 8 and 32.
If A, G and H are respectively arithmetic, geometric and harmonic means between a and b both being unequal and positive. then \(A=\frac { a+b }{ 2 } \Rightarrow a+b=2A,G=\sqrt { ab } \Rightarrow ab={ G }^{ 2 }and\quad H=\frac { 2ab }{ a+b } \Rightarrow { G }^{ 2 }=AH\)
From aboe discussion we can say that a. b are the roots of the equation x2-- 2A x + G2 = 0
Now, quadratic equation x2--Px + Q=0 and quadratic equation a (b - c)x2 + b (c - a)x + c (a - b) = 0 have a root common and satisfy the relation b = \(\frac { 2ac }{ (a+c) } \),where a, b,c are real numbers.
The value of [P] is (where [.] donotes the greatest integer function)
- (a)
-2
- (b)
-1
- (c)
2
- (d)
1
Given a (b - c)x2 + b (c - a)x + c (a - b) = 0 .....(i)
\(\because\)a (b - c) + b(c - a) + c (a - b) = 0
\(\therefore\) x = 1 is a root of Eq. (i)
Let other root is a, then
\(1\times \alpha =\frac { c(a-b) }{ a(b-c) } \)
\(=\frac { c\left( a-\frac { 2ac }{ a+c } \right) }{ a\left( \frac { 2ac }{ a+c } -c \right) } =1\)
\(\therefore \quad \alpha =1\)
Hence, both roots of Eq. (i) are 1, 1
Then roots of x2 - Px + Q = 0 are also 1, 1
Then 1 + 1 =P, 1.1 =Q
\(\therefore\) P = 2, Q =1
\(\therefore\) [P]=[2]=2
If A, G and H are respectively arithmetic, geometric and harmonic means between a and b both being unequal and positive. then \(A=\frac { a+b }{ 2 } \Rightarrow a+b=2A,G=\sqrt { ab } \Rightarrow ab={ G }^{ 2 }and\quad H=\frac { 2ab }{ a+b } \Rightarrow { G }^{ 2 }=AH\)
From above discussion we can say that a. b are the roots of the equation x2-- 2A x + G2 = 0
Now, quadratic equation x2--Px + Q=0 and quadratic equation a ( b - c )x2 + b ( c - a )x + c ( a - b ) = 0 have a root common and satisfy the relation b = \(\frac { 2ac }{ (a+c) } \),where a, b, c are real numbers.
The ratio of the AM, GM and HM of the roots of the given quadratic equation is
- (a)
1 ; 2 : 3
- (b)
1 : 1 : 2
- (c)
2 : 2: 3
- (d)
1 : 1 : 1
AM : GM : HM = 1 : 1 : 1
If a sequence or series is not a direct form of an AP, GP, etc Then its nth term can not be determined. In such cases, we use the following steps to find the nth term (Tn) of the given sequence.
Step - I: Find the differences between the successive terms of the given sequence. If these differences are in AP, then take Tn = an2+ bn + c, where a, b, c are constants.
Step-II: If the successive differences finding in step I are in GP with common ratio r, then take Tn = a + bn + cr n-1, where a, b, c are constants.
Step - Ill : If the second successive differences (Differences of the differences) in step I are in AP, then take Tn = an3 + bn2 + cn + d, where a, b, c, d are constants.
Step-Iv : If the second successive differences (Differences of the differences) in step I are III GP, then take Tn = an2 + bn + c + dr n-1 ,where a, b, c, d are constants.
Now let sequences:
A: 1,b, 18,40,75,126, ...... B : 1,1,6,26,91,291,..... C : In 2 In 4, In 32, In 1024,......
The format of nth term (Tn) of the sequence C is
- (a)
an2 + bn + c
- (b)
an3 + bn2 + cn + d
- (c)
an + b + crn-1
- (d)
an2 + b + c + drn-1
For sequence A :
1,6, 18, 40, 75, 126, ...
5, 12, 22, 35, 51, ... (first successive difference)
7, 10, 13, 16, ... (second successive difference)
For sequence B :
1, 1, 6, 26, 91, 291, ...
0, 5, 20, 65, 200, ... (first successive difference)
5, 15, 45, 135, ... (second successive difference)
For sequence C :
In 2, In 4, In 32, In 1024, ....
or In 2, 2In 2, 5 In 2,10 In 2, .....
In 2, 3In 2, 5 In 2,...... (first successive difference)
\(\because\) First successive difference of sequence C in AP, then Tn=an2 + bn + c
The sum of the squares of three distinct real numbers which are in strictly increasing GP is S2. If their sum is \(\alpha\) S.
\(\alpha\) 2 lies in
- (a)
\(\left( \frac { 1 }{ 3 } ,1 \right) \)
- (b)
(1, 2)
- (c)
\(\left( \frac { 1 }{ 3 } ,3 \right) \)
- (d)
\(\left( \frac { 1 }{ 3 } ,1 \right) \cup (1,3)\)
Let the three numbers in strictly increasing GP are \(\frac { a }{ r } ,a,ar(r>1)\)
According to passage
\(\frac { { a }^{ 2 } }{ { r }^{ 2 } } +{ a }^{ 2 }+{ a }^{ 2 }{ r }^{ 2 }={ S }^{ 2 }\)
or \({ a }^{ 2 }\left( \frac { 1 }{ { r }^{ 2 } } +1+{ r }^{ 2 } \right) ={ S }^{ 2 }\)
or \({ a }^{ 2 }\left( \frac { 1 }{ r } +1+r \right) \left( \frac { 1 }{ r } -1+r \right) ={ S }^{ 2 }\)....(i)
and \(\frac { a }{ r } +a+ar=aS\)
or \(a\left( \frac { 1 }{ r } +1+r \right) =a\quad S\quad \quad \)
\(\therefore \quad { a }^{ 2 }{ \left( \frac { 1 }{ r } +1+r \right) }^{ 2 }={ a }^{ 2 }{ s }^{ 2 }\) ....(ii)
Dividing Eq. (ii) by (i), then
\(\left( \frac { \frac { 1 }{ r } +1+r }{ \frac { 1 }{ r } -1+r } \right) ={ a }^{ 2 }\)
\(\Rightarrow \quad (1+r+{ r }^{ 2 })=a^{ 2 }(1-r+{ r }^{ 2 })\)
\(\\ \Rightarrow \quad ({ a }^{ 2 }-1){ r }^{ 2 }-({ a }^{ 2 }+1)r+{ a }^{ 2 }-1=0\) ....(iii)
or \({ r }^{ 2 }-\left( \frac { { a }^{ 2 }+1 }{ { a }^{ 2 }-1 } \right) r+1=0\quad \quad (\because { a }^{ 2 }\neq 1)\)
\(\because \quad r\quad is\quad real,{ \left( \frac { { a }^{ 2 }+1 }{ { a }^{ 2 }-1 } \right) }^{ 2 }-4.1.1>0\)
(\(\because\) The numbers in GP are distinct)
\(\Rightarrow \quad \left( \frac { { a }^{ 2 }+1 }{ { a }^{ 2 }-1 } +2 \right) \left( \frac { { a }^{ 2 }+1 }{ { a }^{ 2 }{ -1 } } -2 \right) >0\)
\(\Rightarrow \quad (3{ a }^{ 2 }-1)(-a^{ 2 }+3)>0\)
or \(\left( { a }^{ 2 }-\frac { 1 }{ 3 } \right) ({ a }^{ 2 }-3)<0\)
\(\therefore \quad \frac { 1 }{ 3 } <{ a }^{ 2 }<3\)
But a2 \(\neq \) 1
\(\therefore \quad { a }^{ 2 }\in \left( \frac { 1 }{ 3 } ,1 \right) \cup (1,3)\) ......(iv)
Let the sequence an be defined as follows :
a1=1,an=an-1+2 for n \(\ge\)2.
Write the corresponding series.
- (a)
1+3+5+7+9+.......
- (b)
1+5+9+13+17+.....
- (c)
1+4+7+10+13+.........
- (d)
1+6+11+16+21+......
We have, an=an-1+2 n \(\ge\)2 and a1=1
\(\therefore \) a2 =a1+2=1+2=3, a3=a2+2=3+2=5, a4 =a3+2=5+2=7,a5 =a4+2=7+2=9.
\(\therefore \)The corresponding series is 1+3+5+7+9+......
The number of divisors of 72,2025 and 1568 are in
- (a)
A.P
- (b)
G.P.
- (c)
A.G.P.
- (d)
None of these
72=23 x32; number of divisiors of 72 are 12.
2025 =34 x 52; number of divisors of 2025 are 15.
1568=25 x 72 ; number of divisors of 1568 are 18.
Now 12,15,18 are in A.P.
Let tr denotes the rth term of an A.P. Also, suppose that tm=\({1\over n}\) and tn=\({1\over m}\),(m\(\neq\)n), for some positive integers m and n, then which of the following is necessarily a root of the equation (l+m-2n)x2+(m+n-2l)x+(n+l-2m)=0?
- (a)
tn
- (b)
tm
- (c)
tm+n
- (d)
tmn
tm=a+(m-1)d= \({1\over n}\)...(i)
tn=a+(n-1)d= \({1\over m}\)...(ii)
Subtracting (ii) from (i), we get
(m-n)d =\({1\over n}-{1\over m}\Rightarrow (m-n)d={m-n\over mn}\)
\(\Rightarrow d={1\over mn}(\because m\neq n) \).........(iii)
tmn= a+(mn-1)d=a+(mn-1)x\({1\over mn}\)
\(=a-{1\over mn}+1\)
From (i) and (iii)
\(a+(m-1).{1\over mn}={1\over n}\Rightarrow a+{1\over n}-{1\over mn}={1\over n}\Rightarrow a={1\over mn}\)\(\Rightarrow t_{mn}=1\)
Find three arithmetic meansbetween 3 and 19.
- (a)
7,11,15
- (b)
5,10,15
- (c)
6,11,16
- (d)
5,7,9
Let A1,A2,A3,19 are in A.P. whose common difference is d=\({19-3\over 4}=4\)
\(\therefore A_1=3+d \Rightarrow A_1=7,A_2=3+2d \Rightarrow A_2=11;\)
\(A_3=3+3d\Rightarrow A_3=15.\)
Hence, the required A.M.'s are 7,11,15.
The sum of two numbers is \({13\over6}\). An ever number of arithmetic means are being inserted between them and their sum exceeds their number by1. Find the number of means inserred.
- (a)
8
- (b)
10
- (c)
12
- (d)
14
Let a and b be two numbers such that
a+b=\({13\over6}\)...(i)
Let A1, A2,....+A2n be 2n arithmetic means between a and b.
Then, A1+A2+...+A2n=2n \(({a+b\over2})\)
\(\Rightarrow A_1+A_2+..+A_{2n}=n(a+b)={13\over6}n\) [Using (i)]
It is given that A1+A2+..+A2n=2n+1
\(\therefore {13\over6}n=2n+1 \Rightarrow n=6\)
So number of means =2n+12
if a,b,c,d are in G.P., then \({(a^2+b^2+c^2)(b^2+c^2+d^2)\over (ab+bc+cd)^2}=\)
- (a)
1
- (b)
2
- (c)
3
- (d)
5
b=ar, c= ar2,d=ar3
\(\therefore a^2+b^2+c^2=a^2(1+r^2+r^4)\)
\(b^2+c^2+d^2=a^2r^2(1+r^2+r^4)\)
\(ab+bc+cd=a^2r(1+r^2+r^4)\)
\(\therefore {(a^2+b^2+c^2)(b^2+c^2+d^2)\over (ab+bc+cd)^2}=1\)
Find the sum of sequence 7,77,777,7777,.... to n terms.
- (a)
\({7\over9}[{10(10^n-1)\over 9}-n]\)
- (b)
\({7\over9}[{10(10^n+1)\over 9}-n]\)
- (c)
\({10^n-1\over 9}\)
- (d)
None of these
This is not a G.P., however, we can relate it to a G.P. by writing the terms as
Sn=7+77+777+7777+....to n terms
\(={7\over9}[9+99+999+9999+..to \ n \ terms]\)
\(={7\over9}[(10-1)+(10^2-1)+(10^3-1)+(10^4-1)+.. \ n \ terms]\)
\(={7\over9}[(10+10^2+10^3+.. \ n \ terms)-(1+1+1+...n \ terms)]\)
\(={7\over9}[{10(10^n-1)\over10-1}-n]={7\over9}[{10(10^n-1)\over 9}-n]\).
In a G.P. of even number of terms, the sum of all terms is 5 times the sum of the odd terms. The common ratio of the G.P. is
- (a)
\({-4\over 5}\)
- (b)
\({1\over5}\)
- (c)
4
- (d)
-5
Let us consider a G.P. a, ar, ar2,... with 2n terms,
we have \({a(r^{2n}-1)\over r-1}={5a((r^2)^n-1)\over r^2-1}\)
(Since common ratio of odd terms will be r2 and number of therms will be n)
\(\Rightarrow {a(r^{2n}-1)\over r-1}=5{a(r^{2n}-1)\over (r^2-1)}\Rightarrow a(r+1)=5a\Rightarrow r=4\)
The arithmetic mean of two numbers is 3times their geometric mean and the sum of the squares of the two numbers is 34. The two numbers are
- (a)
\(2\sqrt{3}+\sqrt{5},2\sqrt{3}-\sqrt{5}\)
- (b)
\(3+2\sqrt{2},3-2\sqrt{2}\)
- (c)
\(\sqrt{10}+\sqrt{7},\sqrt{10}-\sqrt{7}\)
- (d)
\(\sqrt{5}-2\sqrt{2},\sqrt{5}+2\sqrt{2}\)
a+b=3.2 \(\sqrt{ab} \Rightarrow {a\over b}-6\sqrt{a\over b}+1=0\)
\(\Rightarrow \sqrt{a\over b}=3\pm2\sqrt{2}\Rightarrow{a\over b}=(3\pm2\sqrt{2})^2=17\pm12\sqrt{2}\)
or \({a\over b}={3+2\sqrt{2}\over3-2\sqrt{2}}.{3+2\sqrt{2}\over3+2\sqrt{2}} \ or \ {3-2\sqrt{2}\over 3+2\sqrt{2}}.{3-2\sqrt{2}\over 3-2\sqrt{2}}\).
\(\Rightarrow a=(3\pm2\sqrt{2})k,b=(3\mp2\sqrt{2})k\)
As a2+b2=34\(\Rightarrow\) k=1, thus the two numbers are 3+2\(\sqrt{2}\) and 3-2\(\sqrt{2}\).
If x and y are positive real numbers and m, n are positive intefers, then the maximum value of \({x^my^n\over (1+x^{2m})(1+y^{2n})}\)is
- (a)
2
- (b)
\(1\over4\)
- (c)
\(1\over2\)
- (d)
1
A.M.\(\ge\) G.M.\(\Rightarrow {1+x^{2m}\over 2}\ge x^m\), and \( {1+y^{2n}\over 2}\ge y^n\)
\(\therefore { x^m\over 1+x^{2m}}.{ y^n\over 1+y^{2n}}\le {1\over 2}.{1\over 2}={1\over4}.\)
The A.M. between m and n and the G.M. between a and b are each equal to\({ma+nb\over m+n}\) . Then m=
- (a)
\({a\sqrt{b}\over \sqrt{a}+\sqrt{b}}\)
- (b)
\({b\sqrt{a}\over \sqrt{a}+\sqrt{b}}\)
- (c)
\({2a\sqrt{b}\over \sqrt{a}+\sqrt{b}}\)
- (d)
\({2b\sqrt{a}\over \sqrt{a}+\sqrt{b}}\)
\({m+n\over2}=\sqrt{ab}={ma+nb\over m+n}\)
Taking \(\sqrt{ab}={ma+mb\over m+n}\Rightarrow(m+n)\sqrt{ab}=ma+nb\)
\(\Rightarrow m(\sqrt{ab}-a)=n(b-\sqrt{ab})\)
\(\Rightarrow m(\sqrt{b}-\sqrt{a})\sqrt{a}=n(\sqrt{b}-\sqrt{a})\sqrt{b}\)
\(\Rightarrow {m\over \sqrt{b}}={n\over \sqrt{a}}=k(say)\).....(i)
Further\({m+n\over2}=\sqrt{ab}\)...(ii)
From (i) and (ii), we get
\({k\over 2}(\sqrt{b}+\sqrt{a})=\sqrt{ab}\)
\(\Rightarrow k={2\sqrt{ab}\over \sqrt{a}+\sqrt{b}}\Rightarrow m=k\sqrt{b}={2b\sqrt{a}\over \sqrt{a}+\sqrt{b}}\).
The 10th common term between the series 3+7+11+.. and 1+6+11+..is
- (a)
191
- (b)
193
- (c)
211
- (d)
186
The first common term is 11.
Now, the next common term is obtained by adding L.C.M. of the common difference 4 and 5 i.e.,20.
Therefore, 10th common term=T10 of the A.P. whose a=11 and d=20
\(\therefore T_{10}=a+9d=11+9(20)=191\)
Find the sum to n terms of the series: \({1\over1.3}+{1\over3.5}+{1\over5.7}+...\)
- (a)
\({n\over(2n-1)}\)
- (b)
\({n\over(2n+1)}\)
- (c)
n+1
- (d)
n-1
Let Tr be the rth term of the given series. Then,
\(T_r={1\over (2r-1)(2r+1)},r=1,2,3,...,n\)
\(\Rightarrow T_{ r }=\frac { 1 }{ 2 } \left( \frac { 1 }{ 2r-1 } -\frac { 1 }{ 2r+1 } \right) \)
\(\therefore\) Required sum
\(=\sum^{n}_{r=1}T_r={1\over2}[({1\over1}-{1\over3})+({1\over3}-{1\over5})+......+({1\over2n-1}-{1\over 2n+1})]\)
\(={1\over2}[1-{1\over 2n+1}]={n\over2n+1}\)
The sum of the series \({2\over1.2}+{5\over2.3}.2{10\over3.4}.2^2+{17\over4.5}2^3+.......\) to n terms is
- (a)
\(({n\over n+1})2^{n+1}\)
- (b)
\(({n+1\over n})2^{n+1}\)
- (c)
\(({n\over n+1})2^{n}\)
- (d)
\(({n+1\over n})2^{n}\)
The given series is
\(\sum^{n}_{r=1}{r^2+1\over r(r+1)}2^{r-1}=\sum^{n}_{r=1}({2r\over r+1}-{r-1\over r})2^{r-1}\)
\(=(n+1)[{n(n+1)\over 2}]-{n(n+1)(2n+1)\over 6}={n(n+1)(n+2)\over 6}\)
If 1.3+3.32+5.33+7.34+.. upto n terms is equal to 3+(n-1).3b, then b=
- (a)
n
- (b)
n-1
- (c)
2n-1
- (d)
n+1
S=1.3+3.32+5.33+7.34+....+(2n-1)3n ...........1
3S= 1.32+3.32+......+(2n-3)3n+(2n-1)3n+1...........2
Subtracting 2 from 1, we get
-2S=1.3+2[32+33+...+3n]-(2n-1)3n+1
\(\Rightarrow S=({2n-1\over 2})3^{n+1}-{2.3^2(3^{n-1}-1)\over2(3-1)}-{1.3\over2}\)
=3+(n-1)3n+1
But given S=3+(n-1)3b \(\therefore\) b=n+1
If in an A.P., Sn = qn2 and 5m = qm2, where Sr denotes the sum of r terms of the A.P., then Sq equals
- (a)
\(q^3\over2\)
- (b)
mnq
- (c)
q3
- (d)
(m + n)q2
We have Sn = qn2
\(\Rightarrow{n\over2}[2a+(n-1)d]=qn^2\)
\(\Rightarrow 2a+(n-1)d=2qn\)........(1)
Also, Sm = qm2
\(\Rightarrow 2a+(m-1)d=2qm\)......(2)
(i) - (ii) gives
(n - m)d = 2q(n - m) \(\Rightarrow\) d = 2q
Substituting d = 2q in (i), we get
2a = 2qn - (n - 1) x 2q = 2q
Now,Sq = \({q\over 2}[2a+(q-1)d]={q\over2}[2q+(q-1)\times2q]=q^3\)
Consider first three terms of a sequence Sn is [x - 1], [x - 3],which are in A.P.
Statement-I : The sixth term of Sn is 7 < third term.
Statement-II : If a, a + d, a + 2d, ... are in A.P. (d \(\neq\) 0), then sixth term is (a + 5d).
- (a)
If both Statement-I and Statement-II are true and Statement-II is the correct explanation of Statement -1.
- (b)
If both Statement-I and Statement-II are true but Statement-II is not the correct explanation of Statement -1.
- (c)
If Statement-I is true but Statement-II is false.
- (d)
If Statement-I is false and Statement-II is true.
Case I: If x < 1
Then, 1 - x, 3, 3 - x are in A.P.
\(\Rightarrow\) 6=4-2x \(\Rightarrow\) x=-l
\(\therefore\)Terms are 2, 3, 4. \(\therefore\) Sixth term = 7
Case II : If 1 < x < 3
Then, x - 1,3,3 - x are in A.P. \(\Rightarrow\) 6 = 2 (impossible)
Case III :If x > 3
Then, x-1, 3, x - 3 are in A.P.
\(\Rightarrow\) 6=2x-4 \(\Rightarrow\) x=5
Then, terms are 4, 3, 2.
\(\therefore\) Sixth term is -1.
Statement-I: If a, b, c are in A.P., then 2b = a + c.
Statement-II: If a, b, c are in A.P., then 10a, 10b, 10c are in G.P.
- (a)
If both Statement-I and Statement-II are true and Statement-II is the correct explanation of Statement -1.
- (b)
If both Statement-I and Statement-II are true but Statement-II is not the correct explanation of Statement -1.
- (c)
If Statement-I is true but Statement-II is false.
- (d)
If Statement-I is false and Statement-II is true.
When a, b, c are in A.P., then
b - a = c - b \(\Rightarrow\) 2b = a + c.
So, statement-I is true.
Again, when a, b, c are in A.P. then
10a, 10b, 10c are in G.P.
if \({10^b\over 10^a}={10^c\over10^b}\) i.e.,if 10b-a = 10c-b
i.e., if b - a = c - b i.e., if 2b = a + c, which is true.
For any odd integer n≥1,
\({ n }^{ 3 }-\left( n-1 \right) ^{ 3 }+...+\left( -1 \right) ^{ n-1 }{ 1 }^{ 3 }=\cfrac { 1 }{ A } \left( n+1 \right) ^{ 2 }\left( 2n-B \right) \).Find A-B
- (a)
2
- (b)
3
- (c)
4
- (d)
5
Since n is an odd integer (-1)n-1=1 and n-1,n-3,n-5 etc. are even integers. We have
= n3-(n-1)3+(n-2)3-(n-3)3...+(-1)n-113
= n3+(n-1)3+(n-2)3+....+13-2[(n-1)3+(n-3)3+...+23]
= n3+(n-1)3+(n-2)3+...+13\(-2\times { 2 }^{ 3 }\left[ \left( \cfrac { n-1 }{ 2 } \right) ^{ 3 }+\left( \cfrac { n-3 }{ 2 } \right) ^{ 3 }+...{ 1 }^{ 3 } \right] \)
[∵ n −1, n − 3 are even integers]
= \(\left[ \cfrac { n\left( n+1 \right) ^{ 2 } }{ 2 } \right] -16\left[ \cfrac { 1 }{ 2 } \left( \cfrac { n-1 }{ 2 } \right) \left( \cfrac { n-1 }{ 2 } +1 \right) \right] ^{ 2 }\)
= \(\cfrac { 1 }{ 4 } { n }^{ 2 }\left( n+1 \right) ^{ 2 }-16\cfrac { \left( n-1 \right) ^{ 2 }\left( n+1 \right) ^{ 2 } }{ 16\times 4 } \)
= \(\cfrac { 1 }{ 4 } \left( n+1 \right) ^{ 2 }\left[ { n }^{ 2 }-\left( n-1 \right) ^{ 2 } \right] =\cfrac { 1 }{ 4 } \left( n+1 \right) ^{ 2 }\left( 2n-1 \right) \)
A = 4, B = 1
A – B = 3
11th term of the series
\(\cfrac { { 1 }^{ 3 } }{ 1 } +\cfrac { { 1 }^{ 3 }+{ 2 }^{ 3 } }{ 1+3 } +\cfrac { { 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 } }{ 1+3+5 } +...\) will be
- (a)
25
- (b)
32
- (c)
36
- (d)
24
Obviously \({ T }_{ n }=\cfrac { { 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+...+{ n }^{ 3 } }{ 1+3+5+...upto\quad n\quad term } \)
= \(\cfrac { { \Sigma n }^{ 3 } }{ \frac { n }{ 2 } \left[ 2+\left( n-1 \right) 2 \right] } =\cfrac { 1 }{ 4 } \cfrac { { n }^{ 2 }\left( n+1 \right) ^{ 2 } }{ { n }^{ 2 } } =\cfrac { 1 }{ 4 } \left( { n }^{ 2 }+2n+1 \right) \)
N = 11
Ans = 36