5 Marks Questions

Question 15 Marks

Show that $\frac { 1 \times 2 ^ { 2 } + 2 \times 3 ^ { 2 } + \ldots \ldots + n \times ( n + 1 ) ^ { 2 } } { 1 ^ { 2 } \times 2 + 2 ^ { 2 } \times 3 + \ldots \ldots + n ^ { 2 } ( n + 1 ) } = \frac { 3 n + 5 } { 3 n + 1 }$

Answer

Given: $\frac { 1 \times 2 ^ { 2 } + 2 \times 3 ^ { 2 } + \ldots \ldots + n \times ( n + 1 ) ^ { 2 } } { 1 ^ { 2 } \times 2 + 2 ^ { 2 } \times 3 + \ldots \ldots + n ^ { 2 } ( n + 1 ) } = \frac { 3 n + 5 } { 3 n + 1 }$

$= \frac { \sum n ( n + 1 ) ^ { 2 } } { \sum n ^ { 2 } ( n + 1 ) } = \frac { \sum n \left( n ^ { 2 } + 2 n + 1 \right) } { \sum \left( n ^ { 3 } + n ^ { 2 } \right) }$

$= \frac { \sum n ^ { 3 } + 2 \sum n ^ { 2 } + \sum n } { \sum n ^ { 3 } + \sum n ^ { 2 } }$

$=\frac { \frac { n ^ { 2 } ( n + 1 ) ^ { 2 } } { 4 } + \frac { 2 n ( n + 1 ) ( 2 n + 1 ) } { 6 } + \frac { n ( n + 1 ) } { 2 } } { \frac { n ^ { 2 } ( n + 1 ) ^ { 2 } } { 4 } + \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 } }$

$=\frac { \frac { n ( n + 1 ) } { 2 } \left[ \frac { n ( n + 1 ) } { 2 } + \frac { 2 ( 2 n + 1 ) } { 3 } + 1 \right] } { \frac { n ( n + 1 ) } { 2 } \left[ \frac { n ( n + 1 ) } { 2 } + \frac { ( 2 n + 1 ) } { 3 } \right] }$

$= \frac { 3 n ^ { 2 } + 11 n + 10 } { 3 n ^ { 2 } + 7 n + 2 } = \frac { ( n + 2 ) ( 3 n + 5 ) } { ( n + 2 ) ( 3 n + 1 ) } = \frac { 3 n + 5 } { 3 n + 1 }$

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Question 25 Marks

Find the sum of the following series up to n terms: $\frac { 1 ^ { 3 } } { 1 } + \frac { 1 ^ { 3 } + 2 ^ { 3 } } { 1 + 3 } + \frac { 1 ^ { 3 } + 2 ^ { 3 } + 3 ^ { 3 } } { 1 + 3 + 5 } + \ldots \ldots$

Answer

Given: $\frac { 1 ^ { 3 } } { 1 } + \frac { 1 ^ { 3 } + 2 ^ { 3 } } { 1 + 3 } + \frac { 1 ^ { 3 } + 2 ^ { 3 } + 3 ^ { 3 } } { 1 + 3 + 5 } + \ldots \ldots$ up to n terms$\therefore a _ { n } = \frac { 1 ^ { 3 } + 2 ^ { 3 } + 3 ^ { 3 } + \ldots \ldots + n ^ { 3 } } { 1 + 3 + 5 + \ldots \ldots ( 2 n - 1 ) }$
$= {{\sum {{n^3}} } \over {{n \over 2}\left[ {2 + (n - 1)2} \right]}} = {{\sum {{n^3}} } \over {{n \over 2}(2n)}} = {{\sum {{n^3}} } \over {{n^2}}} = {{{n^2}{{(n + 1)}^2}} \over {4{n^2}}}$
$= \frac { 1 } { 4 } \left( n ^ { 2 } + 2 n + 1 \right)$
$\therefore {S_n} = \sum\limits_{k = 1}^n {{a_{_k}}} = \sum\limits_{k = 1}^n {{{{k^{^2}} + 2k + 1} \over 4}} $
$= \frac { 1 } { 4 } [(1^2 + 2.1 + 1) + (2^2 + 2.2 + 1) + (3^2 + 2.3 + 1) + ...... +(n^2 + 2n + 1)]$
$= {1 \over 4}\left[ {\sum {{n^2} + 2\sum {n + n} } } \right]$
$= \frac { 1 } { 4 } \left[ \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 } + \frac { 2 n ( n + 1 ) } { 2 } + n \right]$
$= \frac { n } { 4 } \left[ \frac { 2 n ^ { 2 } + 3 n + 1 + 6 n + 6 + 6 } { 6 } \right]$
$= \frac { n } { 24 } \left( 2 n ^ { 2 } + 9 n + 13 \right)$

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Question 35 Marks

If $S_1, S_2, S_3$ are the sum of first n natural no. their squares and their cubes respectively, show that $9 S _ { 2 } ^ { 2 } = S _ { 3 } \left( 1 + 8 S _ { 1 } \right)$.

Answer

$S _ { 1 } = \frac { n ( n + 1 ) } { 2 }$
$S _ { 2 } = \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 }$
$S _ { 3 } = \left( \frac { n ( n + 1 ) } { 2 } \right) ^ { 2 }$
R.H.S. $= S_3(1 + 8S_1)$
$= \frac { ( n ( n + 1 ) ) ^ { 2 } } { 2 } \left[ 1 + 8 \frac { n ( n + 1 ) } { 2 } \right]$
$= 9 \left( \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 } \right) ^ { 2 }$
$= 9 \mathrm { S } _ { 2 } ^ { 2 }$

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Question 45 Marks

If $a, b, c$ are in $A.P.; b, c, d$ are in G.P. and $\frac { 1 } { c } , \frac { 1 } { d } , \frac { 1 } { e }$ are in A.P., prove that $a, c, e$ are in G.P.

Answer

Since, $a, b, c$ are in A.P.$\therefore b - a = c - b$
$\Rightarrow 2b = a + c$
$\Rightarrow b = \frac { a + c } { 2 }$
Since, b, c, d are in G.P.
$\therefore \frac { c } { b } = \frac { d } { c }$
$\Rightarrow c^2 = bd……….(i)$
Also $\frac { 1 } { c } , \frac { 1 } { d } , \frac { 1 } { e }$ are in A.P.
$\therefore \frac { 1 } { d } - \frac { 1 } { c } = \frac { 1 } { e } - \frac { 1 } { d }$
$\Rightarrow \frac { 2 } { d } = \frac { 1 } { c } + \frac { 1 } { e }$
$\Rightarrow \frac { 2 } { d } = \frac { c + e } { c e }$
$\Rightarrow d = \frac { 2 c e } { c + e }$
Putting values of

and

in eq. $(i), c ^ { 2 } = \left( \frac { c + a } { 2 } \right) \left( \frac { 2 c e } { c + e } \right)$
$\Rightarrow c ^ { 2 } = \frac { \operatorname { ce } ( c + a ) } { c + e }$
$\Rightarrow c^2(c + e) = ec(c + a)$
$\Rightarrow c^2 + ce = ce + ae$
$\Rightarrow c^2 = ae$ which shows that $a, c, e$ are in G.P.

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Question 55 Marks

The ratio of the A.M. and G.M. of two positive numbers a and b is

Show that $a : b = \left( \begin{array} { c } { m + \sqrt { m ^ { 2 } - n ^ { 2 } } } \end{array} \right) : \left( m - \sqrt { m ^ { 2 } - n ^ { 2 } } \right)$

Answer

Given: $\frac { a + b } { 2 } : \sqrt { a b } = m : n$

$\Rightarrow \frac { a + b } { 2 \sqrt { a b } } = \frac { m } { n }$

By componendo and dividendo,

$\frac { a + b + 2 \sqrt { a b } } { a + b - 2 \sqrt { a b } } = \frac { m + n } { m - n }$

$\Rightarrow \frac { ( \sqrt { a } + \sqrt { b } ) ^ { 2 } } { ( \sqrt { a } - \sqrt { b } ) ^ { 2 } } = \frac { m + n } { m - n }$

$\Rightarrow \frac { \sqrt { a } + \sqrt { b } } { \sqrt { a } - \sqrt { b } } = \frac { \sqrt { m + n } } { \sqrt { m - n } }$

Again by componendo and dividendo,

$\frac { \sqrt { a } + \sqrt { b } + \sqrt { a } - \sqrt { b } } { \sqrt { a } + \sqrt { b } - \sqrt { a } + \sqrt { b } } = \frac { \sqrt { m + n } + \sqrt { m - n } } { \sqrt { m + n } - \sqrt { m - n } }$

$\Rightarrow \frac { 2 \sqrt { a } } { 2 \sqrt { b } } = \frac { \sqrt { m + n } + \sqrt { m - n } } { \sqrt { m + n } - \sqrt { m - n } }$

$\Rightarrow \frac { a } { b } = \frac { ( \sqrt { m + n } + \sqrt { m - n } ) ^ { 2 } } { ( \sqrt { m + n } - \sqrt { m - n } ) ^ { 2 } }$

$\Rightarrow \frac { a } { b } = \frac { m + n + m - n + 2 \sqrt { ( m + n ) ( m - n ) } } { m + n + m - n - 2 \sqrt { ( m + n ) ( m - n ) } }$

$\Rightarrow \frac { a } { b } = \frac { 2 m + 2 \sqrt { ( m + n ) ( m - n ) } } { 2 m - 2 \sqrt { ( m + n ) ( m - n ) } }$

$\Rightarrow \frac { a } { b } = \frac { m + \sqrt { ( m + n ) ( m - n ) } } { m - \sqrt { ( m + n ) ( m - n ) } }$

Therefore, $a : b = \left( \begin{array} { c } { m + \sqrt { m ^ { 2 } - n ^ { 2 } } } \end{array} \right) : \left( m - \sqrt { m ^ { 2 } - n ^ { 2 } } \right)$

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Maths 5 Marks Questions

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