MATHS 3 Marks Question

$\Rightarrow$ r² + r⁴ = 90

$\Rightarrow$ r⁴ + r² - 90 = 0

$\Rightarrow r ^ { 2 } = \frac { - 1 \pm \sqrt { ( 1 ) ^ { 2 } - 4 \times ( - 90 ) \times 1 } } { 2 \times 1 }$

$= \frac { - 1 \pm \sqrt { 1 + 360 } } { 2 } = \frac { - 1 \pm \sqrt { 361 } } { 2 }$

$= \frac { - 1 \pm 19 } { 2 }$=  

$\Rightarrow r ^ { 2 } = \frac { - 1 + 19 } { 2 } = \frac { 18 } { 2 } = 9$or $r ^ { 2 } = \frac { - 1 - 19 } { 2 } = \frac { - 20 } { 2 } = - 10$ which is not possible

Therefore, the common ratio is $r = \pm 3$ 

$\therefore \mathrm { S } _ { n } = \frac { a \left( r ^ { n } - 1 \right) } { r - 1 }$  

$\Rightarrow 315 = \frac { 5 \left( 2 ^ { n } - 1 \right) } { 2 - 1 }$  

$\Rightarrow \frac { 315 } { 5 } = 2 ^ { n } - 1$

$\Rightarrow$ 2^{n - 1} = 63

$\Rightarrow$ 2ⁿ = 64 = 2⁶ 

$\Rightarrow$ n = 6 

$\therefore a _ { 6 } = a r ^ { 6 - 1 } = 5 \times 2 ^ { 5 } = 5 \times 32 = 160$ 

Hence the number of terms=6 and the last term =160

f(1 + 2) = f(1) f(2) = 3 $\times$ 9 = 27
f(1 + 3) = f(1) f(3) = 3 $\times$ 27 = 81
L.H.S.
= f(1) + f(2) + f(3) + ...... + f(n)
= 3 + 9 + 27 + 81 + ... + n terms
= $\frac {3(3^n - 1)}{3-1} = \frac 32 (3^n - 1)$
= f(1) + f(2) + f(3) +----+ f(n)
= 3 + 9 + 27 + 1 +----+n term s
$= \frac { 3 \left( 3 ^ { n } - 1 \right) } { 3 - 1 } = \frac { 3 } { 2 } \left( 3 ^ { n } - 1 \right)$
ATQ
$\frac { 3 } { 2 } \left( 3 ^ { n } - 1 \right) = 120$
3ⁿ - 1 = 80
3ⁿ = 81
n = 4

${S_2} = {{2n} \over 2}\left[ {2a + (2n - 1)d} \right]$…..(ii)

And ${S_3} = {{3n} \over 2}\left[ {2a + (3n - 1)d} \right]$

Now, ${S_2} - {S_1} = {{2n} \over 2}\left[ {2a + (2n - 1)d} \right] - {n \over 2}\left[ {2a + (n - 1)d} \right]$

$\Rightarrow {S_2} - {S_1} = (n - {n \over 2})2a + \left[ {n(2n - 1) - {n \over 2}(n - 1)} \right]d$

$= na + {1 \over 2}\left[ {4{n^2} - 2n - {n^2} + n} \right]d$

$= {n \over 2}\left[ {2a + (3n - 1)d} \right] = {1 \over 3}\left\{ {{{3n} \over 2}\left[ {2a + (3n - 1)d} \right]} \right\} = {1 \over 3}{S_3}$   

$\Rightarrow $ 3(S₂ - S₁) = S₃
Hence proved.

Total letters in the third set = 4³ = 64

$\therefore$ Sequence of letters is 4, 16, 64, ……….  in G.P.

Here a = 4, $r = \frac { 16 } { 4 } = 4$ and n = 8 

$\therefore \quad S _ { n } = \frac { a \left( r ^ { n } - 1 \right) } { r - 1 }$

$= \frac { 4 \left( 4 ^ { 8 } - 1 \right) } { 8 - 1 }$

$= \frac { 4 } { 3 } ( 65536 - 1 )$

$= \frac { 4 } { 3 } \times 65535 = 87380$

 Hence, the total number of letters mailed = 87380

 The amount of postage on each letter = 50 paise

 Therefore total amount spent on postage = $87380 \times 0.50$ = Rs. 43690.

Also S_n = 3 + 7 + 13 + 21 + 31 + ...... + a_n-2 + a_n-1 + a_n  ……….(ii)

Subtracting eq. (i) from eq. (ii), 0 = 3 + ( 4 + 6 + 8 + 10 + ....... up to (n - 1) terms) - a_n 

$\Rightarrow a _ { n } = 3 + \frac { n - 1 } { 2 } [ 2 \times 4 + ( n - 2 ) \times 2 ]$  

$\Rightarrow a _ { n } = 3 + \frac { n - 1 } { 2 } [ 8 + 2 n - 4 ]$

$\Rightarrow$ a_n = 3 + (n - 1) (n + 2)

$\Rightarrow$ an = 3 + n² + n - 2

$\Rightarrow$ a_n = n² + n + 1

$\therefore$ ${S_n} = \sum\limits_{k = 1}^n {{a_{_k}}} = \sum\limits_{k = 1}^n {({k^{^2}}} + k + 1)$

= (1² + 1 + 1) + (2² + 2 + 1) + (3² + 3 + 1) + ...... +(n² + n + 1) 

= (1² + 2² + 3² + ....... + n²) + (1 + 2 + 3 + ...... + n) + n 

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

$= n \left[ \frac { 2 n ^ { 2 } + 3 n + 1 + 3 n + 3 + 6 } { 6 } \right]$ 

$= n \left[ \frac { 2 n ^ { 2 } + 6 n + 10 } { 6 } \right]$ 

$= \frac { n } { 3 } \left( n ^ { 2 } + 3 n + 5 \right)$ 

Therefore, Sum = $\frac{6}{9}\left[n-\frac{1\left(1-10^{-n}\right)}{9}\right]$ 

To prove: (aⁿ + bⁿ), (bⁿ + cⁿ), (cⁿ + dⁿ) are in G.P.

$\Rightarrow \frac { b ^ { n } + c ^ { n } } { a ^ { n } + b ^ { n } } = \frac { c ^ { n } + d ^ { n } } { b ^ { n } + c ^ { n } }$
Let $\frac { b } { a } = \frac { c } { b } = \frac { d } { c } = k$ 

$\therefore \frac { b } { a } = k$

$\Rightarrow$ b = ak 

And $\frac { c } { b } = k$

$\Rightarrow$ c = bk = (ak)k = ak²

Also $\frac { d } { c } = k$

$\Rightarrow$ d = ck = (ak²)k = ak³ 

Now, $\frac { b ^ { n } + c ^ { n } } { a ^ { n } + b ^ { n } } = \frac { c ^ { n } + d ^ { n } } { b ^ { n } + c ^ { n } }$ 

$\Rightarrow \frac { ( a k ) ^ { n } + \left( a k ^ { 2 } \right) ^ { n } } { a ^ { n } + ( a k ) ^ { n } } = \frac { \left( a k ^ { 2 } \right) ^ { n } + \left( a k ^ { 3 } \right) ^ { n } } { ( a k ) ^ { n } + \left( a k ^ { 2 } \right) ^ { n } }$

$\Rightarrow \frac { a ^ { n } k ^ { n } + a ^ { n } k ^ { 2 n } } { a ^ { n } + a ^ { n } k ^ { n } } = \frac { a ^ { n } k ^ { 2 n } + a ^ { n } k ^ { 3 n } } { a ^ { n } k ^ { n } + a ^ { n } k ^ { 2 n } }$

$\Rightarrow \frac { a ^ { n } k ^ { n } \left( 1 + k ^ { n } \right) } { a ^ { n } \left( 1 + k ^ { n } \right) } = \frac { a ^ { n } k ^ { 2 n } \left( 1 + k ^ { n } \right) } { a ^ { n } k ^ { n } \left( 1 + k ^ { n } \right) }$

$\Rightarrow$ kⁿ = kⁿ

Therefore, (aⁿ + bⁿ), (bⁿ + cⁿ), (cⁿ + dⁿ) are in G.P.

Here $\mathrm { S } = \frac { a \left( r ^ { n } - 1 \right) } { r - 1 }$

 P = a.ar.ar² ......... ar^n-1$= {a^n}.{r^{1 + 2 + 3 + ....... + (n - 1)}} = {a^n}.{r^{{{n(n - 1)} \over 2}}}$

and R $= \frac { 1 } { a } + \frac { 1 } { a r } + \frac { 1 } { a r ^ { 2 } } + \ldots \ldots \frac { 1 } { a r ^ { n - 1 } }$$= {{{r^{n - 1}} + {r^{n - 2}} + {r^{n - 3}} + .......... + 1} \over {a{r^{n - 1}}}}$

$= {{1({r^n} - 1)} \over {r - 1}}.{1 \over {a{r^{n - 1}}}}$$= {{{r^n} - 1} \over {a{r^{n - 1}}(r - 1)}}$

Now ${p^2}{R^n} = {{{a^{2n}}.{r^{n(n - 1)}}{{({r^n} - 1)}^n}} \over {{a^n}{r^{n(n - 1)}}{{(r - 1)}^n}}} = {{{a^n}{{({r^n} - 1)}^n}} \over {{{(r - 1)}^n}}} = {a^n}{\left( {{{{r^n} - 1} \over {r - 1}}} \right)^n} = {S^n}$  

Hence proved.

$\Rightarrow S _ { 4 } = \frac { 4 } { 2 } [ 2 \times 11 + ( 4 - 1 ) d ]$ 

$\Rightarrow$ 2(22 + 3d) = 56  

$\Rightarrow$ 22 + 3d = 28

$\Rightarrow$ 3d = 6  

$\Rightarrow$ d = 2 

Also, l + (l - d) + (l - 2d) + (l - 3d) = 112  

$\Rightarrow$ 4l - 6d = 112 

$\Rightarrow$ 4l = 112 + 6d

$\Rightarrow 4 l = 112 + 6 \times 2$  

$\Rightarrow$ 4l = 112 + 12 

$\Rightarrow$ 4l = 124

$\Rightarrow$ l = 31 

$\therefore$ a_n = a+ (n - 1)d  

$\Rightarrow 31 = 11 + ( n - 1 ) \times 2$  

$\Rightarrow$ 2(n - 1) = 20 

$\Rightarrow$ n - 1 = 10

$\Rightarrow$ n = 11 

$\Rightarrow$ a(1 + r + r²) = 56  ......(i)

According to question,  a - 1, ar - 7, ar² - 21  are in A.P.

$\therefore$ (ar - 7) - (a - 1) = (ar² - 21) - (ar - 7) 

$\Rightarrow$ ar - 7 - a + 1 = ar² - 21 - ar + 7   

$\Rightarrow$ ar - a - 6 = ar² - ar - 14

$\Rightarrow$ ar² - 2ar + a = 8

$\Rightarrow$ a(r² - 2r + 1) = 8 ….....(ii)

Dividing eq. (i) by eq. (ii), $\frac { a \left( 1 + r + r ^ { 2 } \right) } { a \left( r ^ { 2 } - 2 r + 1 \right) } = \frac { 56 } { 8 }$

$\Rightarrow$ 1 + r + r² = 7r² - 14r + 7 

$\Rightarrow$ 6r² - 15r + 6 = 0 

$\Rightarrow$ 2r² - 5r + 2 = 0   

$\Rightarrow r = \frac { - ( - 5 ) \pm \sqrt { ( - 5 ) ^ { 2 } - 4 \times 2 \times 2 } } { 2 \times 2 }$ 

$= \frac { 5 \pm \sqrt { 25 - 16 } } { 4 } = \frac { 5 \pm \sqrt { 9 } } { 4 } = \frac { 5 \pm 3 } { 4 }$=  

$\Rightarrow r = \frac { 5 + 3 } { 4 } = \frac { 8 } { 4 } = 2$or $r = \frac { 5 - 3 } { 4 } = \frac { 2 } { 4 } = \frac { 1 } { 2 }$ 

Putting r = 2  in eq. (i), a(1 + 2 + 2²) = 56   

$\Rightarrow a = \frac { 56 } { 7 } = 8$ 

Then the required numbers are 8, 16, 32.

Putting $r = \frac { 1 } { 2 }$ in eq. (i), $a \left( 1 + \frac { 1 } { 2 } + \frac { 1 } { 4 } \right) = 56$   

$\Rightarrow a \times \frac { 7 } { 4 } = 56$  

$\Rightarrow$ a = 32 

Then the required numbers are 32, 16, 8.

and a_m-n = a + (m - n - 1)d  …..(ii)

To prove: a_m+n + a_m-n = 2a_m

Adding eq. (i) and (ii), we get

a_m+n + a_m-n = a + (m + n - 1)d + a +(m - n - 1)d   

$\Rightarrow$ a_m+n + a_m-n = a + md + nd - d + a + md - nd - d 

$\Rightarrow$ a_m+n + a_m-n = 2a + 2md + -2d

$\Rightarrow$ a_m+n + a_m-n = 2(a + md - d)

$\Rightarrow$ a_m+n + a_m-n = 2[a + (m - 1)d] 

$\Rightarrow$ a_m+n + a_m-n = 2a_m

Since the bacteria doubles itself after each hour, then the sequence of bacteria after each hour is a G.P.

Here a = 30 and r = 2

$\therefore$ Bacteria at the end of 2^nd hour = $30 \times 2 ^ { 3 - 1 } = 30 \times 2 ^ { 2 } = 120$

And Bacteria at the end of 4^th hour = $30 \times 2 ^ { 5 - 1 } = 30 \times 2 ^ { 4 } = 480$

And Bacteria at the end of nth hour =${a_{n + 1}} = 30({2^{(n + 1) - 1}}) = 30({2^n})$

Therefore  A = $\frac { a + b } { 2 }$ and G = $\sqrt { a b }$

Now, $\mathrm { A } \pm \sqrt { ( \mathrm { A } + \mathrm { G } ) ( \mathrm { A } - \mathrm { G } ) } = \mathrm { A } \pm \sqrt { \mathrm { A } ^ { 2 } - \mathrm { G } ^ { 2 } }$

= $\frac { a + b } { 2 } \pm \sqrt { \left( \frac { a + b } { 2 } \right) ^ { 2 } - ( \sqrt { a b } ) ^ { 2 } }$

= $\frac { a + b } { 2 } \pm \sqrt { \frac { a ^ { 2 } + b ^ { 2 } + 2 a b } { 4 } - a b }$

= $\frac { a + b } { 2 } \pm \sqrt { \frac { a ^ { 2 } + b ^ { 2 } + 2 a b - 4 a b } { 4 } }$

= $\frac { a + b } { 2 } \pm \sqrt { \frac { ( a - b ) ^ { 2 } } { 4 } } = \frac { a + b } { 2 } \pm \frac { a - b } { 2 }$

= $\frac { a + b } { 2 } + \frac { a - b } { 2 }$ and $\frac { a + b } { 2 } - \frac { a - b } { 2 }$

= $\frac { a + b + a - b } { 2 }$ and $\frac { a + b - a + b } { 2 }$

= $\frac { 2 a } { 2 } = a$ and $\frac { 2 b } { 2 } = b$

$a^{n+1}+b^{n+1}=a^{n+\frac{1}{2}} b^{\frac{1}{2}}+a^{\frac{1}{2}} b^{n+\frac{1}{2}}$
$a^{n+1}-a^{n+\frac{1}{2}} b^{\frac{1}{2}}=a^{\frac{1}{2}} b^{n+\frac{1}{2}}-b^{n+1}$

${a^{n + \frac{1}{2}}}\left( {{a^{\frac{1}{2}}} - {b^{\frac{1}{2}}}} \right) = {b^{n + \frac{1}{2}}}\left( {{a^{\frac{1}{2}}} - {b^{\frac{1}{2}}}} \right)$

$\left(\frac{a}{b}\right)^{n+\frac{1}{2}}=1$

$\left(\frac{a}{b}\right)^{n+\frac{1}{2}}=\left(\frac{a}{b}\right)^{0}$

$n+\frac{1}{2}=0$

$n=\frac{-1}{2}$

Here, first term of G.P. is a

and a_n = b

$\Rightarrow$ ar^n-1 = b ……….(i)

Given: P = a.ar.ar².ar³ ..... ar^n-1

$\Rightarrow$ P = aⁿ.r^{1+2+3+......+ n-1}

$\Rightarrow p = {a^n}{r^{{{n(n - 1)} \over 2}}}$

$\Rightarrow$ ${p^2} = {a^{2n}}{r^{n(n - 1)}} = {\left[ {aa{r^{n - 1}}} \right]^n}$ [Squaring both sides]

$\Rightarrow$ P² = (ab)ⁿ [From eq. (i)]

 Hence proved

$\therefore$ a_p = a

$\Rightarrow$ AR^p-1 = a ……….(i)

a_q = b

$\Rightarrow$ AR^q-1 = b ……….(ii)

a_r = c

$\Rightarrow$ AR^r-1 = c ……….(iii)

Now, L.H.S. = a^q-rb^r-pc^p-q = (AR^p-1)^q-r.(AR^q-1)^r-p.(AR^r-1)^p-q

= A^q-rR^(p-1)(q-r).A^r-pR^(q-1)(r-p).A^p-qR^(r-1)(p-q)

= A^q-r+r-p+p-qR^{pq-pr-q+r+qr-pq-r+p+pr-qr-p+q}

= A⁰R⁰= 1 $\times$ 1 = 1 = R.H.S.

$( a \times \mathrm { A } ) , ( a r \times \mathrm { AR } ) _ { : } \left( a r ^ { 2 } \times \mathrm { AR } ^ { 2 } \right),$........,$\left( a r ^ { n - 1 } \times \mathrm { AR } ^ { n - 1 } \right)$

$\Rightarrow$ (A), (aAr^R), (aAr²R²), ....., (aAr^n-1R^n-1)  are in G.P.

Now ${{{a_2}} \over {{a_1}}} = {{aArR} \over {aA}} = rR$ and ${{{a_3}} \over {{a_2}}} = {{aA{r^2}{R^2}} \over {aArR}} = rR$

Since the ratio of the two succeeding terms are same, the resulting sequence is also in G.P

 and common ratio = $\frac { a \mathrm { A } r \mathrm { R } } { a \mathrm { A } } = r \mathrm { R }$​​​​​​​

$\Rightarrow$ Sn = 8(1 +11 + 111 + 1111 + ...... up to n terms)

$\Rightarrow S _ { n } = \frac { 8 } { 9 }$(9 + 99 + 999 + 9999 + ....... up to n terms)  

$\Rightarrow S _ { n } = \frac { 8 } { 9 } \left[ ( 10 - 1 ) + \left( 10 ^ { 2 } - 1 \right) + \left( 10 ^ { 3 } - 1 \right) + \ldots . \text { up to } n \text { terms } \right]$

$\Rightarrow \mathrm { S } _ { n } = \frac { 8 } { 9 }$[(10 + 10² + 10³ + ..... up to n terms) - (1 + 1 + 1 + ...... up to n terms)]

$\Rightarrow S _ { n } = \frac { 8 } { 9 } \left[ \frac { 10 \times \left( 10 ^ { n } - 1 \right) } { 10 - 1 } - n \right]$

$= \frac { 8 } { 9 } \left[ \frac { 10 } { 9 } \left( 10 ^ { n } - 1 \right) - n \right]$

$= \frac { 80 } { 81 } \left( 10 ^ { n } - 1 \right) - \frac { 8 } { 9 } n$

Given: a + ar = -4 

$\Rightarrow$ a(1 + r) = -4 ……..(i)

And a₅ = 4a₃

$\Rightarrow$ ar⁴ = 4ar²

$\Rightarrow$ r² = 4

$\Rightarrow r = \pm 2$

Putting r = 2 in eq. (i), we get a(1 + 2) = -4 

$\Rightarrow a = \frac { - 4 } { 3 }$

Therefore, required G.P. is $\frac { - 4 } { 3 } , \frac { - 8 } { 3 } , \frac { - 16 } { 3 } , \dots$

Putting r = -2  in eq. (i), we get a(1 - 2) = -4 

$\Rightarrow$ a = 4

Therefore, required G.P. is 4, -8, 16, -32,....... 

$\Rightarrow a r ^ { 6 } = 64$

$\Rightarrow 729{r^6} = 64$

$\Rightarrow {r^6} = {{64} \over {729}} = {\left( {{2 \over 3}} \right)^6}$

$\Rightarrow r = \frac { 2 } { 3 }$

$\Rightarrow S _ { n } = \frac { a \left( 1 - r ^ { n } \right) } { 1 - r }$ when r < 1

$\Rightarrow S _ { 7 } = \frac { 729 \left[ 1 - \left( \frac { 2 } { 3 } \right) ^ { 7 } \right] } { 1 - \frac { 2 } { 3 } } = \frac { 729 \left[ 1 - \frac { 128 } { 2187 } \right] } { \frac { 3 - 2 } { 3 } }$

$\Rightarrow S _ { 7 } = 729 \times 3 \left( \frac { 2187 - 128 } { 2187 } \right)$

$\Rightarrow \mathrm { S } _ { 7 } = \frac { 729 \times 3 \times 2059 } { 2187 } = 2059$

According to question,$\frac { a } { l ^ { r } } + a + a r = \frac { 39 } { 10 }$……….(i)

And$\frac { a } { r } \times a \times r = 1$

$\Rightarrow a ^ { 3 } = 1$
$\Rightarrow a = 1$

Putting value of a in eq. (i),
$\Rightarrow$ 10+ 10r + 10r² = 39r
$\Rightarrow$ 10r² - 29r +10  = 0
$\Rightarrow r = \frac { - ( - 29 ) \pm \sqrt { ( - 29 ) ^ { 2 } - 4 \times 10 \times 10 } } { 2 \times 10 }$
$\Rightarrow r = \frac { 29 \pm \sqrt { 841 - 400 } } { 20 }$
$\Rightarrow r = \frac { 29 \pm 21 } { 20 }$
Taking $r = \frac { 29 + 21 } { 20 } = \frac { 50 } { 120 } = \frac { 5 } { 2 }$ and

then the first three terms are $\frac { 1 } { 5 / 2 } , 1,1 \times \frac { 5 } { 2 }$

$\Rightarrow \frac { 2 } { 5 } , 1 , \frac { 5 } { 2 }$
Taking $r = \frac { 29 - 21 } { 20 } = \frac { 8 } { 20 } = \frac { 2 } { 5 }$

 then first three terms are $\frac { 1 } { 2 / 5 } , 1,1 \times \frac { 2 } { 5 }$

$\Rightarrow\frac { 5 } { 2 } , 1 , \frac { 2 } { 5 }$

= 22 + (3 + 3² + 3³ +....... +3¹¹) ……….(i)

Here 3, 3²,3³ ....... ,3¹¹is in G.P.

$\therefore$a = 3 and r = $\frac { 3 ^ { 2 } } { 3 } = 3$
$\mathrm { S } _ { n } = \frac { 3 \left( 3 ^ { 11 } - 1 \right) } { 3 - 1 } = \frac { 3 } { 2 } \left( 3 ^ { 11 } - 1 \right)$

Putting the value of S_n in eq. (i), we get $\sum _ { k = 1 } ^ { 11 } \left( 2 + 3 ^ { k } \right) = 22 + \frac { 3 } { 2 } \left( 3 ^ { 11 } - 1 \right)$