Question 13 Marks
The first term of a G.P. is 1. The sum of the third term and fifth term is $90.$ Find the common ratio of G.P.
AnswerGiven: $a = 1$ and$ a_3 + a_5 = 90$
$\Rightarrow ar^2 + ar^4 = 90$
$\Rightarrow a(r^2 + r4) = 90$
$\Rightarrow 1 \times \left( r ^ { 2 } + r ^ { 4 } \right) = 90\Rightarrow r^2 + r^4 = 90$
$\Rightarrow r^4 + r^2 - 90 = 0$
$\Rightarrow r ^ { 2 } = \frac { - 1 \pm \sqrt { ( 1 ) ^ { 2 } - 4 \times ( - 90 ) \times 1 } } { 2 \times 1 }$
$= \frac { - 1 \pm \sqrt { 1 + 360 } } { 2 } = \frac { - 1 \pm \sqrt { 361 } } { 2 }$
$= \frac { - 1 \pm 19 } { 2 }=$
$\Rightarrow r ^ { 2 } = \frac { - 1 + 19 } { 2 } = \frac { 18 } { 2 } = 9or r ^ { 2 } = \frac { - 1 - 19 } { 2 } = \frac { - 20 } { 2 } = - 10$ which is not possible
Therefore, the common ratio is $r = \pm 3$
View full question & answer→Question 23 Marks
The sum of some terms of G.P. is $315$ whose first term and the common ratio are $5$ and $2$ respectively. Find the last term and the number of terms.
AnswerGiven: $a = 15, r = 2$ and $S_n = 315 \therefore \mathrm { S } _ { n } = \frac { a \left( r ^ { n } - 1 \right) } { r - 1 }$
$\Rightarrow 315 = \frac { 5 \left( 2 ^ { n } - 1 \right) } { 2 - 1 }$
$\Rightarrow \frac { 315 } { 5 } = 2 ^ { n } - 1$
$\Rightarrow 2^{n - 1} = 63$
$\Rightarrow 2^n = 64 = 2^6$
$\Rightarrow n = 6$
$\therefore a _ { 6 } = a r ^ { 6 - 1 } = 5 \times 2 ^ { 5 } = 5 \times 32 = 160$
Hence the number of terms=6 and the last term $=160$
View full question & answer→Question 33 Marks
If $f$ is a function satisfying $f (x+y) = f (x) f (y)$ for all $x, y \in N $ such that $f (1) = 3$ and $\sum\limits_{x = 1}^n f (x) = 120$ find the value of $n.$
Answer$f (1) = 3f(1 + 2) = f(1) f(2) = 3 \times 9 = 27$
$f(1 + 3) = f(1) f(3) = 3 \times 27 = 81$
L.H.S.
$= f(1) + f(2) + f(3) + ...... + f(n)$
$= 3 + 9 + 27 + 81 + ... + n$ terms
$= \frac {3(3^n - 1)}{3-1} = \frac 32 (3^n - 1)$
$= f(1) + f(2) + f(3) +----+ f(n)$
$= 3 + 9 + 27 + 1 +----+n$ terms
$= \frac { 3 \left( 3 ^ { n } - 1 \right) } { 3 - 1 } = \frac { 3 } { 2 } \left( 3 ^ { n } - 1 \right)$
ATQ
$\frac { 3 } { 2 } \left( 3 ^ { n } - 1 \right) = 120$
$3^n - 1 = 80$
$3^n = 81$
$n = 4$
View full question & answer→Question 43 Marks
Find the sum of all two digit numbers which when divided by $4,$ yield $1$ as remainder.
AnswerGiven: A.P. $13, 17, 21, .... , 97$
Here $a = 13, d = 17 - 13 = 4$ and $a_n = 97$
Now, $a_n = a + (n - 1)d$
$\Rightarrow 97 = 13 + ( n - 1 ) \times 4$
$\Rightarrow 97 - 13 = ( n - 1 ) \times 4$
$\Rightarrow 84 = ( n - 1 ) \times 4$
$\Rightarrow n - 1 = \frac { 84 } { 4 } = 21$
$\Rightarrow n = 22$
$\therefore$ $S _ { n } = \frac { n } { 2 } \left( a + a _ { n } \right)$
$\Rightarrow \mathrm { S } _ { 22 } = \frac { 22 } { 2 } ( 13 + 97 )$
$\Rightarrow \mathrm { S } _ { 22 } = \frac { 22 } { 2 } \times 110=1210$
View full question & answer→Question 53 Marks
Find the sum of integers from $1$ to $100$ that are divisible by $2$ or $5.$
AnswerDivisible by $2$
$2, 4, 6, ... 100$
$a = 2, d = 2, a_n = 100$
$100 = 2 + (n-1) \times 2$
$n = 50$
$S _ { 50 } = \frac { 50 } { 2 } [ 2 + 100 ] = 2550$
divisible by 5
$a = 5, d = 5, a_n = 100$
$5 + ( n - 1 ) \times 5 = 100$
$n = 20$
$S_{20} = 1050$
divisible by both $2$ or $5$
$10, 20, 30, ...... 100$
$a = 10, d = 10, a_n = 100$
$100 = 10 + (n - 1) \times 10$
$n = 10$
$S _ { 10 } = \frac { 10 } { 2 } [ 10 + 100 ]$
$= 550$
According to question, Sum $= 2550 + 1050 - 550$
$= 3050$
View full question & answer→Question 63 Marks
Find the sum of all numbers between $200$ and $400$ which are divisible by $7.$
AnswerGiven: A.P. $203, 210, 217, ………., 399$
Here $a = 203, d = 210 - 203 = 7$ and $a_n = 399$
Now, $a_n = a + (n - 1)d$
$\Rightarrow 399 = 203 + ( n - 1 ) \times 7$
$\Rightarrow 399 - 203 = ( n - 1 ) \times 7$
$\Rightarrow 196 = ( n - 1 ) \times 7$
$\Rightarrow n - 1 = \frac { 196 } { 7 } = 28$
$\Rightarrow n = 29$
$\therefore S _ { n } = \frac { n } { 2 } \left( a + a _ { n } \right)$
$\Rightarrow S _ { 29 } = \frac { 29 } { 2 } ( 203 + 399 )$
$\Rightarrow \mathrm { S } _ { 29 } = \frac { 29 } { 2 } \times 602=8729$
View full question & answer→Question 73 Marks
150 workers were engaged to finish a job in a certain number of days 4 workers dropped out on the second day, 4 more workers dropped out on the third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.
AnswerGiven, a = 150, d = -4
$S _ { n } = \frac { n } { 2 } [ 2 \times 150 + ( n - 1 ) ( - 4 ) ]$
If total works who would have worked all n days 150(n-8)
According to question, $ \frac { n } { 2 } [ 300 + ( n - 1 ) ( - 4 ) ] = 150 ( n - 8 )$
Hence, n = 25
View full question & answer→Question 83 Marks
A man deposited ₹ $10,000$ in a bank at the rate of $5\%$ simple interest annually. Find the amount in $15^{th} $ year since he deposited the amount and also calculate the total amount after $20$ year.
AnswerTotal amount deposited $= ₹\ 10000,$ Rate of interest $= 5\%$ per annum
Interest of first year $= \frac { 10000 \times 5 \times 1 } { 100 } = ₹ 500$
Here $a = 1000, d = 500$
$\therefore$ Amount in $15^{th} $ year $= a_{15} = 10000 + (15 - 1)\times500 = 10000 + 7000 = ₹ 17000$
Total amount after $20$ years $=$ Amount in the $21^{st} $ year $= a_{21} = 1000 + (21 - 1)500$
$= 10000 + 10000 = ₹20000$
View full question & answer→Question 93 Marks
Let sum of $n, 2n, 3n$ terms of an A.P. be $S_1, S_2$_ and $S_3$_respectively, show that $S_3 = 3(S_2 - S_1).$
AnswerGiven: ${S_1} = {n \over 2}\left[ {2a + (n - 1)d} \right]$ …..(i)${S_2} = {{2n} \over 2}\left[ {2a + (2n - 1)d} \right]…..(ii)$
And ${S_3} = {{3n} \over 2}\left[ {2a + (3n - 1)d} \right]$
Now, ${S_2} - {S_1} = {{2n} \over 2}\left[ {2a + (2n - 1)d} \right] - {n \over 2}\left[ {2a + (n - 1)d} \right]$
$\Rightarrow {S_2} - {S_1} = (n - {n \over 2})2a + \left[ {n(2n - 1) - {n \over 2}(n - 1)} \right]d$
$= na + {1 \over 2}\left[ {4{n^2} - 2n - {n^2} + n} \right]d$
$= {n \over 2}\left[ {2a + (3n - 1)d} \right] = {1 \over 3}\left\{ {{{3n} \over 2}\left[ {2a + (3n - 1)d} \right]} \right\} = {1 \over 3}{S_3}$
$\Rightarrow 3(S_2 - S_1) = S_3$
Hence proved.
View full question & answer→Question 103 Marks
A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when $8^{th}$ set of letter is mailed.
AnswerTotal letters in the first set $= 4,$ Total letters in the second set $= 4^2 = 16$ Total letters in the third set $= 4^3 = 64$
$\therefore$ 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 $= 87380 \times 0.50 = Rs. 43690.$
View full question & answer→Question 113 Marks
Shamshad Ali buys a scooter for $₹\ 22000.$ He pays $₹\ 4000$ cash and agrees to pay the balance in annual installment of $₹\ 1000$ plus $10\%$ interest on the unpaid amount. How much will the scooter cost him$?$
AnswerTotal cost of the scooter $= ₹ 22000$, Cash paid $= ₹ 4000$
Balance to be paid $= 22000 – 4000 = ₹ 18000$
Annual installment $= ₹ 1000$
\therefore Number of installment $= {{18000} \over {1000}} = 18$
Interest of $1^{st}$ installment $= \frac { 18000 \times 10 \times 1 } { 100 } = ₹ 1800$
Amount of $1^{st}$ installment $= 1000 + 1800 = ₹ 2800$
Interest of $2^{nd}$ installment $= \frac { 17000 \times 10 \times 1 } { 100 } = ₹ 1700$
Amount of $2^{nd}$ installment = 1000 + 1700 = ₹ 2700
Interest of $3^{rd}$ installment $= \frac { 16000 \times 10 \times 1 } { 100 } = ₹ 1600$
Amount of $3^{rd}$ installment $= 1000 + 1600 = ₹ 2600$
$\therefore$ Sequence of installments is $2800, 2700, 2600, …$ in A.P
Here, $a = 2800, d = 2700 - 2800 = -100$ and $n = 18$
$\therefore S_n= \frac n2[2a + (n - 1) d] = \frac {18}2 [2 \times 2800 + (18 - 1) \times (-100)]$
$= 9 [5600 – 1700] = ₹ 35100$
Therefore, the total cost of tractor is $(35100 + 4000) = ₹ 39100$
View full question & answer→Question 123 Marks
A farmer buys a used tractor for $₹\ 12000.$ He pays $₹\ 6000$ cash and agrees to pay the balance in annual installments of $₹\ 500$ plus $12\ %$ interest on the unpaid amount. How much will the tractor cost him$?$
AnswerTotal cost of the tractor $= ₹\ 12000,$ Cash paid $= ₹\ 6000$
Balance to be paid $= 12000 – 6000 = ₹ 6000$
Annual installment $= ₹\ 500$
$\therefore$ Number of installment = ${{6000} \over {500}} = 12$
Interest of $1^{st}$ installment $= \frac { 6000 \times 12 \times 1 } { 100 } = ₹ 720$
Amount of $1^{st}$ installment $= 500 + 720 = ₹ 1220$
Interest of $2^{nd}$ installment $= \frac { 5500 \times 12 \times 1 } { 100 } = ₹ 660$
Amount of $2^{nd}$ installment $= 500 + 660 = ₹ 1160$
Interest of $3^{rd}$ installment $= \frac { 5000 \times 12 \times 1 } { 100 } = ₹ 600$
Amount of $3^{rd}$ installment $= 500 + 600 = ₹ 1100$
$\therefore$ Sequence of installments is $1220, 1160, 1100, ... $ which is in A.P
Here, $a = 1220, d = 1160 - 1220 = -60$ and $n = 12$
$\therefore S_n = \frac n2[2a + (n - 1) d]$
$= \frac {12}2 [ 2 \times 1220 + (12 - 1) \times (-60)]$
$= 6 [2440 - 660] = ₹ 10680$
Therefore, the total cost of tractor is $(10680 + 6000) = ₹ 16680$
View full question & answer→Question 133 Marks
Find the sum of the first n terms of the series: $3 + 7 + 13 + 21 + 31 + ……….$
AnswerGiven: $S_n = 3 + 7 + 13 + 21 + 31 + ...... + a_{n-1} + a_n……….(i)$
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$) - a_n$
$\Rightarrow a _ { n } = 3 + \frac { n - 1 } { 2 } [ 2 \times 4 + ( n - 2 ) \times 2 ]$
$\Rightarrow a _ { n } = 3 + \frac { n - 1 } { 2 } [ 8 + 2 n - 4 ]$
$\Rightarrow a_n = 3 + (n - 1) (n + 2)$
$\Rightarrow an = 3 + n^2 + n - 2$
$\Rightarrow a_n = n^2 + n + 1$
$\therefore {S_n} = \sum\limits_{k = 1}^n {{a_{_k}}} = \sum\limits_{k = 1}^n {({k^{^2}}} + k + 1)$
$= (1^2 + 1 + 1) + (2^2 + 2 + 1) + (3^2 + 3 + 1) + ...... +(n^2 + n + 1)$
$= (1^2 + 2^2 + 3^2 + ....... + n^2) + (1 + 2 + 3 + ...... + n) + n$
$= \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 } + \frac { n ( n + 1 ) } { 2 } + n$
$= n \left[ \frac { 2 n ^ { 2 } + 3 n + 1 + 3 n + 3 + 6 } { 6 } \right]$
$= n \left[ \frac { 2 n ^ { 2 } + 6 n + 10 } { 6 } \right]$
$= \frac { n } { 3 } \left( n ^ { 2 } + 3 n + 5 \right)$
View full question & answer→Question 143 Marks
Find the $20^{th}$ term of the series $2 \times 4 + 4 \times 6 + 6 \times 8 + \ldots \ldots \ldots +$ n terms.
AnswerGiven: $2 \times 4 + 4 \times 6 + 6 \times 8 + \ldots \ldots \ldots + n$ terms
$\therefore a_n = (n^{th}$ term of $2, 4, 6, ......) (n^{th}$ term of $4, 6, 8, ........)$
$\Rightarrow a_n = [2 + (n - 1)2] [4 + (n - 1)2] = 2n(2n + 2)$
$\therefore a _ { 20 } = 2 \times 20 ( 2 \times 20 + 2 ) = 40 \times 42 = 1680$
View full question & answer→Question 153 Marks
Find the sum of the series up to n terms .$6 +. 66 + .666+…$
AnswerThe given sum is not in GP but we can write it as follows: -
Sum $= .6 + .66 + .666 + …$to $n$ terms
$= 6(0.1) + 6(0.11) + 6(0.111) + …$to $n$ terms
taking 6 common
$= 6[0.1 + 0.11 + 0.111 + …$to $n$ terms$]$
divide & multiply by 9,we get
$=(\frac{6}{9})[9(0.1 + 0.11 + 0.111 + …$to $n$ terms$)]$
$= (\frac{6}{9})[0.9 + 0.99 + 0.999 + …$to $n$ terms$]$
$= \frac{6}{9}\left[\left(\frac{9}{10}\right)+\left(\frac{99}{100}\right)+\left(\frac{999}{1000}\right)+\ldots \text { to } n \text { terms }\right]$
$= \frac{6}{9}\left[\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{100}\right)+\left(1-\frac{1}{1000}\right)+\ldots \text { to } n \text { terms }\right]$
$= \frac{6}{9}[\{1+1+1+\ldots \text { to } n \text { terms }\}$ - $\left.\left\{\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\ldots \text { to } n \text { terms }\right\}\right]$
$= \frac{6}{9}\left[n-\left\{\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\ldots \text { to } n \text { terms }\right\}\right]$
Since $\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\ldots \text { to } n \text { terms }$ is in GP with
first term(a) = $\frac{1}{10}$
common ratio(r) = $\frac{10^{-2}}{10^{-1}} = 10 ^{- 1} = \frac{1}{10}$
We know that
Sum of n terms = $\frac{a\left(1-r^{n}\right)}{1-r} [$As $r < 1]$
Substituting valuevalue of a & r
$\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\ldots \text { to } n \text { terms }=\frac{a\left(1-r^{n}\right)}{1-r}$
= $\frac{\frac{1}{10}\left(1-\left(\frac{1}{10}\right)^{n}\right)}{\left(1-\frac{1}{10}\right)}$
= $\frac{\frac{1}{10}\left(1-\left(\frac{1}{10}\right)^{n}\right)}{\frac{9}{10}}$
= $\frac{1\left(1-10^{-n}\right)}{9}$Therefore, Sum = $\frac{6}{9}\left[n-\frac{1\left(1-10^{-n}\right)}{9}\right]$
View full question & answer→Question 163 Marks
Find the sum of the series up to $n$ terms $5 + 55 + 555 + …$
AnswerNow,to find sum $= 5 + 55 + 555 + …. n$ terms.
$= \frac{5}{9} [9 + 99 + 999 + …. n$ terms$]$
$= \frac{5}{9} [(10 - 1) + (100 - 1) + (1000 - 1) + … n$ terms$]$
$= \frac{5}{9} [10 + 100 + 1000 ….. – (1 + 1 + … 1)]$
$= \frac{5}{9} [10(10^n - 1)/(10 - 1) + (1 + 1 + … n$ times$)]$
$= \frac{50}{81}(10^n – 1) - \frac{5n}{9}$
View full question & answer→Question 173 Marks
If the sum of three numbers in A.P., is $24$ and their product is $440$. Find the numbers.
AnswerLet $(a - d), a, (a + d)$ be three numbers in A.P.
According to question, $(a - d) + a + (a + d) = 24$
$\Rightarrow 3a = 24$
$\Rightarrow a = 8$
And $(a - d) (a) (a + d) = 440$
$\Rightarrow (a^2 - d^2)a = 440$
$\Rightarrow (64 - d^2)8 = 440$
$\Rightarrow 64 - d^2 = 55$
$\Rightarrow d^2 = 64 - 55$
$\Rightarrow d^2 = 9$
$\Rightarrow d = \pm 3$
Taking $d = 3, A.P. is (8 – 3), 8, (8 + 3)$
$\Rightarrow 5, 8, 11$
Taking $d = -3, A.P. is (8 + 3), 8, (8 – 3)$
$\Rightarrow 11, 8, 5$
View full question & answer→Question 183 Marks
If a and b are the roots $x^2 - 3x + p = 0$ and $c, d$ are roots of $x^2 - 12x + q = 0$ where $a, b, c, d$ form a G.P. Prove that$ (q + p):(q - p) = 17:15.$
AnswerLet $ \frac { b } { a } = \frac { c } { b } = \frac { d } { c } = k$
$\therefore \frac ba = k$
$\Rightarrow b = ak$
And $\frac cb = k$
$\Rightarrow c = bk = (ak)k = ak^2$
Also $\frac dc = k$
$\Rightarrow d = ck = (ak^2)k = ak^3$
$\because a^a and b^b are the roots x^2 - 3x + p = 0$
$\therefore a + b = \frac {-(-3)} {1} = 3$
$\Rightarrow a + ak = 3$
$\Rightarrow a(1 + k) = 3 ...(i)$
And $ab = \frac p 1$
$\Rightarrow a(ak) = p$
$\Rightarrow a^2k = p ...(ii)$
Also $c, d$ are roots of $x^2 - 12x + q = 0$
$\therefore c + d = \frac { - ( - 12 ) } { 1 } = 12$
$\Rightarrow ak^2 + ak^3 = 12$
$\Rightarrow ak^2(1 + k) = 12 ...(iii)$
And $c d = \frac q1$
$\Rightarrow ak^2(ak^3) = q$
$\Rightarrow a^2k^5 = q ...(iv)$
Dividing eq. $(iii)$ by eq. $(i), \frac { a k ^ { 2 } ( 1 + k ) } { a ( 1 + k ) } = \frac { 12 } { 3 }$
$\Rightarrow k^2 = 4$
$\Rightarrow k = \pm 2$
Now $\frac { q + p } { q - p } = \frac { a ^ { 2 } k ^ { 5 } + a ^ { 2 } k } { a ^ { 2 } k ^ { 5 } - a ^ { 2 } k } = \frac { a ^ { 2 } k \left( k ^ { 4 } + 1 \right) } { a ^ { 2 } k \left( k ^ { 4 } - 1 \right) }$
$= \frac { ( \pm 2 ) ^ { 4 } + 1 } { ( \pm 2 ) ^ { 4 } - 1 } = \frac { 16 + 1 } { 16 - 1 } = \frac { 17 } { 15 }$
Therefore, $(q + p):(q - p) = 17:15$
View full question & answer→Question 193 Marks
If $a, b, c, d$ are in G.P., prove that $(a^n + b^n), (b^n + c^n), (c^n + d^n)$ are in G.P.
AnswerGiven: $a, b, c, d$ are in G.P.To prove: $(a^n + b^n), (b^n + c^n), (c^n + d^n)$ 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 $\frac { c } { b } = k$
$\Rightarrow c = bk = (ak)k = ak^2$
Also $\frac { d } { c } = k$
$\Rightarrow d = ck = (ak^2)k = ak^3$
Now, $\frac { b ^ { n } + c ^ { n } } { a ^ { n } + b ^ { n } } = \frac { c ^ { n } + d ^ { n } } { b ^ { n } + c ^ { n } }$
$\Rightarrow \frac { ( a k ) ^ { n } + \left( a k ^ { 2 } \right) ^ { n } } { a ^ { n } + ( a k ) ^ { n } } = \frac { \left( a k ^ { 2 } \right) ^ { n } + \left( a k ^ { 3 } \right) ^ { n } } { ( a k ) ^ { n } + \left( a k ^ { 2 } \right) ^ { n } }$
$\Rightarrow \frac { a ^ { n } k ^ { n } + a ^ { n } k ^ { 2 n } } { a ^ { n } + a ^ { n } k ^ { n } } = \frac { a ^ { n } k ^ { 2 n } + a ^ { n } k ^ { 3 n } } { a ^ { n } k ^ { n } + a ^ { n } k ^ { 2 n } }$
$\Rightarrow \frac { a ^ { n } k ^ { n } \left( 1 + k ^ { n } \right) } { a ^ { n } \left( 1 + k ^ { n } \right) } = \frac { a ^ { n } k ^ { 2 n } \left( 1 + k ^ { n } \right) } { a ^ { n } k ^ { n } \left( 1 + k ^ { n } \right) }$
$\Rightarrow k^n = k^n$^
Therefore, $(a^n + b^n), (b^n + c^n), (c^n + d^n)$ are in G.P.
View full question & answer→Question 203 Marks
If a$(\frac 1b + \frac 1c)$, b$(\frac 1c + \frac 1a)$, c$(\frac 1a + \frac 1b)$ are in A.P., prove that a, b, c are in A.P.
AnswerGiven: a$(\frac 1b + \frac 1c)$, b$(\frac 1c + \frac 1a)$, c$(\frac 1a + \frac 1b)$ are in A.P.
$\Rightarrow$ a$\left( \frac {b + c} {bc} \right)$, b$\left( \frac {c + a} {ca} \right) $, c$\left( \frac {a + b} {ab} \right)$ are in A.P.
$\Rightarrow \frac { a b + a c } { b c } , \frac { b c + a b } { c a } , \frac { a c + b c } { a b }$ are in A.P.
$\Rightarrow \frac {ab + ac} {bc}$ + 1, $\frac {bc + ab} {ca}$ + 1, $\frac {ac + bc} {ab}$ + 1 are in A.P. [Adding 1 to each term in the sequence]
$\Rightarrow \frac { a b + a c + b c } { b c } , \frac { b c + a b + c a } { c a } , \frac { a c + b c + a b } { a b }$ are in A.P.
$\Rightarrow \frac { 1 } { b c } , \frac { 1 } { c a } , \frac { 1 } { a b }$ are in A.P.[Dividing each fraction by ab + bc + ca]
$\Rightarrow \frac { a b c } { b c } , \frac { a b c } { c a } , \frac { a b c } { a b }$ are in A.P.[Multiplying each fraction by abc]
$\Rightarrow$ a, b, c are in A.P.
View full question & answer→Question 213 Marks
The $p^{th}, q^{th}$ and $r^{th}$ terms of an A.P. are $a, b, c$ respectively. Show that $(q - r)a + (r - p)b + (p - q)c = 0.$
AnswerAccording to question, $a_p = a +(p - 1)d = a, a_q = a + (q - 1)d = b$ and $a_r = a + (r - 1)d = c$
Now $(q - r)a + (r - p)b + (p - q)c = 0$
Putting values of a, b and c, we get
$(q - r) [a + (p - 1)d] + (r - p) [a + (q - 1)d] + (p - q) [a + (r - 1)d] = 0$
$\Rightarrow (q - r) [a + pd - d] + (r - p)[a + qd - d] + (p - q) [a + rd - d] = 0$
$\Rightarrow aq + pqd - qd - ra - rpd + rd + ar + qrd - dr - pa - pqd + pd + pa + prd - pd - qa - qrd + qd = 0$
$\Rightarrow 0 = 0$ Proved.
View full question & answer→Question 223 Marks
Let $S$ be the sum, $P$ the product and $R$ the sum of reciprocals of n terms in a G.P. Prove that $P^2R^n = S^n.$
AnswerLet the G.P be $a, ar, ar^2, ar^3 ..........,ar^{n - 1}$
Here $\mathrm { S } = \frac { a \left( r ^ { n } - 1 \right) } { r - 1 }$
$P = a.ar.ar^2 ......... ar^{n-1}= {a^n}.{r^{1 + 2 + 3 + ....... + (n - 1)}} = {a^n}.{r^{{{n(n - 1)} \over 2}}}$
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.
View full question & answer→Question 233 Marks
If $\frac { a + b x } { a - b x } = \frac { b + c x } { b - c x } = \frac { c + d x } { c - d x }$ (x $\neq$ 0), then show that a, b, c and d are in G.P.
AnswerTaking $\frac { a + b x } { a - b x } = \frac { b + c x } { b - c x }$
$\Rightarrow (a + bx)(b - cx) = (b + cx) (a - bx)$
$\Rightarrow ab - acx + b^2x - bcx^2 = ab - b^2x + acx - bcx^{2}$
$\Rightarrow 2b^2x = 2acx$
$\Rightarrow b^2 = ac$
$\Rightarrow \frac { b } { a } = \frac { c } { b } ...(i)$
Taking $\frac { b + c x } { b - c x } = \frac { c + d x } { c - d x }$
$\Rightarrow (b + cx) (c - dx) = (c + dx) (b - cx)$
$\Rightarrow 2c^2x = 2bdx$
$\Rightarrow c^2 = bd$
$\Rightarrow \frac { c } { b } = \frac { d } { c } ...(ii)$
From eq. $(i)$ and $ (ii), \frac { b } { a } = \frac { c } { b } = \frac { d } { c }$
View full question & answer→Question 243 Marks
The sum of the first four terms of an A.P. is $56.$ The sum of the last four terms is $112.$ If its first term is $11,$ then find the number of terms.
AnswerGiven: $a = 11$ and $S_4 = 56\Rightarrow S _ { 4 } = \frac { 4 } { 2 } [ 2 \times 11 + ( 4 - 1 ) d ]$
$\Rightarrow 2(22 + 3d) = 56$
$\Rightarrow 22 + 3d = 28$
$\Rightarrow 3d = 6$
$\Rightarrow d = 2$
Also, $l + (l - d) + (l - 2d) + (l - 3d) = 112$
$\Rightarrow 4l - 6d = 112$
$\Rightarrow 4l = 112 + 6d$
$\Rightarrow 4 l = 112 + 6 \times 2$
$\Rightarrow 4l = 112 + 12$
$\Rightarrow 4l = 124$
$\Rightarrow l = 31$
$\therefore a_n = a+ (n - 1)d$
$\Rightarrow 31 = 11 + ( n - 1 ) \times 2$
$\Rightarrow 2(n - 1) = 20$
$\Rightarrow n - 1 = 10$
$\Rightarrow n = 11$
View full question & answer→Question 253 Marks
A G.P. consists of an even number of terms. If the sum of all the terms is $5$ times the sum of terms occupying odd places, then find its common ratio.
AnswerLet the number of terms be $2n$ then we have the number of odd terms is $n$
Let the G.P be $a, ar, ar^2, .... ar^{2n-1}$
Then the odd terms $a, ar^2, ar^4, ar^6, ....$ form a G.P
$\therefore S_{2n} = \frac {a \left(r ^ {2n} - 1\right)} {r - 1}$ and $S_n = a\left[ {{{{{({r^2})}^n} - 1} \over {{r^2} - 1}}} \right]$
According to question, $S_{2n} = 5S_n$
}$_\Rightarrow a\left[ {{{{r^{2n}} - 1} \over {r - 1}}} \right] = 5a\left[ {{{{{({r^2})}^n} - 1} \over {{r^2} - 1}}} \right]$
$\Rightarrow {1 \over {r - 1}} = {5 \over {{r^2} - 1}}$
$\Rightarrow r + 1 = 5$
$\Rightarrow r = 4$
View full question & answer→Question 263 Marks
The sum of three numbers in G.P. is $56.$ If we subtract $1, 7, 21$ from these numbers in that order, we obtain an arithmetic progression. Find the numbers.
AnswerLet $a, ar, ar^2$^ be three numbers in G.P., therefore,$ a + ar + ar^2 = 56 \Rightarrow a(1 + r + r^2) = 56 ......(i)$
According to question, $a - 1, ar - 7, ar^2 - 21$ are in A.P.
$\therefore (ar - 7) - (a - 1) = (ar^2 - 21) - (ar - 7)$
$\Rightarrow ar - 7 - a + 1 = ar^2 - 21 - ar + 7$
$\Rightarrow ar - a - 6 = ar^2 - ar - 14$
$\Rightarrow ar^2 - 2ar + a = 8$
$\Rightarrow a(r^2 - 2r + 1) = 8 ….....(ii)$
Dividing eq. $(i)$ by eq. $(ii), \frac { a \left( 1 + r + r ^ { 2 } \right) } { a \left( r ^ { 2 } - 2 r + 1 \right) } = \frac { 56 } { 8 }$
$\Rightarrow 1 + r + r^2 = 7r^2 - 14r + 7$
$\Rightarrow 6r^2 - 15r + 6 = 0$
$\Rightarrow 2r^2 - 5r + 2 = 0$
$\Rightarrow r = \frac { - ( - 5 ) \pm \sqrt { ( - 5 ) ^ { 2 } - 4 \times 2 \times 2 } } { 2 \times 2 }$
$= \frac { 5 \pm \sqrt { 25 - 16 } } { 4 } = \frac { 5 \pm \sqrt { 9 } } { 4 } = \frac { 5 \pm 3 } { 4 }=$
$\Rightarrow r = \frac { 5 + 3 } { 4 } = \frac { 8 } { 4 } = 2or r = \frac { 5 - 3 } { 4 } = \frac { 2 } { 4 } = \frac { 1 } { 2 }$
Putting $r = 2$ in eq. $(i), a(1 + 2 + 2^2) = 56$
$\Rightarrow a = \frac { 56 } { 7 } = 8$
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 $32, 16, 8.$
View full question & answer→Question 273 Marks
Show that the sum of $(m + n)^{th}$ and $(m - n)^{th}$ terms of an A.P. is equal to twice the $m^{th}$ terms.
AnswerHere, $a_{m+n} = a + (m + n - 1)d …..(i)$
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
$a_{m+n} + a_{m-n} = a + (m + n - 1)d + a +(m - n - 1)d$
$\Rightarrow a_{m+n} + a_{m-n} = a + md + nd - d + a + md - nd - d$
$\Rightarrow a_{m+n} + a_{m-n} = 2a + 2md + -2d$
$\Rightarrow a_{m+n} + a_{m-n} = 2(a + md - d)$
$\Rightarrow a_{m+n} + a_{m-n} = 2[a + (m - 1)d]$
$\Rightarrow a_{m+n} + a_{m-n} = 2a_m$
View full question & answer→Question 283 Marks
Find the sum of n terms of the geometric progression $1, -a, a^2, -a^3 ... n$ terms $($if a $\ne -1).$
AnswerHere, $a = 1$ and $r = \frac {-a} {1} = -a$
$\therefore ^ { \mathrm { S } _ { n } = \frac { a \left( 1 - r ^ { n } \right) } { 1 - r } } when r < 1$
$\Rightarrow S_n= \frac { 1 \left[ 1 - (-a) ^ { n } \right] } { 1 - (-a) }$
$\Rightarrow S_n = \frac { 1 } { 1 + a }[1 - (-a)^n]$
View full question & answer→Question 293 Marks
Find the sum to indicated number of terms of the geometric progression $ \sqrt { 7 } , \sqrt { 21 } , 3 \sqrt { 7 }$, ... n terms.
AnswerHere,a = $ \sqrt 7$ and r =$ \frac { \sqrt { 21 } } { \sqrt { 7 } } = \sqrt { 3 }$
$ \therefore S_ { n } = \frac { a \left( r ^ { n } - 1 \right) } { r - 1 }$ when r > 1
$ \Rightarrow \mathrm { S } _ { n } = \frac { \sqrt { 7 } \left[ ( \sqrt { 3 } ) ^ { n } - 1 \right] } { \sqrt { 3 } - 1 }$
$ S _ { n } = \frac { \sqrt { 7 } } { \sqrt { 3 } - 1 } \times \frac { \sqrt { 3 } + 1 } { \sqrt { 3 } + 1 } \left[ ( 3 ) ^ { \frac { n } { 2 } } - 1 \right]$
$ \Rightarrow \mathrm { S } _ { n } = \frac { \sqrt { 7 } ( \sqrt { 3 } + 1 ) } { 2 } \left[ ( 3 ) ^ { \frac { n } { 2 } } - 1 \right]$
View full question & answer→Question 303 Marks
Find the sum to indicated number of terms of the geometric progression 0.15, 0.015, 0.0015, ... 20 terms.
AnswerHere,a = 0.15 and r = $\frac { 0.015 } { 0.15 } = \frac { 1 } { 10 }$
$\mathrm { S } _ { n } = \frac { a \left( 1 - r ^ { n } \right) } { 1 - r }$ when r < 1
$\Rightarrow S _ { 20 } = \frac { 0.15 \left[ 1 - \left( \frac { 1 } { 10 } \right) ^ { 20 } \right] } { 1 - \frac { 1 } { 10 } }$
$\Rightarrow \mathrm { S } _ { 20 } = \frac { 15 } { 100 } \times \frac { 10 } { 9 } \left[ 1 - ( 0.1 ) ^ { 20 } \right]$
$\Rightarrow \mathrm { S } _ { 20 } = \frac { 1 } { 6 } \left[ 1 - ( 0.1 ) ^ { 20 } \right]$
View full question & answer→Question 313 Marks
Which term of the sequence $\sqrt { 3 } , 3,3 \sqrt { 3 } , .....$ is $729?$
AnswerHere a = $\sqrt3$ , r = $\frac { 3 } { \sqrt { 3 } } = \sqrt { 3 }$ and $a_n = 729$
$\therefore a_n = ar^{n-1}$
$\Rightarrow 729 = \sqrt { 3 } \times ( \sqrt { 3 } ) ^ { n - 1 }$
$\Rightarrow ( \sqrt { 3 } ) ^ { 12 } = ( \sqrt { 3 } ) ^ { n }$
$\Rightarrow n = 12$
Therefore, $12^{th}$ term of the given G.P. is $729.$
View full question & answer→Question 323 Marks
If A.M. and G.M. of roots of a quadratic equation are $8$ and $5$ respectively then obtain the quadratic equation.
AnswerLet a and b be the roots of required quadratic equation.
Then A.M. $= \frac { a + b } { 2 } = 8$
$\Rightarrow$
$a + b = 16$
And G.M. $= \sqrt { a b } = 5$
$\Rightarrow ab = 25$
Now, Quadratic equation $x^2 - ($Sum of roots$) x + ($Product of roots$) = 0$
$\Rightarrow x^2 - (a + b)x + ab = 0$
$\Rightarrow x^2 - 16x + 25 = 0$
Therefore, required equation is $x^2 - 16x + 25 = 0$
View full question & answer→Question 333 Marks
What will $₹\ 500$ amount to 10 years after its deposit in a bank which pays annual interest rate of $10\%$ compounded annually?
AnswerOriginal amount $= ₹\ 500$, Rate of interest $= 10\%$ compounded annually
\therefore Interest of one year $= \frac { 500 \times 10 \times 1 } { 100 } = ₹\ 50$
And Amount after one year $= 500 + 50 = ₹\ 550$
Here $a = 500$ and $r = \frac { 550 } { 500 } = 1.1$
Therefore, amount after $10$ years = Amount in the $11^{th}$ year
$= 5.5 \times (1.1)^{11 - 1} = ₹\ 500(1.1)^{10}$
View full question & answer→Question 343 Marks
The number of bacteria in a certain culture doubles every hour. If there were $30$ bacteria present in the culture originally, how many bacteria will be present at the end of $2^{nd}$ hour, $4^{th}$ hour and $n^{th}$ hour$?$
AnswerBacteria present in the culture originally $= 30$
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 =${a_{n + 1}} = 30({2^{(n + 1) - 1}}) = 30({2^n})$
View full question & answer→Question 353 Marks
If A and G be A.M. and G.M. respectively between two positive numbers, prove that the numbers are $\mathrm { A } \pm \sqrt { ( \mathrm { A } + \mathrm { G } ) ( \mathrm { A } - \mathrm { G } ) }$.
AnswerLet the two positive numbers be a and b 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 $\frac { 2 b } { 2 } = b$
View full question & answer→Question 363 Marks
The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio $( 3 + 2 \sqrt { 2 } ):( 3 - 2 \sqrt { 2 } )$.
AnswerLet the numbers be a and b
Given: a + b = 6$\sqrt { a b } \Rightarrow \frac { a + b } { 2 \sqrt { a b } } = \frac { 3 } { 1 }$
Applying componendo and dividendo, we get
$\frac { a + b + 2 \sqrt { a b } } { a + b - 2 \sqrt { a b } } = \frac { 3 + 1 } { 3 - 1 }$
$\Rightarrow \frac { ( \sqrt { a } + \sqrt { b } ) ^ { 2 } } { ( \sqrt { a } - \sqrt { b } ) ^ { 2 } } = \frac { 4 } { 2 }$
$\Rightarrow \frac { \sqrt { a } + \sqrt { b } } { \sqrt { a } - \sqrt { b } } = \frac { \sqrt { 2 } } { 1 }$
Again applying componendo and dividendo, we get
${{\sqrt a + \sqrt b + \sqrt a - \sqrt b } \over {\sqrt a + \sqrt b - \sqrt a + \sqrt b }} = {{\sqrt 2 + 1} \over {\sqrt 2 - 1}}$
$\Rightarrow \frac { \sqrt { a } } { \sqrt { b } } = \frac { \sqrt { 2 } + 1 } { \sqrt { 2 } - 1 }$
Squaring both sides, $\frac { a } { b } = \frac { 2 + 1 + 2 \sqrt { 2 } } { 2 + 1 - 2 \sqrt { 2 } }$
$\Rightarrow \frac { a } { b } = \frac { 3 + 2 \sqrt { 2 } } { 3 - 2 \sqrt { 2 } }$
Therefore, the numbers are in the ratio $( 3 + 2 \sqrt { 2 } ) : ( 3 - 2 \sqrt { 2 } )$
View full question & answer→Question 373 Marks
Find the value of n so that $\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}$ may be the geometric mean between a and b.
Answer$\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}=\sqrt{a b}$
$\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}=\frac{a^{\frac{1}{2}} b^{\frac{1}{2}}}{1}$
$a^{n+1}+b^{n+1}=a^{\frac{1}{2}} b^{\frac{1}{2}}\left(a^{n}+b^{n}\right)$ $a^{n+1}+b^{n+1}=a^{n+\frac{1}{2}} b^{\frac{1}{2}}+a^{\frac{1}{2}} b^{n+\frac{1}{2}}$
$a^{n+1}-a^{n+\frac{1}{2}} b^{\frac{1}{2}}=a^{\frac{1}{2}} b^{n+\frac{1}{2}}-b^{n+1}$
${a^{n + \frac{1}{2}}}\left( {{a^{\frac{1}{2}}} - {b^{\frac{1}{2}}}} \right) = {b^{n + \frac{1}{2}}}\left( {{a^{\frac{1}{2}}} - {b^{\frac{1}{2}}}} \right)$
$\left(\frac{a}{b}\right)^{n+\frac{1}{2}}=1$
$\left(\frac{a}{b}\right)^{n+\frac{1}{2}}=\left(\frac{a}{b}\right)^{0}$
$n+\frac{1}{2}=0$
$n=\frac{-1}{2}$
View full question & answer→Question 383 Marks
Insert two numbers between $3$ and $81,$ so that resulting sequence is GP.
AnswerLet the two numbers be a and b, then $3, a, b, 81$ are in GP.
$\because n^{th}$^ term, $T_n = Ar^{n-1}$
$\therefore T_4 =81 = 3r^{4-1} \Rightarrow r^3 = \frac { 81 } { 3 }$
$\Rightarrow r^{3}= 27 \Rightarrow r^3 = 3^3$
On comparing the base of power 3 from both sides, we get
$r = 3$
$\therefore a = Ar = 3 \times 3 = 9$
and $b = Ar^2 = 3 \times 3^2 = 27$
View full question & answer→Question 393 Marks
If $a, b, c$ and d are in G.P show that $(a^2+ b^2+ c^2) (b^2+ c^2+ d^2) = (ab + bc + cd)^2.$
Answer$a, b, c, d$ are in G. P
$b = ar$
$c = ar^2$
$d = ar^3$
L. H. S $ = (a^2 + b^2 + c^2) (b^2 + c^2 + d^2)$
$= (a^2 + a^2r^2 + a^2r^4) (a^2r^2 + a^2r^4 + a^2r^6)$
$= a^4r^2 (1 + r^2 + r^4)^2$^
R. H. S. $= (ab + bc + cd)^2$
$= (a^2r + a^2r^3 + a^2r^5)^2$
$= a^4r^2 (1 + r^2 + r^4)^2$
H.p
View full question & answer→Question 403 Marks
Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from $(n + 1)^{th}$ to $(2n)^{th}$ term is $\frac { 1 } { r ^ { n } }$.
AnswerLet a be the first term and r be the common ratio of given G.P.
Then $\frac { \text { Sum of first } n \text { terms } } { \text { Sum of terms from } ( n + 1 ) ^ { \text { th } } \text { to } ( 2 n ) ^ { \frac { k } { n ^ { 2 } } } }$
= ${{a + ar + a{r^2} + .... + a{r^{n - 1}}} \over {a{r^n} + a{r^{n + 1}} + ... + a{r^{2n - 1}}}}$
= $\frac { a + a r + a r ^ { 2 } + \ldots + a r ^ { n - 1 } } { r ^ { n } \left[ a + a r + a r ^ { 2 } + \ldots + a r ^ { n - 1 } \right] } = \frac { 1 } { r ^ { n } }$
View full question & answer→Question 413 Marks
If the first and the $n^{th}$ term of a G.P. are a and $b$ respectively and if $P$ is the product of n terms, prove that $P^2 = (ab)^n.$
AnswerLet r be the common ratio of the given G.PHere, first term of G.P. is a
and $a_n = b$
$\Rightarrow ar^{n-1} = b ……….(i)$
Given: $P = a.ar.ar^2.ar^3 ..... ar^{n-1}$
$\Rightarrow P = a^n.r^{1+2+3+......+ n-1}$
$\Rightarrow p = {a^n}{r^{{{n(n - 1)} \over 2}}}$
$\Rightarrow {p^2} = {a^{2n}}{r^{n(n - 1)}} = {\left[ {aa{r^{n - 1}}} \right]^n} [$Squaring both sides$]$
$\Rightarrow P^2 = (ab)^n[$From eq. $(i)]$
Hence proved
View full question & answer→Question 423 Marks
If the pth, qth and rth terms of a G.P. are $a, b$ and $c$ respectively. Prove that $a^{q -r} b^{r- p} c^{p - q}= 1.$
AnswerLet A be the first term and R be the common ratio of given G.P.
$\therefore a_p = a$
$\Rightarrow AR^{p-1} = a ……….(i)$
$a_q = b$
$\Rightarrow AR^{q-1} = b ……….(ii)$
$a_r = c$
$\Rightarrow AR^{r-1} = c ……….(iii)$
Now, L.H.S. $= a^{q-r}b^{r-p}c^{p-q} = (AR^{p-1})^{q-r}.(AR^{q-1})^{r-p}.(AR^{r-1})^{p-q}$
$= A^{q-r}R^{(p-1)(q-r)}.A^{r-p}R^{(q-1)(r-p)}.A^{p-q}R^{(r-1)(p-q)}$
$= A^{q-r+r-p+p-q}R^{pq-pr-q+r+qr-pq-r+p+pr-qr-p+q}$
$= A^0R^0= 1 \times 1 = 1 =$ R.H.S.
View full question & answer→Question 433 Marks
Find four numbers forming a geometric progression in which the third term is greater than the first term by $9$ and the second term is greater than by $4^{th}$ by $18.$
AnswerLet the four numbers in G.P. be $a, ar, ar^2, ar^3$
$\therefore ar^2 = a + 9$ and $ar = ar^3 + 18$
Now, $ar^2 - a = 9$
$\Rightarrow a(r^2 - 1) = 9 ...(i)$
And $ar - ar^3 = 18$
$\Rightarrow ar(1 - r^2) = 18+$
$\Rightarrow -ar(r^2 - 1) = 18 ...(ii)$
Dividing eq. $(ii)$ by eq. $(i),$ we have
$\frac { - a r \left( r ^ { 2 } - 1 \right) } { a \left( r ^ { 2 } - 1 \right) } = \frac { 18 } { 9 }$
$\Rightarrow r = -2$
Putting value of r in eq. (i), we get
$a(4 - 1) = 9$
$\Rightarrow a = 3$
$\therefore ar = 3 \times (-2) = -6$
$ar^2 = 3 \times (-2)^2 = 12a r ^ { 3 }$
$= 3 \times (-2)^3 = -24$
Therefore, the required numbers are $3, -6, 12, -24$
View full question & answer→Question 443 Marks
Show that the products of the corresponding terms of the sequences $a, ar, ar^2 ,........ar^{n-1} $ and $A, AR, AR^2......AR^{n-1}$ form a G.P. and find the common ratio.
AnswerMultiplying the corresponding terms of the given sequences, we have
$( a \times \mathrm { A } ) , ( a r \times \mathrm { AR } ) _ { : } \left( a r ^ { 2 } \times \mathrm { AR } ^ { 2 } \right), ........, \left( a r ^ { n - 1 } \times \mathrm { AR } ^ { n - 1 } \right)$
$\Rightarrow (A), (aAr^R), (aAr^2R^2), ....., (aAr^{n-1}R^{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 = $\frac { a \mathrm { A } r \mathrm { R } } { a \mathrm { A } } = r \mathrm { R }$
View full question & answer→Question 453 Marks
Find the sum of the product of the corresponding terms of the sequences $2, 4, 8,16, 32$ and $128, 32, 8, 2, \frac 12$.
AnswerMultiplying the corresponding terms of the given sequences $2, 4, 8,16, 32$ and $128, 32, 8, 2, \frac 12$
$(2 \times 128), (4 \times 32), (8 \times 8), (16 \times 2), (32 \times \frac 12)$
$\Rightarrow 256, 128, 64, 32, 16$ are in G.P.
Here $a = 256, r = \frac { 128 } { 256 } = \frac { 1 } { 2 }$ and $n = 5$
$\therefore \ \mathrm { S } _ { n } = \frac { a \left( 1 - r ^ { n } \right) } { 1 - r }$ when $r < 1$
$S_5 = \frac { 256 \left[ 1 - \left( \frac { 1 } { 2 } \right) ^ { 5 } \right] } { 1 - \frac { 1 } { 2 } } = 256 \times 2 \left( 1 - \frac 1{32} \right)$
$\Rightarrow S_5 = 256 \times 2 \times \frac {31}{32} = 496$
View full question & answer→Question 463 Marks
Find the sum to $n$ terms of the sequences $8, 88, 888, 8888, ……$
AnswerHere $Sn = 8 + 88 + 888 + 8888 + .......$ up to n terms
$\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$) $
$\Rightarrow S _ { n } = \frac { 8 } { 9 } \left[ ( 10 - 1 ) + \left( 10 ^ { 2 } - 1 \right) + \left( 10 ^ { 3 } - 1 \right) + \ldots . \text { up to } n \text { terms } \right]$
$\Rightarrow \mathrm { S } _ { n } = \frac { 8 } { 9 }[(10 + 10^2 + 10^3 + .....$ up to n terms$) - (1 + 1 + 1 + ......$ up to n terms$)]$
$\Rightarrow S _ { n } = \frac { 8 } { 9 } \left[ \frac { 10 \times \left( 10 ^ { n } - 1 \right) } { 10 - 1 } - n \right]$
$= \frac { 8 } { 9 } \left[ \frac { 10 } { 9 } \left( 10 ^ { n } - 1 \right) - n \right]$
$= \frac { 80 } { 81 } \left( 10 ^ { n } - 1 \right) - \frac { 8 } { 9 } n$
View full question & answer→Question 473 Marks
If the $4^{th}, 10^{th} $ and $16^{th} $ terms of a G.P. are $x, y$ and $z$ respectively. Prove that $x, y , z$ are in G.P.
AnswerLet a be the first term and r be the common ratio of given G.P.
$\therefore a_4 = x$
$\Rightarrow ar^3 = x ...(i)$
$a_{10} = y$
$\Rightarrow ar^9 = y ...(ii)$
$a_{16} = z$
$\Rightarrow ar^{15} = z ...(iii)$
From eq. $(ii), ar^9 = y$
$\Rightarrow (ar^9)^2 = y^2$
$\Rightarrow y^2 = (ar^3)(ar^{15})$
$\Rightarrow y^2 = xz [$From eq. $(i)$ and $(iii)]$
$\therefore x, y, z$ are in G.P.
View full question & answer→Question 483 Marks
Find a G.P. for which sum of the first two terms is $-4$ and the fifth term is $4$ times the third term.
AnswerLet a be the first term and r be the common ratio of given G.P.Given: $a + ar = -4$
$\Rightarrow a(1 + r) = -4 ……..(i)$
And $a_5 = 4a_3$
$\Rightarrow ar^4 = 4ar^2$
$\Rightarrow r^2 = 4$
$\Rightarrow r = \pm 2$
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 $4, -8, 16, -32,.......$
View full question & answer→Question 493 Marks
Given a G.P. with $a = 729$ and $7^{th} $ term $64,$ determine $S_7 .$
AnswerGiven: $a = 729$ and $ a_7 = 64\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 $r < 1$
$\Rightarrow S _ { 7 } = \frac { 729 \left[ 1 - \left( \frac { 2 } { 3 } \right) ^ { 7 } \right] } { 1 - \frac { 2 } { 3 } } = \frac { 729 \left[ 1 - \frac { 128 } { 2187 } \right] } { \frac { 3 - 2 } { 3 } }$
$\Rightarrow S _ { 7 } = 729 \times 3 \left( \frac { 2187 - 128 } { 2187 } \right)$
$\Rightarrow \mathrm { S } _ { 7 } = \frac { 729 \times 3 \times 2059 } { 2187 } = 2059$
View full question & answer→Question 503 Marks
The Sum of first three terms of a G.P. is $16$ and the Sum of the next three term is $128.$ determine the first term, the common ratio and the Sum to n terms of the G.P.
Answer$s_3 = 16$
$\frac{a\left(1-r^{3}\right)}{1-r} = 16 (1)$
$s_6 - s_3 = 128$
$\frac{a\left(1-r^{6}\right)}{1-r} - 16 = 128$
$\frac{a\left(1-r^{6}\right)}{1-r},= 144 (2)$
$(2) \div (1)$
$\frac{1-r^{6}}{1-r^{3}}=\frac{144}{16}$
$1 + r^3 = 9$
$r^3 = 8$
$r = 2$
$s_{3}=\frac{a\left(r^{3}-1\right)}{r-1} = 16$
$a = 16 / 7$
$s_{n}=\frac{a\left(r^{n}-1\right)}{r-1}$$=\frac{16}{7} \frac{\left(2^{n}-1\right)}{2-1}=\frac{16}{7}\left(2^{n}-1\right)$
View full question & answer→Question 513 Marks
How many terms of G.P. $3, 3^2 , 3^3 ......$ are needed to give the sum $120?$
AnswerHere, $a = 3$ and r = $ \frac { 3 ^ { 2 } } { 3 } = 3$
$\therefore ^ { \mathrm { S } _ { n } = \frac { a \left( r ^ { - } - 1 \right) } { r - 1 } }$ when $r > 1$
$\Rightarrow 120 = \frac { 3 \left( 3 ^ { n } - 1 \right) } { 3 - 1 }$
$\Rightarrow 120 = \frac { 3 } { 2 } \left( 3 ^ { n } - 1 \right)$
$ \Rightarrow 120 \times \frac { 2 } { 3 } = 3 ^ { n } - 1$
$ \Rightarrow 3^n = 81$
$\Rightarrow 3^n = (3)^4$
$\Rightarrow n = 4$
Therefore, the sum of $4$ terms of the given G.P. is $120$.
View full question & answer→Question 523 Marks
The sum of first three terms of a G.P. is $\frac{39}{10}$ and their product is $1$. Find the common ratio and the terms.
AnswerLet $\frac ar , a, r$ be first three terms of the given G.P.According to question$,\frac { a } { l ^ { r } } + a + a r = \frac { 39 } { 10 }……….(i)$
And$\frac { a } { r } \times a \times r = 1$
$\Rightarrow a ^ { 3 } = 1$
$\Rightarrow a = 1$
$Putting value of a in eq. (i),$
$\Rightarrow 10+ 10r + 10r^2 = 39r$
$\Rightarrow 10r^2- 29r +10 = 0$
$\Rightarrow r = \frac { - ( - 29 ) \pm \sqrt { ( - 29 ) ^ { 2 } - 4 \times 10 \times 10 } } { 2 \times 10 }$
$\Rightarrow r = \frac { 29 \pm \sqrt { 841 - 400 } } { 20 }$
$\Rightarrow r = \frac { 29 \pm 21 } { 20 }$
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 $\frac { 1 } { 2 / 5 } , 1,1 \times \frac { 2 } { 5 }$
$\Rightarrow\frac { 5 } { 2 } , 1 , \frac { 2 } { 5 }$
View full question & answer→Question 533 Marks
Evaluate:$\sum \limits_ { k = 1 } ^ { 11 } \left( 2 + 3 ^ { k } \right)$
AnswerGiven:$\sum _ { k = 1 } ^ { 11 } \left( 2 + 3 ^ { k } \right)$
$= (2 + 3^1) + (2 + 3^2) + (2 + 3^3) + (2 + 3^{11})$
$= ( 2 + 2 + 2 +........11\ \text{times}) + (3 + 3^2 + 3^3 +....... +3^{11})= 22 + (3 + 3^2 + 3^3 +....... +3^{11}) ……….(i)$
Here $3, 3^2,3^3 ....... ,3^{11}$ is 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 $\sum _ { k = 1 } ^ { 11 } \left( 2 + 3 ^ { k } \right) = 22 + \frac { 3 } { 2 } \left( 3 ^ { 11 } - 1 \right)$
View full question & answer→Question 543 Marks
Find the sum to indicated number of terms of the geometric progression $x^3, x^5, x^7 ... n$ terms $($if $x \ne \pm1)$.
AnswerHere,$a = x^3$ and $r = \frac { x ^ { 5 } } { x ^ { 3 } } = x^2$
$S _ { n } = \frac { a \left( 1 - r ^ { n } \right) } { 1 - r }$ when $r < 1$
$\Rightarrow S _ { n } = \frac { x ^ { 3 } \left[ 1 - \left( x ^ { 2 } \right) ^ { n } \right] } { 1 - x ^ { 2 } }$
$\Rightarrow \mathrm { S } _ { n } = \frac { x ^ { 3 } } { 1 - x ^ { 2 } } \left[ 1 - x ^ { 2 n } \right]$
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