Question 513 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 523 Marks
If the $4^{\text {th }}, 10^{\text {th }}$ and $16^{\text {th }}$ 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.
$\therefore a_4=x$
$\Rightarrow a r^3=x \ldots \text { (i) }$
$a_{10}=y$
$\Rightarrow a r^9=y \ldots \text { (ii) }$
$a_{16}=z$
$\Rightarrow a r^{15}=z \ldots \text { (iii) }$
From eq. (ii), $a r^9=y$
$\Rightarrow\left(a r^9\right)^2=y^2$
$\Rightarrow y^2=\left(a r^3\right)\left(a r^{15}\right)$
$\Rightarrow y^2=x z[\text { From eq. (i) and (iii)] }$
$\therefore x, y, z \text { are in G.P. }$
View full question & answer→Question 533 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: $a+a r=-4$
$\Rightarrow a(1+r)=-4$……..(i)
$\text { And } a_5=4 a_3$
$\Rightarrow a r^4=4 r^2$
$\Rightarrow r^2=4$
$\Rightarrow r= \pm 2$$\qquad$
Putting $\mathrm{r}=2 r=2$ in eq. (i), we get $\mathrm{a}(1+2)=-4$
$\Rightarrow a=\frac{-4}{3}$
Therefore, required G.P. is $\frac{-4}{3}, \frac{-8}{3}, \frac{-16}{3}, \ldots$
Putting $r=-2$ in eq. (i), we get $a(1-2)=-4$
$\Rightarrow a=4$
Therefore, required G.P. is $4,-8,16,-32, \ldots . . .$.
View full question & answer→Question 543 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 553 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+\mathrm{r}^3=9$
$\mathrm{r}^3=8$
$\mathrm{r}=2$
$s_3=\frac{a\left(r^3-1\right)}{r-1}=16$
$\mathrm{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 563 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 573 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 583 Marks
Evaluate:$\sum \limits_ { k = 1 } ^ { 11 } \left( 2 + 3 ^ { k } \right)$
AnswerGiven:$\sum _ { k = 1 } ^ { 11 } \left( 2 + 3 ^ { k } \right)$
$=\left(2+3^1\right)+\left(2+3^2\right)+\left(2+3^3\right)+\left(2+3^{11}\right)$
$=(2+2+2+\ldots \ldots . .11 \text { times })+\left(3+3^2+3^3+\ldots \ldots . .+3^{11}\right)$
$=22+\left(3+3^2+3^3+\ldots \ldots . .+3^{11}\right) \ldots \ldots \ldots . \text { (i) }$
Here $3,3^2, 3^3 \ldots \ldots . ., 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 593 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]$
View full question & answer→Question 603 Marks
The sums of n terms of two arithmetic progressions are on the ratio 5n + 4:9n + 6. Find the ratio of their $18^{th}$ terms.
AnswerLet $a_1, a_2$ and $d_1, d_2$ be the first terms and common differences of two A.P’s respectively.$\therefore \frac { \frac { n } { 2 } \left[ 2 a _ { 1 } + ( n - 1 ) d _ { 1 } \right] } { \frac { n } { 2 } \left[ 2 a _ { 2 } + ( n - 1 ) d _ { 2 } \right] } = \frac { 5 n + 4 } { 9 n + 6 }$
$\Rightarrow \frac { 2 a _ { 1 } + ( n - 1 ) d _ { 1 } } { 2 a _ { 2 } + ( n - 1 ) d _ { 2 } } = \frac { 5 n + 4 } { 9 n + 6 }$
$\Rightarrow \frac { a _ { 1 } + \left( \frac { n - 1 } { 2 } \right) d _ { 1 } } { a _ { 2 } + \left( \frac { n - 1 } { 2 } \right) d _ { 2 } } = \frac { 5 n + 4 } { 9 n + 6 }$
Now, to get $18^{th}$ term, $\frac { n - 1 } { 2 } = 17$
$\Rightarrow$ n = 35
$\therefore \frac { a _ { 1 } + 17 d _ { 1 } } { a _ { 2 } + 17 d _ { 2 } } = \frac { 5 \times 35 + 4 } { 9 \times 35 + 6 }$
$\Rightarrow \frac { a _ { 1 } + 17 d _ { 1 } } { a _ { 2 } + 17 d _ { 2 } } = \frac { 179 } { 321 }$
Therefore, the ratio of $18^{th}$ terms of two A.P.’s is 179: 321.
View full question & answer→Question 613 Marks
If the sum of n terms of an A.P. is $(pn + qn^2)$, where p and q are constants, find the common difference.
AnswerGiven: $S_n=p n+q n^2$
Put $n=1$ we get, $S_1=p+p \Rightarrow a=p+q \ldots$... $(i)$
Now $S_n=p n+q n^2$
$\Rightarrow \frac{n}{2}[2 a+(n-1) d]=p n+q n^2$
$\Rightarrow \frac{n}{2}[2(p+q)+(n-1) d]=p n+q n^2 \text { (using.(i)) }$
$\Rightarrow 2 n(p+q)+n(n-1) d=2 p n+2 q n^2$
$\Rightarrow 2 n q+n(n-1) d=2 q n^2$
$\Rightarrow n(n-1) d=2 q n(n-1)$
$\Rightarrow d=2 q$
View full question & answer→Question 623 Marks
Find the sum of $n$ terms of an A.P. whose $k^{\text {th }}$ term is $5 k+1$.
AnswerGiven: $a_k=5 k+1$
Putting $\mathrm{k}=1$ and $\mathrm{k}=\mathrm{n}$, we get
$a=5 \times 1+1=6$ and $a_n=5 n+1$
$\therefore S_{n} = \frac{n}{2}(a + l)$
$\Rightarrow S_{n}=\frac{n}{2}(6+5 n+1)=\frac{n}{2}(5 n+7)$
View full question & answer→Question 633 Marks
If the sum of a certain number of terms of the A.P. $25, 22, 19$, ….. is $116$, find the last term.
AnswerHere $\mathrm{a}=25, \mathrm{~d}=22-25=-3$ and $\mathrm{S}_{\mathrm{n}}=116$
We have $S_n=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow 116=\frac{n}{2}[2 \times 25+(n-1) \times(-3)]$
$\Rightarrow 232=n[50-3 n+3]$
$\Rightarrow 232=53 n-3 n^2$
$\Rightarrow 3 n^2-53 n+232=0$
$\Rightarrow$ n = $ \frac { - ( - 53 ) \pm \sqrt { ( - 53 ) ^ { 2 } - 4 \times 3 \times 232 } } { 2 \times 3 }$
$\Rightarrow$ n = $\frac { 53 \pm \sqrt { 2809 - 2784 } } { 6 }$
$\Rightarrow$ n = $\frac { 53 \pm \sqrt { 25 } } { 6 } = \frac { 53 \pm 5 } { 6 }$
$\Rightarrow$ n $\frac { 53 + 5 } { 6 }$ or n = $\frac { 53 - 5 } { 6 }$
$\Rightarrow$ n = $ \frac { 58 } { 6 }$ or n = $\frac { 48 } { 6 }$ = 8
But n = $ \frac { 58 } { 6 }$ is not possible as n $\in$ N. Therefore, n = 8
Now, $a_n = a + (n - 1)d$
$\Rightarrow a_8 = 25 + (8 - 1) \times (-3)$
View full question & answer→Question 643 Marks
In an A.P., if $p^{th}$ term is $\frac { 1 } { q }$ and $q^{th}$ term is $\frac { 1 } { p },$ prove that the sum of first pq terms is $\frac { 1 } { 2 } ( p q + 1 ),$ where $p \neq q.$
AnswerLet a be the first term and d be the common difference of given A.P.And $a _ { p } = \frac { 1 } { q }$ and $a _ { q } = \frac { 1 } { p }$
$\therefore a + ( p - 1 ) d = \frac { 1 } { q }$ and $a + ( q - 1 ) d = \frac { 1 } { p }$
$\Rightarrow a + p d - d = \frac { 1 } { q }$ …..(i) and $a + q d - d = \frac { 1 } { p }$…..(ii)
Subtracting eq. (ii) from eq. (i), we get
a + pd - d - (a + qd - d) $= \frac { 1 } { q } - \frac { 1 } { p }$
$\Rightarrow$ pd - d - a - qd + d $= \frac { p - q } { p q }$
$\Rightarrow ( p - q ) d = \frac { p - q } { p q }$
$\Rightarrow d = \frac { p - q } { p q } \times \frac { 1 } { p - q } = \frac { 1 } { p q }$
Putting value of d in eq. (i), we get
$a + p{1 \over {pq}} - d = {1 \over q}$
$\Rightarrow a + {1 \over q} - d = {1 \over q}$
$\Rightarrow a = {1 \over q} + d - {1 \over q} = d = {1 \over {pq}}$
Now, $S _ { n } = \frac { n } { 2 } [ 2 a + ( n - 1 ) d ]$
$\Rightarrow S _ { p q } = \frac { p q } { 2 } \left[ 2 \times \frac { 1 } { p q } + ( p q - 1 ) \times \frac { 1 } { p q } \right]$
$\Rightarrow S _ { p q } = \frac { p q } { 2 } \left[ \frac { 2 } { p q } + \frac { p q - 1 } { p q } \right]$
$\Rightarrow S _ { p q } = \frac { p q } { 2 } \left[ \frac { 2 + p q - 1 } { p q } \right]$
$\Rightarrow S _ { p q } = \frac { p q } { 2 } \left[ \frac { 1 + p q } { p q } \right] _ { = } \frac { p q + 1 } { 2 }$
$\Rightarrow S _ { p q } = \frac { 1 } { 2 } ( p q + 1 )$
View full question & answer→Question 653 Marks
How many terms of the A.P., $- 6 , \frac { - 11 } { 2 } , - 5 , \ldots$are needed to give the sum -25?
AnswerHere, a = -6, $d = \frac { - 11 } { 2 } - ( - 6 ) = \frac { - 11 } { 2 } + 6 = \frac { 1 } { 2 }$$\therefore {{\text{S}}_n} = \frac{n}{2}[2a + (n - 1)d]$
$\Rightarrow - 25 = \frac { n } { 2 } \left[ 2 \times ( - 6 ) + ( n - 1 ) \times \frac { 1 } { 2 } \right]$
$\Rightarrow - 50 = n \left[ - 12 + \frac { n - 1 } { 2 } \right]$
$\Rightarrow - 50 = n \left[ \frac { - 24 + n - 1 } { 2 } \right]$
$\Rightarrow-100=n^2-25 n$
$\Rightarrow n^2-25 n+100=0$
$\Rightarrow(n-20)(n-5)=0$
$\Rightarrow n=20 \text { or } n=5$
View full question & answer→Question 663 Marks
In an A.P. the first term is $2$ and the sum of the first five terms is one-fourth of the next five terms. Show that $20th$ term is $-112$.
AnswerAccording to question, $a=2$ and $\mathrm{S}_5=\frac{1}{4}\left[\mathrm{~S}_{10}-\mathrm{S}_5\right]$
$\Rightarrow 4 \mathrm{~S}_5=\mathrm{S}_{10}-\mathrm{S}_5$
$\Rightarrow 5 \mathrm{~S}_5=\mathrm{S}_{10}$
$\left.\Rightarrow \sqrt[5]{\left.\frac{5}{2}\{2 \times 2+(5-1) d\}\right]}=\frac{10}{2}[2 \times 2+(10-1) \mathrm{d}] \text { since } \mathrm{S}_{\mathrm{n}}=\frac{n}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}\}\right]$
$\Rightarrow \frac{25}{2}[4+4 \mathrm{~d}]=5[4+9 \mathrm{~d}]$
$\Rightarrow 25[4+4 \mathrm{~d}]=10[4+9 \mathrm{~d}]$
$\Rightarrow 100+100 \mathrm{~d}=40+90 \mathrm{~d}$
$\Rightarrow 10 \mathrm{~d}=-60$
$\Rightarrow \mathrm{~d}=-6$
Now, $a_n=a+(n-1) d$
$\Rightarrow a_{20}=2+(20-1) \times(-6)$
$\Rightarrow a_{20}=2-114=-112$
View full question & answer→Question 673 Marks
Find the sum of all natural numbers lying between 100 and 1000 which are multiples of 5.
AnswerAccording to question, series is 105, 110, 115, 120, ………, 995
Here a = 105, d = 110 - 105 = 5 and $a_n = 995$
$\therefore a_n = a + (n - 1)d$
$\Rightarrow 995 = 105 + ( n - 1 ) \times 5$
$\Rightarrow 995 - 105 = ( n - 1 ) \times 5$
$\Rightarrow \frac { 890 } { 5 } = ( n - 1 )$
$\Rightarrow$ n = 178 + 1 = 179
Now, $\mathrm { S } _ { n } = \frac { n } { 2 } ( a + l )$
$\Rightarrow S _ { 199 } = \frac { 179 } { 2 } ( 105 + 995 )$
$\Rightarrow S _ { 179 } = \frac { 179 } { 2 } \times 1100=98450$
View full question & answer→Question 683 Marks
The difference between any two consecutive interior angles of a polygon is $5^{\circ}$. If the smallest angle is $120^{\circ}$. find the number of the sides of the polygon.
AnswerLet the number of sides of polygon be $n$. The interior angles of the polygon form an A.P.
Here, $a=120^{\circ}$ and $d=5^{\circ}$
Since, Sum of interior angles of a polygon with $n$ sides is $(n-2) \times 180^{\circ}$
$\therefore S_n=(n-2) \times 180^{\circ}$
$\Rightarrow \frac{n}{2}[2 \times 120+(n-1) \times 5]=180 n-360$
$\Rightarrow 120 n+\frac{5 n^2-5 n}{2}=180 n-360$
$\Rightarrow 240 n+5 n^2-5 n=360 n-720$
$\Rightarrow 5 n^2-125 n+720=0$
divide by 5 , we get
$\Rightarrow n^2-25 n+144=0$
$\Rightarrow(n-16)(n-9)=0$
$\Rightarrow n=16 \text { or } n=9$
But $\mathrm{n}=16$ not possible because $\mathrm{a}_{16}=\mathrm{a}+15 \mathrm{~d}=120+15 \times 5=195^{\circ}>180^{\circ}$
Therefore, number of sides of the polygon are 9
View full question & answer→Question 693 Marks
A man starts repaying a loan as first installment of ₹ 100. If he increases the installment by ₹ 5 every month, what amount he will pay in the $30^{th}$ installment?
AnswerAmount of $1^{st}$ installment = Rs. 100 and Amount of $2^{nd}$ installment = Rs. 105
The monthly installments 100,105,110, .............. form an A.P
$\therefore$ a = 100, d = 105 - 100 =5 and n = 30
Now $a_n = a + (n - 1)d$
$\Rightarrow a _ { 30 } = 100 + ( 30 - 1 ) \times 5$
$\Rightarrow a _ { 30 } = 100 + 29 \times 5$
$\Rightarrow a_{30} = 100 + 145 = Rs. 245$
Therefore, the amount of $30^{th}$ installment is Rs. 245.
View full question & answer→Question 703 Marks
Between 1 and $31, \mathrm{~m}$ numbers have been inserted in such a way that resulting sequence is an A.P. and the ratio of $7^{\text {th }}$ and $(m-1)^{\text {th }}$ numbers is $5: 9$. Find the value of $m$.
AnswerLet $A_1, A_2, A_3, A_4$, …….., $A_m$ be $m$ numbers between 1 and 31 . Here, $a=1$ and let the common difference be d .
$\therefore a_{m+2}=31$
$\Rightarrow a+(m+2-1) d=31$
$\Rightarrow 1+(m+1) d=31$
$\Rightarrow d=\frac{30}{m+1}$
Now, $\mathrm{A}_7=a+7 d=1+7 \times\left(\frac{30}{m+1}\right)=\frac{m+1+210}{m+1}=\frac{m+211}{m+1}$
And $\mathrm{A}_{\mathrm{m}-1}=\mathrm{a}+(\mathrm{m}-1) \mathrm{d}=1+(m-1) \times\left(\frac{30}{m+1}\right)=\frac{m+1+30 m-30}{m+1}=\frac{31 m-29}{m+1}$
According to question, $\frac{A_7}{A_{m-1}}=\frac{5}{9}$
$\Rightarrow \frac{\frac{m+211}{m+1}}{\frac{31 m+29}{m+1}}=\frac{5}{9}$
$\Rightarrow \frac{m+211}{31 m-29}=\frac{5}{9}$
$\Rightarrow 9 m+1899=155 m-145$
$\Rightarrow 146 m=2044$
$\Rightarrow m=14$
View full question & answer→Question 713 Marks
If the sum of $n$ terms of an A.P. is $3 n^2+5 n$ and its $m^{\text {th }}$ term is 164 , find the value of $m$.
AnswerGiven: $S_n=3 n^2+5 n$ and $a_m=164$
Put $\mathrm{n}=1$ and $\mathrm{n}=2$ in $\mathrm{S}_n$ we get,
$\mathrm{S}_1=3+5=8$
$\Rightarrow a=8 \ldots$
and $S_2=3(4)+5(2)=12+10=22$
We have $a_n=S_n-S_{n-1}$
$\therefore a_2=S_2-S_1$
$\Rightarrow a_2=22-8=14$
Now $d=a_2-a_1=14-8=6$
$\therefore a_m=164$
$\Rightarrow a+(m-1) d=164$
$\Rightarrow 8+(m-1) 6=164$
$\Rightarrow 8+6 m-6=164$
$\Rightarrow 6 m=162$
$\Rightarrow m=\frac{162}{6}=27$
View full question & answer→Question 723 Marks
The ratio of the sum of $m$ and $n$ terms of an A.P. is $m^2: n^2$ Show that the ratio of $m^{\text {th }}$ and $n^{\text {th }}$ term is $(2 m-1):(2 n-1)$.
AnswerLet a be the first term and d be the common difference of given A.P.
$\therefore S _ { m } = \frac { m } { 2 } [ 2 a + ( m - 1 ) d ]$ and $\mathrm { S } _ { n } = \frac { n } { 2 } [ 2 a + ( n - 1 ) d ]$
According to question, $\frac { S _ { m } } { S _ { n } } = \frac { \frac { m } { 2 } [ 2 a + ( m - 1 ) d ] } { \frac { n } { 2 } [ 2 a + ( n - 1 ) d ] } = \frac { m ^ { 2 } } { n ^ { 2 } }$
$\Rightarrow \frac { 2 a + ( m - 1 ) d } { 2 a + ( n - 1 ) d } = \frac { m ^ { 2 } } { n ^ { 2 } } \times \frac { n } { 2 } \times \frac { 2 } { m }$
$\Rightarrow \frac { 2 a + ( m - 1 ) d } { 2 a + ( n - 1 ) d } = \frac { m } { n }$
$\Rightarrow 2 a n+n(m-1) d=2 a m+m(n-1) d$
$\Rightarrow 2 a n-2 a m=(m n-m) d-(m n-n) d$
$\Rightarrow 2 a(n-m)=(m n-m-m n+n) d$
$\Rightarrow 2 a(n-m)=(n-m) d$
$\Rightarrow d=2 a$
Now, $\frac { a _ { m } } { a _ { n } } = \frac { a + ( m - 1 ) d } { a + ( n - 1 ) d } = \frac { a + ( m - 1 ) 2 a } { a + ( n - 1 ) 2 a }$
$= {{a(1 + 2m - 2)} \over {a(1 + 2n - 2)}} = {{2m - 1} \over {2n - 1}}$
$\therefore a_m:a_n = (2m - 1):(2n - 1)$
View full question & answer→Question 733 Marks
Sum of the first p, q and r terms of an A.P. are a, b and c respectively. Prove that $\frac { a } { p } ( q - r ) + \frac { b } { q } ( r - p ) + \frac { c } { r } ( p - q ) = 0.$
AnswerLet A be the first term and d be the common difference of given A.P. $\therefore S _ { p } = \frac { p } { 2 } [ 2 \mathrm { A } + ( p - 1 ) d ] = a$
$\Rightarrow A + \frac { p - 1 } { 2 } d = \frac { a } { p }$……….(i)
$\mathrm { S } _ { q } = \frac { q } { 2 } [ 2 \mathrm { A } + ( q - 1 ) d ] = b$
$\Rightarrow A + \frac { q - 1 } { 2 } d = \frac { b } { q }$……….(ii)
$\mathrm { S } _ { r } = \frac { r } { 2 } [ 2 \mathrm { A } + ( r - 1 ) d ] = c$
$\Rightarrow A + \frac { r - 1 } { 2 } d = \frac { c } { r }$……….(iii)
Now $\frac { a } { p } ( q - r ) + \frac { b } { q } ( r - p ) + \frac { c } { r } ( p - q ) = 0$
Putting the values of $\frac { a } { p } , \frac { b } { q }$ and $\frac { c } { r }$ from eq. (i), (ii) and (iii), we get
$\left[ \mathrm { A } + \frac { p - 1 } { 2 } d \right] ( q - r ) + \left[ \mathrm { A } + \frac { q - 1 } { 2 } d \right] ( r - p ) + \left[ \mathrm { A } + \frac { r - 1 } { 2 } d \right] ( p - q ) = 0$
$\Rightarrow A ( q - r + r - p + p - q ) + d \left[ \frac { p - 1 } { 2 } ( q - r ) + \frac { q - 1 } { 2 } ( r - p ) + \frac { r - 1 } { 2 } ( p - q ) \right] = 0$
$\Rightarrow A ( 0 ) + d \left[ \frac { p q - p r - q + r } { 2 } + \frac { q r - p q - r + p } { 2 } + \frac { p r - q r - p + q } { 2 } \right] = 0$
$\Rightarrow 0 + d \left[ \frac { p q - p r - q + r + q r - p q - r + p + p r - q r - p + q } { 2 } \right] = 0$
$\Rightarrow 0 + d \left[ \frac { 0 } { 2 } \right] = 0$
$\Rightarrow$ 0 = 0 = 0
$\Rightarrow$ 0 = 0
$\Rightarrow$ L.H.S. = R.H.S. Proved.
View full question & answer→Question 743 Marks
If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first $(p + q)$ terms.
AnswerLet a be the first term and d be the common difference of given A.P.
$\therefore$ $S _ { p } = \frac { p } { 2 } [ 2 a + ( p - 1 ) d ]$ and $\mathrm { S } _ { q } = \frac { q } { 2 } [ 2 a + ( q - 1 ) d ]$
According to question, $S_p = S_q$
$\Rightarrow \frac { p } { 2 } [ 2 a + ( p - 1 ) d ] = \frac { q } { 2 } [ 2 a + ( q - 1 ) d ]$
$\Rightarrow 2 a p+p^2 d-p d=2 a q+q^2 d-q d$
$\Rightarrow 2 a p-2 a q=q^2 d-p^2 d+p d-q d$
$\Rightarrow 2 a(p-q)=\left[-\left(p^2-q^2\right) d+(p-q) d\right]$
$\Rightarrow 2 a(p-q)=[-(p-q)(p+q) d+(p-q) d]$
$\Rightarrow 2 a(p-q)=(p-q)[1-p-q] d$
$ \Rightarrow$ a = $\frac { ( 1 - p - q ) d } { 2 }$
Now $\mathrm { S } _ { p + q } = \frac { p + q } { 2 } \left[ \frac { 2 ( 1 - p - q ) d } { 2 } + ( p + q - 1 ) d \right]$
$= \frac { p + q } { 2 } [ d - p d - q d + p d + q d - d ]$
$\Rightarrow S _ { p + q } = \frac { p + q } { 2 } \times 0 = 0$
View full question & answer→Question 753 Marks
Find the sum of odd integers from 1 to 2001.
AnswerOdd integers from 1 to 2001 are $1,3,5,7, \ldots, 2001$.
Here, $a=1, d=3-1=2$ and $a_n=2001$
$\therefore a_n=a+(n-1) d$
$\therefore 2001=1+(n-1) \times 2$
$\Rightarrow 2001-1=(n-1) \times 2$
$\Rightarrow \frac{2000}{2}=(n-1)$
$\Rightarrow n=1000+1=1001$
Now, $\mathrm{S}_{\mathrm{n}}=\frac{n}{2}(\mathrm{a}+\mathrm{I})$
$\Rightarrow S_{1001}=\frac{1001}{2}(1+2001)$
$\Rightarrow S_{1001}=\frac{1001}{2} \times 2002=1002001$
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