5 Marks Questions

Question 15 Marks

Find the sum of the series, $1.3 .4+5.7 .8+9.11 .12+\ldots$ upto $n$ terms.

Answer

Let $T_n$ be the $n^{\text {th }}$ term of the given series, Then$
\begin{aligned}
T_n= & \left(n^{\text {th }} \text { term of } 1,5,9 \ldots\right) \times\left(n^{\text {th }} \text { term } 3,7,11 \ldots\right) \\
& \times\left(n^{\text {th }} \text { term of } 4,8,12 \ldots\right) \\
= & {[1+(n-1) 4] \times[3+(n-1) 4] \times[4+(n-1) 4] } \\
= & (4 n-3) \times(4 n-1) \times(4 n) \\
= & \left(16 n^2-16 n+3\right) 4 n \\
= & 64 n^3-64 n^2+12 n
\end{aligned}
$
Let $S_n$ denote the sum to $n$ terms of the given series. Then,$
\begin{aligned}
S_n =\sum_{k=1}^n T_k=\sum_{k=1}^n\left(64 n^3-64 n^2+12 n\right) \\
=64 \sum_{k=1}^n n^3-64 \sum_{k=1}^n n^2+12 \sum_{k=1}^n n
\end{aligned}
$
$\begin{array}{l}=64\left[\frac{n(n+1)}{2}\right]^2-64\left[\frac{n(n+1)(2 n+1)}{6}\right]+12\left[\frac{n(n+1)}{2}\right] \\ =\frac{64\{n(n+1)\}^2}{4}-\frac{64\{n(n+1)(2 n+1)\}}{6}+6 n(n+1) \\ =16\{n(n+1)\}^2-\frac{32\{n(n+1)(2 n+1)\}}{3}+6 n(n+1) \end{array}$
$\begin{array}{l}=\frac{n(n+1)}{3}[48\{n(n+1)\}-32(2 n+1)+18] \\ =\frac{n(n+1)}{3}\left[48\left(n^2+n\right)-(64 n+32)+18\right]\end{array}$
$\begin{array}{l}=\frac{n(n+1)}{3}\left[48 n^2+48 n-64 n-32+18\right] \\ =\frac{n(n+1)}{3}\left[48 n^2-16 n-14\right]\end{array}$

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Question 25 Marks

If $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of an A.P. and G.P. are both $a, b$ and $c$ respectively, then show that $a^{b-c}$. $b^{c-a} \cdot c^{a-b}=1$.

Answer

Let $A, d$ are the first term and common difference of A.P. and $x . R$ are the first term and common ratio of G.P., respectively.
According to the given condition,
$
\begin{array}{l}
A+(p-1) d=a \quad \ldots(i) \\
A+(q-1) d=b \quad \ldots(ii) \\
A+(r-1) d=c \quad \ldots(iii) \\
and \quad\quad\quad a=x R^{p-1} \quad \ldots(iv) \\
\quad\quad\quad\quad b=x R^{q-1} \quad \ldots(v) \\
\quad\quad\quad\quad c=x R^{r-1} \quad \ldots(vi)
\end{array}
$
On subtracting Eq. (ii) from Eq. (i). we get
$
\begin{aligned}
& & d(p-1-q+1) & =a-b \quad \ldots(vii) \\
\Rightarrow & & a-b & =d(p-q)
\end{aligned}
$
On subtracting Eq. (iii) from Eq. (ii). we get
$
\begin{aligned}
& & d(q-1-r+1) & =b-c \\
\Rightarrow & & b-c & =d(q-r)\quad \ldots(viii)
\end{aligned}
$
On subtracting Eq. (i) from Eq. (iii). we get
$
\begin{aligned}
& & d(r-1-p+1) & =c-a \\
\Rightarrow & & c-a & =d(r-p)\quad \ldots(ix)
\end{aligned}
$
Now, we have to prove $a^{b-c} b^{c-a} c^{a-b}=1$
Taking LHS $=a^{b-c} b^{c-a} c^{a-b}$
Using Eqs. (iv), (v), (vi) and (vii), (viii), (ix),
$\begin{array}{rlr}\text { LHS } & =\left(x R^{p-1}\right)^{d(q-r)}\left(x R^{q-1}\right)^{d(r-p)}\left(x R^{r-1}\right)^{d(p-q)} \\ & =x^{d(q-r)+d(r-p)+d(p-q)} \\ & =R^{d(q-r+p+p-q) d(q-r)+(q-1) d(r-p)+(r-1) d(p-q)} \\ & \quad R^{d(p q-p r-q+r+q r-p q-r+p+r p-r q-p+q)} \\ & =x^0 R^0=1 & 1 \\ & =\text { RHS } & \end{array}$
Hence Proved

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Question 35 Marks

Let $S$ be the sum, $P$ be the product and $R$ be the sum of reciprocals of $n$ terms in a G.P. Prove that $P^2 R^n=S^n$.

Answer

Let the G.P. is $a, a r, a r^2, a r^3 \ldots \ldots . a r^{n-1}$
Given, $S=$ sum of $n$ terms $=a, a r, a r^2, a r^3 \ldots \ldots a r^{n-1}$
$=\frac{a\left(r^n-1\right)}{r-1} \quad(\text { let } r>1)\quad \ldots(i) $
and $R=$ sum of the reciprocals of $n$ terms
$
=\frac{1}{a}+\frac{1}{a r}+\frac{1}{a r^2}+\ldots . .+\frac{1}{a r^{n-1}} \quad\left(\frac{1}{r}<1\right)
$
$=\frac{\frac{1}{a}\left[1-\left(\frac{1}{r}\right)^n\right]}{1-\frac{1}{r}}=\frac{1}{a}\left[\frac{1}{1}-\frac{1}{r^n}\right] \times \frac{1}{\frac{r-1}{r}}$
$=\frac{1}{a}\left[\frac{r^n-1}{r^n}\right] \times \frac{r}{r-1}$
$\Rightarrow \quad R=\frac{\left(r^n-1\right) r}{a r^n(r-1)} \quad \ldots(ii) $
and
$
\quad \begin{aligned}
P= & \text { Product } n \text { terms } \\
= & a \times a r \times a r^2 \times a r^3 \times \ldots \ldots \times a r^{n-1} \\
= & a^{1+1+1 \ldots . . .+n \text { terms }} \\
& r^{1+2+3+\ldots . .+(n-1) \text { terms }} \\
= & a^n r^{\frac{n(n-1)}{2}} \quad\left[\because \sum n=\frac{n(n+1)}{2}\right] \\
\Rightarrow \quad P^2= & a^{2 n} r^{n(n-1)} \quad \ldots(iii)
\end{aligned}
$
Now, we have to prove $P^2 R^n=S^n$
or $
P^2=\frac{S^n}{R^n} \text { or } P^2=\left(\frac{S}{R}\right)^n
$
$
\begin{aligned}
\text { Taking RHS } & =\left(\frac{S}{R}\right)^n=\left[\frac{a\left(r^n-1\right)}{r-1} \times \frac{a r^n(r-1)}{\left(r^n-1\right) r}\right]^n 1 \\
& \quad \text [{ using \ eqs. \ (u) \ and \ (ii) }] \\
& =\left[a^2 r^n r^{-1}\right]^n=\left(a^2 r^{n-1}\right)^n=\left[a^{2 n} r^{n(n-1)}\right] \\
& =P^2=\text { LHS } \quad\quad\quad [from eq. (iii)]
\end{aligned}
$
Hence Proved.

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Question 45 Marks

If $p, q, r$ are in G.P. and the equations $p x^2+2 q x+$ $r=0$ and $d x^2+2 e x+f=0$ have a common root, then show that $\frac{d}{p}, \frac{e}{q}, \frac{f}{r}$ are in A.P.

Answer

It is given that $p, q, r$ are in G.P.
$
\begin{array}{rlrl}\therefore q^2 =p r \\
\text { Now, } \quad p x^2+2 q x+r & =0 \\
\Rightarrow\quad p x^2+2 x \cdot \sqrt{p r}+r & =0 \\
\Rightarrow \quad\quad (x \cdot \sqrt{p}+\sqrt{r})^2 & =0 \\
\Rightarrow\quad\quad\quad\quad \sqrt{p} x+\sqrt{r} & =0 \\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad x & =-\sqrt{\frac{r}{p}}
\end{array}
$
It is given that the equations $p x^2+2 q x+r=0$ and $d x^2+2 e x+f=0$ have a common root and the equation $p x^2+2 q x+r=0$ has equal roots equal to $-\sqrt{\frac{r}{p}}$
$\therefore-\sqrt{\frac{r}{p}}$ is a root of the equation $d x^2+2 e x+f=0$
$
\Rightarrow \quad d \frac{r}{p}-2 e \sqrt{\frac{r}{p}}+f=0
$
$\begin{array}{lrr}\Rightarrow & \frac{d}{p}-2 e \sqrt{\frac{1}{p r}}+\frac{f}{r}=0 & \text { [Dividing by } r] \\ \Rightarrow & \frac{d}{p}-\frac{2 e}{q}+\frac{f}{r}=0 & {\left[\because q^2=p r\right]}\end{array}$
$\begin{array}{l}\Rightarrow \quad\quad\quad\quad\quad\quad 2 \frac{e}{q}=\frac{d}{p}+\frac{f}{r} \\ \Rightarrow \frac{d}{p}, \frac{e}{q}, \frac{f}{r} \text { are in A.P. }\end{array}$

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Question 55 Marks

The sum of an infinite G.P. is 57 and the sum of the cubes of its term is 9747 , find the G.P.

Answer

Let the first term of G.P. be 'a' and the common ratio be'r' where - 1 < r < 1
The G.P. is a, ar, ar²....
Therefore the sum of the infinite terms of the G.P. is
$\frac{a}{1-r}=57$....(i)
If taking the cube of each terms, the new G.P. is $a^3$, $a^3 r^3, a^3 r^6, \ldots$
Therefore the sum of the infinite terms of their cube is $\frac{a^3}{1-r^3}=9747$......(ii)
Taking the cube of the (i)
$\frac{a^3}{(1-r)^3}=(57)^3$
$\Rightarrow \quad a^3=(57)^3(1-r)^3$.....(iii)
Substituting the value of r in eq (ii), we get $\frac{(57)^3(1-r)^3}{1-r^3}=9747$
$\frac{(1-r)^3}{(1-r)\left(1+r^2+r\right)}=\frac{9747}{(57)^3}$
$\frac{(1-r)^2}{\left(1+r^2+r\right)}=\frac{1}{19}$
$\frac{1+r^2-2 r}{1+r^2+r}=\frac{1}{19}$
$\begin{aligned} 19\left(1+r^2-2 r\right) & =1+r^2+r \\ 18 r^2-39 r+18 & =0 \\ 6 r^2-13 r+6 & =0 \\ 6 r^2-9 r-4 r+6 & =0 \\ (2 r-3)(3 r-2) & =0\end{aligned}$
Therefore, $r=\frac{3}{2}$ or $r=\frac{2}{3}$ since $r<1$
$\therefore$ $r=\frac{2}{3}$
Substitute in eq. (i).
$\frac{a}{1-\frac{2}{3}}=57$
$\Rightarrow$ 3a = 57
$\Rightarrow$ $a=\frac{57}{3}=19$
Thus, the first term of the G.P. is 19 and the common ratio is $\frac{2}{3}$.
The G.P. is 19, $\frac{38}{3}, \frac{76}{9}$ and so on.

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Question 65 Marks

Sum of the first $p, q$ and $r$ terms of an A.P. are $a, b$ and $c$ respectively. Prove that
\[\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0\]

Answer

Let the first term is $A$ and common difference is $d$.
$
\begin{array}{l}
Given,\ S_p= \frac{p}{2}[2 A+(p-1) d]=a \\
\quad\quad\quad\left(\because S_n=\frac{n}{2}[2 a+(n-1) d]\right) \quad \ldots(i) \\
\quad\quad\quad S_q= \frac{q}{2}[2 A+(q-1) d]=b \quad \ldots(ii) \\
and\quad S_r= \frac{r}{2}[2 A+(r-1) d]=c \quad \ldots(iii)
\end{array}
$
Now, we are to prove
$
\begin{array}{l}
\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0 \\
\text { LHS }=\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)
\end{array}
$
Putting the values of $a, b$ and $c$ from Eqs. (i), (ii) and (iii) respectively, we get$
\begin{aligned}
\therefore LHS= & \frac{1}{p} \times \frac{p}{2}[2 A+(p-1) d](q-r)+\frac{1}{q} \times \frac{q}{2} \\
{[2 A+} & (q-1) d](r-p)+\frac{1}{r} \times \frac{r}{2}[2 A+(r-1) d](p-q) \\
= & \frac{1}{2}\{\{2 A+(p-1) d\}(q-r)+\{2 A+(q-1) d\} \\
& (r-p)+[2 A+(r-1) d\}(p-q)]
\end{aligned}
$
$\begin{aligned}\quad\quad\quad= & \frac{1}{2}[2 A(q-r)+(p-1) d(q-r)+2 A(r-p) \\ + & (q-1) d(r-p)+2 A(p-q)+(r-1) d(p-q)] \\ = & \frac{1}{2} 2 A(q-r+r-p+p-q)+d \\ & {[(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)] }\end{aligned}$
$
\begin{array}{l}
\quad\quad\quad=\frac{1}{2}[2 A \times(0)+d(p q-p r-q+r+q r-p q \\
\quad\quad\quad-r+p+r p-r q-p+q)] \\
\quad\quad\quad=\frac{1}{2}(0+d \times 0)=0 \\
\therefore \text { LHS }=\text { RHS } \\
\end{array}
$
Hence proved.

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Question 75 Marks

Find the sum of integers from 1 to 100 that are divisible by 2 or 5 .

Answer

The number from 1 to 100 which are divisible by 2 are
$2,4,6,8, \ldots . ., 100$.
Clearly they are in A.P., where $a=2, d=4-2=2$
$
\begin{array}{l}
\therefore T_n=a+(n-1) d \Rightarrow 100=2+(n-1) 2 \\
\Rightarrow 100-2=(n-1) 2 \Rightarrow 98=(n-1) 2 \\
\Rightarrow 49=n-1 \Rightarrow n=50
\end{array}
$
Therefore, sum of 50 numbers,
$
\begin{array}{l}
S_{50}=\frac{50}{2}[2 \times 2+(50-1) 2] \\
\qquad\left[S_n=\frac{n}{2}[2 a+(n-1) d]\right] \\
=25[4+49 \times 2] \\
=25[4+98]=25 \times 102 \quad \ldots(i) \\
S_{50}=2550
\end{array}
$
Now, the numbers from 1 to 100 which are divisible by 5 are$
5,10,15,20, \ldots \ldots . .100
$
Clearly, they are in A.P. where $a=5, d=10-5=5$
$
\begin{aligned}
\therefore \quad T_n & =a+(n-1) d \\
100 & =5+(n-1) 5 \\
\Rightarrow 100-5 & =(n-1) 5
\end{aligned}
$
$
\begin{array}{l}
\Rightarrow \quad(n-1)=\frac{95}{5} \\
\Rightarrow \quad n-1=19 \\
\Rightarrow \quad n=19+1=20 \\
\text { Now, } \quad S_n=\frac{n}{2}[2 a+(n-1) d] \\
\Rightarrow \quad S_{20}=\frac{20}{2}[2 \times 5+(20-1) 5] \\
=10[10+19 \times 5] \\
=10[10+95]=10(105) \\
\Rightarrow \quad S_{20}=1050 \quad \ldots(ii)
\end{array}
$
Now, the number from 1 to 100 which are divisible by 10 are $10,20,30, \ldots . ., 100$
Clearly, they are in A.P. where $a=10, d=20-10=$ 10 and $n=10$
$
\begin{array}{l}
\because \quad S_n=\frac{n}{2}[2 a+(n-1) d] \\
\therefore \quad S_{10}=\frac{10}{2}[2 \times 10+(10-1) 10] \\
=5[20+9 \times 10] \\
=5[20+90] \\
=5 \times 110=550 \quad \ldots(iii)
\end{array}
$
Hence, required sum of integers from 1 to 100 which are divisible by 2 or $5=2550+1050-550$ (using Eqs. (i), (ii) and (iii)) $=3600-550=3050 $

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Question 85 Marks

Let the sum of $n, 2 n, 3 n$ terms of an A.P. be $S_1, S_2$ and $S_3$. respectively. Show that $S_3=3\left(S_2-S_1\right)$.

Answer

Let $a$ and $d$ be the first term and common difference respectively of an A.P.
Given,
$
\begin{aligned}
S_1 & =\text { Sum of } n \text { terms } \\
& =\frac{n}{2}[2 a+(n-1) d] \quad \ldots(i) \\
S_2 & =\text { Sum of } 2 n \text { terms } \\
& =\frac{2 n}{2}[2 a+(2 n-1) d] \quad \ldots(ii)
\end{aligned}
$
and
$
\begin{aligned}
S_3 & =\text { Sum of } 3 n \text { terms } \\
& =\frac{3 n}{2}[2 a+(3 n-1) d] \quad \ldots(iii)
\end{aligned}
$
Now, we have to prove $S_3=3\left(S_2-S_1\right)$
Now, RHS $=3\left(S_2-S_1\right)$
$
\begin{array}{l}
=3\left[\frac{2 n}{2}[2 a+(2 n-1) d]-\frac{n}{2}[2 a+(n-1) d] \right. \\
\quad \quad[\text { from Eqs. (i) and (ii)] } \\
=\frac{3 n}{2}[2\{2 a+(2 n-1) d\}-\{2 a+(n-1) d\}] \\
=\frac{3 n}{2}[4 a+2(2 n-1) d-2 a-(n-1) d] \\
=\frac{3 n}{2}[(4 a-2 a)+d(4 n-2-n+1)] \\
=\frac{3 n}{2}[2 a+(3 n-1) d] \\
=S_3=\text { LHS } \quad[\text { from Eq. (iii) }]
\end{array}
$
$\because LHS = RHS$

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Question 95 Marks

If the A.M. between $p^{\text {th }}$ and $q^{\text {th }}$ terms of an A.P. be equal to A.M. between $r^{\text {th }}$ and $s^{\text {th }}$ terms of the A.P., then show that $p+q=r+s$.

Answer

Let $a$ be the first term and $d$ be the common difference of the given A.P. Then
$
\begin{array}{l}
a_p=p^{\text {th }} \text { term }=a+(p-1) d ; \\
a_q=q^{\text {th }} \text { term }=a+(q-1) d ; \\
a_r=r^{\text {th }} \text { term }=a+(r-1) d ; \\
\end{array}
$
and $a_{s}=s^{\text {th }} \text { term }=a+(s-1) d ;$
It is given that
A.M. between $a_p$ and $a_q=$ A.M. between $a_r$ and $a_s$
$
\begin{array}{ll}
\Rightarrow \frac{1}{2}\left(a_p+a_q\right)=\frac{1}{2}\left(a_r+a_s\right) \\
\Rightarrow a_p+a_q=a_r+a_{s} \\
\Rightarrow a+(p-1) d+a+(q-1) d
\end{array}
$
$\begin{array}{rlrl} & =a+(r-1) d+a+(s-1) d \\ \Rightarrow & (p+q-2) d =(r+s-2) d \\ \Rightarrow & p+q =r+s\end{array}$

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Question 105 Marks

Find the sum of first ' $n$ ' terms of the series $0.7+$ $0.77+0.777+\ldots$

Answer

We have, $0.7+0.77+0.777+\ldots$ to $n$ terms
$=7 \times 0.1+7 \times 0.11+7 \times 0.111+\ldots$. to $n$ terms
$=7\{0.1+0.11+0.111+\ldots$ to $n$ terms $\}$
$\begin{array}{l}=\frac{7}{9}\{0.9+0.99+0.999+\ldots \text { to } n \text { terms }\} \\ =\frac{7}{9}\left\{\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+\ldots \text { to } n \text { terms }\right\}\end{array}$
$=\frac{7}{9}\left\{\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{100}\right)+\right.$$\left(1-\frac{1}{1000}\right)+\ldots$ to $n$ terms $\}$
$\begin{array}{l}=\frac{7}{9}\left\{\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{10^2}\right)+\left(1-\frac{1}{10^3}\right)+\ldots\left(1-\frac{1}{10^n}\right)\right\} \\ =\frac{7}{9}\left\{n-\left(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^2}+\ldots+\frac{1}{10^n}\right)\right\}\end{array}$
$=\frac{7}{9}\left\{n-\frac{1}{10} \frac{\left\{1-\left(\frac{1}{10}\right)^n\right\}}{\left(1-\frac{1}{10}\right)}\right\}$
$\begin{array}{l}=\frac{7}{9}\left\{n-\frac{1}{9}\left(1-\frac{1}{10^n}\right)\right\} \\ =\frac{7}{81}\left\{9 n-1+\frac{1}{10^n}\right\}\end{array}$

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Question 115 Marks

If $x=a+\frac{a}{r}+\frac{a}{r^2}+\ldots+\infty, y=b-\frac{b}{r}+\frac{b}{r^2}-\ldots+\infty$ and $z=c+\frac{c}{r^2}+\frac{c}{r^4}+\ldots+\infty$ Prove that $\frac{x . y}{z}=\frac{a b}{c}$

Answer

Given series $x=a+\frac{a}{r}+\frac{a}{r^2}+\ldots+\infty$ is in G.P. with common ratio $\frac{1}{r}$.
and series $y=b-\frac{b}{r}+\frac{b}{r^2}-\ldots+\infty$ is in G.P. with common ratio $-\frac{1}{r}$.
and series $z=c+\frac{c}{r^2}+\frac{c}{r^4}+\ldots+\infty$ is in G.P. with common ratio $\frac{1}{r^2}$
Sum of infinite terms of series$
x=a+\frac{a}{r}+\frac{a}{r^2}+\ldots+\infty \text { is } x=\frac{a}{1-\frac{1}{r}}
$
Sum of infinite term of series $y=b-\frac{b}{r}+\frac{b}{r^2}-\ldots+\infty$ is
$
y=\frac{b}{1-\left(-\frac{1}{r}\right)}=\frac{b}{1+\frac{1}{r}}
$
Sum of infinite terms of series $z=c+\frac{c}{r^2}+\frac{c}{r^4}+\ldots+\infty$ is
$
z=\frac{c}{1-\frac{1}{r^2}}
$
Now, LHS $=\frac{x y}{z}$
$=\frac{\left\{\frac{a}{\left(1-\frac{1}{r}\right)}\right\}\left\{\frac{b}{\left(\left(1+\frac{1}{r}\right)\right.}\right\}}{\frac{c}{1-\frac{1}{r^2}}}$
$\begin{array}{l}=\frac{\frac{a b}{\left(1-\frac{1}{r^2}\right)}}{\frac{c}{\left(1-\frac{1}{r^2}\right)}} \\ =\frac{a b}{c}=\text { R.H.S }\end{array}$
Hence Proved

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Applied Maths 5 Marks Questions

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