Question 1: Find:

i) ${10}^{th}$ term of A.P $1, 4, 7, 10, \ldots$          ii) ${18}^{th}$ term of A.P $\sqrt{2}, 3\sqrt{2}, 5\sqrt{2}, \ldots$

iii) $n^{th}$ term of A.P $13, 8, 3, -2, \ldots$

i)       Given A.P. series $1, 4, 7, 10, \ldots$

Therefore  $a = 1$

Common difference $(d) = 4-1 = 3$

We know, $a_n = a + (n-1) d$

$\therefore a_{10} = 1+ ( 10 - 1) (3) = 1 + 27 = 28$

ii)      Given A.P. series $\sqrt{2}, 3\sqrt{2}, 5\sqrt{2}, \ldots$

Therefore  $a = \sqrt{2}$

Common difference $(d) = 3\sqrt{2} - \sqrt{2} = 2\sqrt{2}$

We know, $a_n = a + (n-1) d$

$\therefore a_{18} = \sqrt{2} + ( 18 - 1) (2\sqrt{2}) = \sqrt{2} + 34\sqrt{2} = 35\sqrt{2}$

iii)    Given A.P. series $13, 8, 3, -2, \ldots$

Therefore  $a = 13$

Common difference $(d) = 8-13 = -5$

We know, $a_n = a + (n-1) d$

$\therefore a_{n} = 13+ ( n - 1) (-5) = -5n+18$

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Question 2: In an A.P., show that $a_{m + n} + a_{m - n} = 2a_m$.

Let the first term be $a$ and the common difference be $d$

LHS $= a_{m + n} + a_{m - n}$

$= [a+ (m+n-1)d ] + [a+(m-n-1)d ]$

$= 2a + ( m+n-1+m-n-1)d$

$= 2a + 2 (m-1) d$

$= 2 [ a+ ( m-1)d ]$

$= 2 a_m$. Hence proved.

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Question 3:

i) Which term of the A.P. $3, 8, 13, \ldots$ is $248$ ?

ii) Which term of the A.P. $84, 80,76, \ldots$ is $0$ ?

iii) Which term of the A.P. $4, 9,14, \ldots$ is $254$ ?

i)       Given A.P. series $3, 8, 13, \ldots$ is $248$ ?

$\Rightarrow a = 3$           $d = (8-3) = 5$          $a_n = 248$

We know $a_n = a + (n-1) d$

$\Rightarrow 248 = 3 + (n-1)(5)$

$\Rightarrow 49 = n - 1$

$\Rightarrow n = 50$

Hence $248$ is the $50^{th}$ term.

ii)      Given A.P. series $84, 80,76, \ldots$ is $0$ ?

$\Rightarrow a = 84$           $d = (80 - 84) = -4$          $a_n = 0$

We know $a_n = a + (n-1) d$

$\Rightarrow 0 = 84 + (n-1)(-4)$

$\Rightarrow 21 = n - 1$

$\Rightarrow n = 22$

Hence $0$ is the $22^{nd}$ term.

iii)     Given A.P. series $4, 9,14, \ldots$ is $254$ ?

$\Rightarrow a = 4$           $d = (9-4) = 5$          $a_n = 254$

We know $a_n = a + (n-1) d$

$\Rightarrow 254 = 4 + (n-1)(5)$

$\Rightarrow 50 = n - 1$

$\Rightarrow n = 51$

Hence $254$ is the $51^{st}$ term.

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Question 4:

(i) Is $58$ a term of the A.P. $7,10,13, \ldots$?

(ii) Is $302$ a term of the A.P. $3,8,13, \ldots$ ?

(i)      Given A.P. series $7,10,13, \ldots$?

$\Rightarrow a = 7$           $d = (10-7) = 3$          $a_n = 68$

We know $a_n = a + (n-1) d$

$\Rightarrow 68 = 7 + (n-1)(3)$

$\Rightarrow \frac{61}{3} = n - 1$

$\Rightarrow n = \frac{61}{3} + 1$

Since $n$ is not a natural number, $68$ is NOT a term in the given A.P.

(ii)    Given A.P. series $3,8,13, \ldots$ ?

$\Rightarrow a = 3$           $d = (8-3) = 5$          $a_n = 302$

We know $a_n = a + (n-1) d$

$\Rightarrow 302 = 3 + (n-1)(5)$

$\Rightarrow$ $\frac{299}{5}$ $= n - 1$

$\Rightarrow n =$ $\frac{299}{5}$ $+ 1$

Since $n$ is not a natural number, $302$ is NOT a term in the given A.P.

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Question 5:

i) Which term of the sequence $24, 23$ $\frac{1}{4}$ $, 22$ $\frac{1}{2}$ $, 21$ $\frac{3}{4}$ $, \ldots$  is the first negative term?

ii) Which term of  the sequence $12+8i, 11,+6i, 10+4i , \ldots$  is (a) purely real (b) purely imaginary ?

i)      Given series $24, 23$ $\frac{1}{4}$ $, 22$ $\frac{1}{2}$ $, 21$ $\frac{3}{4}$ $, \ldots$

$\Rightarrow a = 24$           $d = (23$ $\frac{1}{4}$ $-24) = -$ $\frac{3}{4}$

Let the first negative term is $a_n$.

Therefore $a_n < 0$

$\Rightarrow a + (n-1) d < 0$

$\Rightarrow 24 + (n-1) ( -$ $\frac{3}{4}$ $) < 0$

$\Rightarrow 24 +$ $\frac{3}{4}$ $<$ $\frac{3n}{4}$

$\Rightarrow$ $\frac{99}{4}$ $<$ $\frac{3n}{4}$

$\Rightarrow 99 < 3n$

Therefore $n > 33$

Thus the $34^{th}$ term is the first negative term of the given AP.

ii)      Given series $12+8i, 11,+6i, 10+4i , \ldots$

$\Rightarrow a = 12 + 8i$

$d = ( 11+6i) - ( 12 + 8i) = -1 -2i$

Let the real term be $a_n$

$a_n = a + (n-1)d$

$= ( 12 + 8i) + ( n-1) ( -1 -2i)$

$= 12 + 8i + 1 + 2i - n - 2in$

$= ( 12 + 1 - n) + (10i - 2in)$

$= (13-n) + i ( 10-2n)$

a) for $a_n$ to be real, $10-2n = 0 \Rightarrow n = 5$

b) for $a_n$ to be imaginary, $13-n = 0 \Rightarrow n = 13$

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Question 6:

i) How many terms are there in the A.P. $7, 10, 13, \ldots , 43$ ?

ii) How many terms are there in the A.P. $-1,$ $\frac{-5}{6}$ $,$ $\frac{-2}{3}$ $,$ $\frac{-1}{2}$ $, \ldots ,$ $\frac{10}{3}$ ?

i)       Given A.P. series $7, 10, 13, \ldots , 43$

Therefore $a = 7$     $d= ( 10 - 7) = 3$       $a_n = 43$

We know $a_n = a + (n-1)d$

$\Rightarrow 43 = 7 + (n-1) (3)$

$\Rightarrow 36 = (n-1) (3)$

$\Rightarrow n -1 = 12$

$\Rightarrow n = 13$

Therefore there are $13$ terms in the given A.P.

ii)     Given A.P. series $-1,$ $\frac{-5}{6}$ $,$ $\frac{-2}{3}$ $,$ $\frac{-1}{2}$ $, \ldots ,$ $\frac{10}{3}$ ?

Therefore $a = -1$     $d= \Big[$ $\frac{-5}{6}$ $- (-1) \Big] =$ $\frac{1}{6}$        $a_n =$ $\frac{10}{3}$

We know $a_n = a + (n-1)d$

$\Rightarrow$ $\frac{10}{3}$ $= -1 + (n-1) ($ $\frac{1}{6}$ $)$

$\Rightarrow$ $\frac{13}{3}$ $= (n-1) ($ $\frac{1}{6}$ $)$

$\Rightarrow n -1 = 26$

$\Rightarrow n = 27$

Therefore there are $27$ terms in the given A.P.

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Question 7: The first term of an A.P. is $5$, the common difference is $3$ and the last term is $80$; find the number of terms.

Given: $a = 5$         $d = 3$          $a_n = 80$

We know $a_n = a + (n-1)d$

$\Rightarrow 80 = 5 + ( n-1) (3)$

$\Rightarrow (n-1) (3) = 75$

$\Rightarrow n-1 = 25$

$\Rightarrow n = 26$

Therefore there are $26$ terms in the given A.P.

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Question 8: The $6^{th}$ and $17^{th}$ terms of an A.P. are $19$ and $41$ respectively, find the $40^{th}$ term.

Given $a_6 = 19$   and $a_{17} = 41$

We know $a_n = a + (n-1)d$

$\Rightarrow a + ( 6-1) d = 19$

$\Rightarrow a + 5d = 19$     … … … … … i)

$\Rightarrow a + ( 17-1) d = 41$

$\Rightarrow a + 16d = 41$     … … … … … ii)

Solving i) and ii)

$19-5d = 41 - 16d$

$\Rightarrow 11d = 22$

$\Rightarrow d = 2$

Therefore $a = 19 - 5( 2) = 9$

Hence $40^{th}$ is $a_{40} = 9 + (40-1) (2) = 87$

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Question 9: If $9^{th}$ term of an A.P. is zero, prove that its $29^{th}$ term is double the $19^{th}$ term.

Given: $a_9 = 0$

$\therefore a + ( 9-1) d = 0$

$\Rightarrow a + 8d = 0$

$\Rightarrow a = -8d$

To prove: $a _{29} = 2 a_{19}$

$\Rightarrow a + ( 29-1) d = 2 [ a + (19-1) d]$

$\Rightarrow a + 28d = 2a + 36d$

Substituting $a = -8d$

$\Rightarrow -8d + 28d = 2 (-8d) + 36d$

$\Rightarrow 20 d = 20 d$.    Hence proved.

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Question 10: If $10$ times the $10^{th}$ term of an A.P. is equal to $15$ times the $15^{th}$ term, show that $25^{th}$ term of the A.P. is zero.

Given $10a_{10} = 15 a_{15}$

We know $a_n = a + (n-1)d$

$\Rightarrow 10 [ a+ ( 10 - 1) d] = 15 [ a+ ( 15 - 1) d]$

$\Rightarrow 10 a + 90d = 15a + 210 d$

$\Rightarrow 5a + 120 d = 0$

$\Rightarrow a + 24 d = 0$

$\Rightarrow a = - 24 d$

Now $a_{25} = a + ( 25 - 1) d = -24d + 24 d = 0$. Hence proved.

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Question 11: The $10^{th}$ and $18^{th}$ terms of an A.P. are $41$ and $73$ respectively. Find $26^{th}$ term.

Given $a_{10} = 41$ and $a_{18} = 73$

$\therefore a + ( 10 -1) D = 41$

$\Rightarrow a + 9d = 41$     … … … … … i)

Also $a + ( 18 - 1) d = 73$

$\Rightarrow a + 17d = 73$      … … … … … ii)

Solving i) and ii) we get

$41- 9d = 73 - 17d$

$\Rightarrow 8d = 32$

$\Rightarrow d = 4$

Substituting from i) we get

$a+ 9(4) = 41 \Rightarrow a = 5$

Therefore $a_{26} = 5 + ( 26-1) (4) = 105$

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Question 12: In a certain A.P. the $24^{th}$ term is twice the $10^{th}$ term. Prove that the $72^{nd}$ term is twice the $34^{th}$ term.

Given $a_{24} = 2a_{10}$

$\Rightarrow a + (24-1) d = 2 [ a + (10-1) d]$

$\Rightarrow a + 23 d = 2a + 18 d$

$\Rightarrow a = 5d$     … … … … … i)

To prove: $a_{72} = 2 a_{34}$

$\Rightarrow a + ( 72 -1) d = 2 [ a + ( 34 - 1) d]$

$\Rightarrow a + 71 d = 2a + 66d$

Substituting from i) we get

$\Rightarrow 5d + 71 d = 2 ( 5d) + 66d$

$\Rightarrow 76d = 76 d$. Hence proved.

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Question 13: lf $(m + 1)^{th}$ term of an A.P. is twice the $(n + 1)^{th}$ term, prove that $(3m+ 1)^{th}$ term is twice the $(m+n+1)^{th}$ term.

Given $a_{m+1} = 2a_{n+1}$

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

$\Rightarrow a + md = 2a + 2nd$

$\Rightarrow a = (m-2n) d$      … … … … … i)

To prove $a_{3m+1} = 2 a_{m+n+1}$

$\Rightarrow a + (3m+1-1)d = 2 [ a + (m+n+1-1) d$

$\Rightarrow a + 3md = 2a + 2(m+n) d$

Substituting from i)

$\Rightarrow (m-2n) d + 3md = 2 [(m-2n)d] + 2(m-n)d$

$\Rightarrow 4md - 2nd = 2md - 4nd + 2md + 2nd$

$\Rightarrow 4md - 2nd = 4md - 2nd$. Hence proved.

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Question 14: If the $n^{th}$ term of the A.P. $9, 7, 5, \ldots$ is same as the $n^{th}$ term of the A.P. $15, 12, 9, \ldots$ find $n$ .

$n^{th}$ term of sequence $9, 7, 5, \ldots$

$\Rightarrow a = 9, d = (7-9) = -2$

$\therefore a_n = a + (n-1) d$

$= 9 + ( n-1) ( -2)$

$= 11 - 2n$

$n^{th}$ term of sequence $15, 12, 9, \ldots$

$\Rightarrow a = 15, d = (12-15) = -3$

$\therefore a_n = a + (n-1) d$

$= 15 + ( n-1) ( -3)$

$= 18 - 3n$

Given: $11 - 2n = 18-3n \Rightarrow n = 7$

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Question 15: Find the $12^{th}$ term from the end of the following arithmetic progressions: (i) $3,5,7,9, \ldots , 201$ (ii) $3,8,13 , \ldots , 253$ (iii) $1,4,7,10, \ldots ,88$

(i)      Given series $3,5,7,9, \ldots , 201$

$a = 3$        $d = 5 - 3 = 2$          $l = 201$

$n^{th}$ term from the end $= l - ( n-1) d$

Therefore $12^{th}$ term from the end $= 201 - ( 12-1)(2) = 201 - 22 = 179$

(ii)     Given series $3, 8,13 , \ldots , 253$

$a = 3$        $d = 8 - 3 = 5$          $l = 253$

$n^{th}$ term from the end $= l - ( n-1) d$

Therefore $12^{th}$ term from the end $= 252 - ( 12-1)(5) = 253 - 55 = 198$

(iii)    Given series $1,4,7,10, \ldots ,88$

$a = 1$        $d = 4 - 1 = 3$          $l = 88$

$n^{th}$ term from the end $= l - ( n-1) d$

Therefore $12^{th}$ term from the end $= 88 - ( 12-1)(3) = 88 - 33 = 55$

$\$

Question 16: The $4^{th}$ term of an A.P. is three times the first and the $7^{th}$ term exceeds twice the third term by $1$. Find the first term and the common difference.

Given $a_4 = 3a$

$\Rightarrow a+ ( 4 - 1) d = 3a$

$\Rightarrow 3d = 2a$      … … … … … i)

Also $a_7 - 2a_3 = 1$

$\Rightarrow a+(7-1)d - 2 [ a + ( 3-1) d ] = 1$

$\Rightarrow a + 6d - 2 ( a + 2d) = 1$

$\Rightarrow a + 6d - 2a - 4d = 1$

$\Rightarrow 2d - a = 1$      … … … … … ii)

Substituting in i) we get

$3d = 2 ( 2d -1 )$

$\Rightarrow 3d = 4d - 2$

$\Rightarrow d = 2$

$\therefore a = 2(2) - 1 = 4 -1 = 3$

$\$

Question 17: Find the second term and $n^{th}$ term of an A.P. whose $6^{th}$ term is $12$ and the $8^{th}$ term is $22$.

We know $a_n = a + (n-1)d$

Given, $a_6 = 12$

$\Rightarrow a + ( 6-1) d = 12$

$\Rightarrow a + 5d = 12$      … … … … … i)

Also $a_8 = 22$

$\Rightarrow a + ( 8-1) d = 22$

$\Rightarrow a + 7d = 22$      … … … … … ii)

Solving i) and ii) we get

$12-5d = 22-7d$

$2d = 10$

$d = 5$

$\therefore a = 12 - 5 ( 5) = 12 - 25 = - 13$

Hence $a_2 = -13 + ( 2-1) (5) = -13 + 5 = -8$

$a_n = -13 + ( n-1)(5) = -13 + 5n - 5 = -18 + 5n$

$\$

Question 18: How many numbers of two digit are divisible by $3$ ?

The sequence would be $12, 15, 18, \ldots , 96, 99$

Therefore $a = 12 , \ \ \ \ d = ( 15-12) = 3 \ \ \ \ \ a_n = 99$

We know $a_n = a + (n-1)d$

$\therefore 99 = 12 + (n-1) (3)$

$\Rightarrow 90 = 3n$

$\Rightarrow n = 30$

Therefore there are $30$ such numbers which are two digit are divisible by $3$

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Question 19: An A.P. consists of $60$ terms. If the first and the last terms be $7$ and $125$ respectively, find $32^{nd}$ term.

Given $a = 7, \ \ \ \ n = 60 , \ \ \ \ \ l = 125$

We know $l = a + ( n-1) d$

$\Rightarrow 125 = 7 + ( 60-1) d$

$\Rightarrow 118 = 59 d$

$\Rightarrow d = 2$

We know $a_n = a + (n-1)d$

$\therefore a_{32} = 7 + ( 32-1) (2) = 69$

$\$

Question 20: The sum of $4^{th}$ and $8^{th}$ terms of an A.P. is $24$ and the sum of the $6^{th}$ and $10^{th}$ terms is $34$. Find the first term and the common difference of the A.P.

Given $a_4 + a_8 = 24$

$\Rightarrow a + ( 4-1) d + a + ( 8-1) d = 24$

$\Rightarrow 2a + 10d = 24$

$\Rightarrow a + 5d = 12$

Also $a_6 + a_{10} = 34$

$\Rightarrow a + ( 6-1) d + a + ( 10-1) d = 34$

$\Rightarrow 2a + 14d = 34$

$\Rightarrow a + 7d = 17$

Solving i) and ii) we get

$12-5d = 17 - 7d$

$\Rightarrow 2d = 5$

$\Rightarrow d =$ $\frac{5}{2}$

Substituting in i) we get

$a = 2 - 5 ($ $\frac{5}{2}$ $) = 12 -$ $\frac{25}{2}$ $=$ $\frac{-1}{2}$

$\$

Question 21: How many numbers are there between $1$ and $1000$ which when divided by $7$ leave remainder $4$?

If a number $N$ is divided by $7$ and leaves $4$ as remainder, it can be represented as $N = 7q + 4 , \ \ q \in N$

Therefore the sequence would be $4, 11, 18, \ldots, 998$

Hence $a = 4, \ \ \ \ d = ( 11- 4) = 7, \ \ \ \ \ l = 998$

We know $l = a + ( n-1) d$

$\Rightarrow 998 = 4 + ( n-1) (7)$

$\Rightarrow 1001 = 7n$

$\Rightarrow n = 143$

Hence there are $143$ numbers between $1$ and $1000$ which when divided by $7$ leave remainder $4$

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Question 22: The first and the last terms of an A.P. are $a$ and $l$ respectively. Show that the sum of $n^{th}$ term from the beginning and $n^{th}$ term from the end is $a + l$.

Given: first term $= a$ and last term $= l$

$a_n = a + (n-1) d$

$n^{th}$ term from end $= l - (n-1)d$

Therefore the sum of the two terms $= a + (n-1) d + l - (n-1)d = a+l$

$\$

Question 23: If an A.P. is such that $\frac{a_4}{a_7}$ $=$ $\frac{2}{3}$, find $\frac{a_6}{a_8}$.

Given $\frac{a_4}{a_7}$ $=$ $\frac{2}{3}$

$\Rightarrow$ $\frac{a + (4-1)d}{a + (7-1) d}$ $=$ $\frac{2}{3}$

$\Rightarrow 3(a+3d) = 2(9+6d)$

$\Rightarrow 3a + 9d = 2a +12d$

$\Rightarrow a = 3d$

Therefore $\frac{a_6}{a_8}$ $=$ $\frac{a+(6-1)d}{a+(8-1)d}$  $=$ $\frac{a+5d}{a+7d}$ $=$ $\frac{3d+5d}{3d+7d}$ $=$ $\frac{8}{10}$ $=$ $\frac{4}{5}$

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Question 24: If  $\theta_1, \theta_2, \theta_3, \ldots \theta_n$ are in AP, whose common difference is $d$, show that, $\sec \theta_1 \ \sec \theta_2 + \sec \theta_2 \ \sec \theta_3 + \ldots + \sec \theta_{n-1} \ \sec \theta_n =$ $\frac{\tan \theta_n - \tan \theta_1}{\sin d}$

Given $\theta_1, \theta_2, \theta_3, \ldots \theta_n$ are in AP

$\therefore d = \theta_2 - \theta_1= \theta_3 - \theta_2= \ldots = \theta_{n} - \theta_{n-1}$

LHS $= \sec \theta_1 \ \sec \theta_2 + \sec \theta_2 \ \sec \theta_3 + \ldots + \sec \theta_{n-1} \ \sec \theta_n$

$= \frac{1}{\cos \theta_1 \cdot \cos \theta_2} + \frac{1}{\cos \theta_2 \cdot \cos \theta_3} + \ldots + \frac{1}{\cos \theta_{n-1} \cdot \cos \theta_n}$

$= \frac{1}{\sin d} \Big[ \frac{\sin d}{\cos \theta_1 \cdot \cos \theta_2} + \frac{\sin d}{\cos \theta_2 \cdot \cos \theta_3} + \ldots + \frac{\sin d}{\cos \theta_{n-1} \cdot \cos \theta_n} \Big]$

$= \frac{1}{\sin d} \Big[ \frac{\sin (\theta_2 - \theta_1)}{\cos \theta_1 \cdot \cos \theta_2} + \frac{\sin (\theta_3 - \theta_2)}{\cos \theta_2 \cdot \cos \theta_3} + \ldots + \frac{\sin (\theta_n - \theta_{n-1})}{\cos \theta_{n-1} \cdot \cos \theta_n} \Big]$

$= \frac{1}{\sin d} \Big[ \frac{\sin \theta_2 \cdot \cos \theta_1 - \cos \theta_2 \cdot \sin \theta_1 }{\cos \theta_1 \cdot \cos \theta_2} + \frac{\sin \theta_3 \cdot \cos \theta_2 - \cos \theta_3 \cdot \sin \theta_2}{\cos \theta_2 \cdot \cos \theta_3} + \ldots + \frac{\sin \theta_n \cdot \cos \theta_{n-1} - \cos \theta_n \cdot \sin \theta_{n-1}}{\cos \theta_{n-1} \cdot \cos \theta_n} \Big]$

$= \frac{1}{\sin d}$ $[ \tan \theta_2 - \tan \theta_1 + \tan \theta_3 - \tan \theta_2 + \ldots + \tan \theta_n - \tan \theta_{n-1} ]$

$= \frac{1}{\sin d}$ $[ \tan \theta_n - \tan \theta_1 ]$

$= \frac{ \tan \theta_n - \tan \theta_1}{\sin d}$ $=$ RHS. Hence proved.