Question 1: Find:

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

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

$\displaystyle \text{i) } \text{Given A.P. series } 1, 4, 7, 10, \ldots$

$\displaystyle \text{Therefore } a = 1$

$\displaystyle \text{Common difference } (d) = 4-1 = 3$

$\displaystyle \text{We know, } a_n = a + (n-1) d$

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

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

$\displaystyle \text{Therefore } a = \sqrt{2}$

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

$\displaystyle \text{We know, } a_n = a + (n-1) d$

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

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

$\displaystyle \text{Therefore } a = 13$

$\displaystyle \text{Common difference } (d) = 8-13 = -5$

$\displaystyle \text{We know, } a_n = a + (n-1) d$

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

$\displaystyle \\$

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

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

$\displaystyle \text{LHS } = a_{m + n} + a_{m - n}$

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

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

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

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

$\displaystyle = 2 a_m \text{ Hence proved. }$

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Question 3:

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

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

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

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

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

$\displaystyle \text{We know } a_n = a + (n-1) d$

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

$\displaystyle \Rightarrow 49 = n - 1$

$\displaystyle \Rightarrow n = 50$

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

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

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

$\displaystyle \text{We know } a_n = a + (n-1) d$

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

$\displaystyle \Rightarrow 21 = n - 1$

$\displaystyle \Rightarrow n = 22$

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

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

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

$\displaystyle \text{We know } a_n = a + (n-1) d$

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

$\displaystyle \Rightarrow 50 = n - 1$

$\displaystyle \Rightarrow n = 51$

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

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Question 4:

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

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

$\displaystyle \text{i) } \text{Given A.P. series } 7,10,13, \ldots$?

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

$\displaystyle \text{We know } a_n = a + (n-1) d$

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

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

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

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

$\displaystyle \text{ii) } \text{Given A.P. series } 3,8,13, \ldots$ ?

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

$\displaystyle \text{We know } a_n = a + (n-1) d$

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

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

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

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

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Question 5:

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

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

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

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

Let the first negative term is $\displaystyle a_n$.

$\displaystyle \text{Therefore } a_n < 0$

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

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

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

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

$\displaystyle \Rightarrow 99 < 3n$

$\displaystyle \text{Therefore } n > 33$

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

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

$\displaystyle \Rightarrow a = 12 + 8i$

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

Let the real term be $\displaystyle a_n$

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

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

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

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

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

a) for $\displaystyle a_n$ to be real, $\displaystyle 10-2n = 0 \Rightarrow n = 5$
b) for $\displaystyle a_n$ to be imaginary, $\displaystyle 13-n = 0 \Rightarrow n = 13$

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Question 6:

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

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

$\displaystyle \text{i) } \text{Given A.P. series } 7, 10, 13, \ldots , 43$

$\displaystyle \text{Therefore } a = 7 d= ( 10 - 7) = 3 a_n = 43$

$\displaystyle \text{We know } a_n = a + (n-1)d$

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

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

$\displaystyle \Rightarrow n -1 = 12$

$\displaystyle \Rightarrow n = 13$

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

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

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

$\displaystyle \text{We know } a_n = a + (n-1)d$

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

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

$\displaystyle \Rightarrow n -1 = 26$

$\displaystyle \Rightarrow n = 27$

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

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

$\displaystyle \text{Given: } a = 5 d = 3 a_n = 80$

$\displaystyle \text{We know } a_n = a + (n-1)d$

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

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

$\displaystyle \Rightarrow n-1 = 25$

$\displaystyle \Rightarrow n = 26$

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

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

$\displaystyle \text{Given } a_6 = 19 \text{ and } a_{17} = 41$

$\displaystyle \text{We know } a_n = a + (n-1)d$

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

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

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

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

Solving i) and ii)

$\displaystyle 19-5d = 41 - 16d$

$\displaystyle \Rightarrow 11d = 22$

$\displaystyle \Rightarrow d = 2$

$\displaystyle \text{Therefore } a = 19 - 5( 2) = 9$

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

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

$\displaystyle \text{Given: } a_9 = 0$

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

$\displaystyle \Rightarrow a + 8d = 0$

$\displaystyle \Rightarrow a = -8d$

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

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

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

Substituting $\displaystyle a = -8d$

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

$\displaystyle \Rightarrow 20 d = 20 d \text{ Hence proved. }$

$\displaystyle \\$

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

$\displaystyle \text{Given } 10a_{10} = 15 a_{15}$

$\displaystyle \text{We know } a_n = a + (n-1)d$

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

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

$\displaystyle \Rightarrow 5a + 120 d = 0$

$\displaystyle \Rightarrow a + 24 d = 0$

$\displaystyle \Rightarrow a = - 24 d$

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

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

$\displaystyle \text{Given } a_{10} = 41 \text{ and } a_{18} = 73$

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

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

$\displaystyle \text{Also } a + ( 18 - 1) d = 73$

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

Solving i) and ii) we get

$\displaystyle 41- 9d = 73 - 17d$

$\displaystyle \Rightarrow 8d = 32$

$\displaystyle \Rightarrow d = 4$

Substituting from i) we get

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

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

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

$\displaystyle \text{Given } a_{24} = 2a_{10}$

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

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

$\displaystyle \Rightarrow a = 5d$ … … … … … i)

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

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

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

Substituting from i) we get

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

$\displaystyle \Rightarrow 76d = 76 d \text{ Hence proved. }$

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

$\displaystyle \text{Given } a_{m+1} = 2a_{n+1}$

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

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

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

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

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

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

Substituting from i)

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

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

$\displaystyle \Rightarrow 4md - 2nd = 4md - 2nd \text{ Hence proved. }$

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

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

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

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

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

$\displaystyle = 11 - 2n$

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

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

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

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

$\displaystyle = 18 - 3n$

$\displaystyle \text{Given: } 11 - 2n = 18-3n \Rightarrow n = 7$

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

$\displaystyle \text{i) } \text{Given series } 3,5,7,9, \ldots , 201$

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

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

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

$\displaystyle \text{ii) } \text{Given series } 3, 8,13 , \ldots , 253$

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

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

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

$\displaystyle \text{iii) } \text{Given series } 1,4,7,10, \ldots ,88$

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

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

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

$\displaystyle \\$

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

$\displaystyle \text{Given } a_4 = 3a$

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

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

$\displaystyle \text{Also } a_7 - 2a_3 = 1$

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

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

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

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

Substituting in i) we get

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

$\displaystyle \Rightarrow 3d = 4d - 2$

$\displaystyle \Rightarrow d = 2$

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

$\displaystyle \\$

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

$\displaystyle \text{We know } a_n = a + (n-1)d$

Given, $\displaystyle a_6 = 12$

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

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

$\displaystyle \text{Also } a_8 = 22$

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

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

Solving i) and ii) we get

$\displaystyle 12-5d = 22-7d$

$\displaystyle 2d = 10$

$\displaystyle d = 5$

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

$\displaystyle \text{Hence } a_2 = -13 + ( 2-1) (5) = -13 + 5 = -8$

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

$\displaystyle \\$

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

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

$\displaystyle \text{Therefore } a = 12 , d = ( 15-12) = 3 a_n = 99$

$\displaystyle \text{We know } a_n = a + (n-1)d$

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

$\displaystyle \Rightarrow 90 = 3n$

$\displaystyle \Rightarrow n = 30$

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

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

$\displaystyle \text{Given } a = 7, n = 60 , l = 125$

$\displaystyle \text{We know } l = a + ( n-1) d$

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

$\displaystyle \Rightarrow 118 = 59 d$

$\displaystyle \Rightarrow d = 2$

$\displaystyle \text{We know } a_n = a + (n-1)d$

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

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

$\displaystyle \text{Given } a_4 + a_8 = 24$

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

$\displaystyle \Rightarrow 2a + 10d = 24$

$\displaystyle \Rightarrow a + 5d = 12$

$\displaystyle \text{Also } a_6 + a_{10} = 34$

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

$\displaystyle \Rightarrow 2a + 14d = 34$

$\displaystyle \Rightarrow a + 7d = 17$

Solving i) and ii) we get

$\displaystyle 12-5d = 17 - 7d$

$\displaystyle \Rightarrow 2d = 5$

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

Substituting in i) we get

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

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

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

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

$\displaystyle \text{Hence } a = 4, d = ( 11- 4) = 7, l = 998$

$\displaystyle \text{We know } l = a + ( n-1) d$

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

$\displaystyle \Rightarrow 1001 = 7n$

$\displaystyle \Rightarrow n = 143$

Hence there are $\displaystyle 143$ numbers between $\displaystyle 1 \text{ and } 1000$ which when divided by $\displaystyle 7$ leave remainder $\displaystyle 4$

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

Given: first term $\displaystyle = a \text{ and last term } = l$

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

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

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

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

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

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

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

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

$\displaystyle \Rightarrow a = 3d$

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

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

$\displaystyle \text{Given } \theta_1, \theta_2, \theta_3, \ldots \theta_n$ are in AP

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

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

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

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

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

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

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

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

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