CBSE Class 10 Arithmetic Progressions Sure Shot Questions Set 14

Question. Find how many integers between 200 and 500 are divisible by 8.
Answer: Smallest number divisible by 8 in the given range = 208
Last number divisible by 8 in the given range = 496
So, \( a = 208, d = 8, n = ?, T_n = 496 \)
We know, \( T_n = a + (n - 1)d \)
\( \Rightarrow 496 = 208 + (n - 1) 8 \)
or \( 8n + 208 - 8 = 496 \)
or \( 8n = 496 - 200 \)
\( \Rightarrow 8n = 296 \)
\( \Rightarrow n = \frac{296}{8} = 37 \)
So number of terms between 200 and 500 divisible by 8 are 37.

Question. Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.
Answer: As the numbers that are divisible by both 2 and 5, have their last digits as always 0. Hence, the series is 110, 120, 130, …, 990.
Here \( a = 110, d = 10 \) and \( t_n = 990 \)
We know that,
\( t_n = a + (n - 1)d \)
\( \Rightarrow 990 = 110 + (n - 1)10 \)
\( \Rightarrow (n - 1)10 = 880 \)
\( \Rightarrow n - 1 = 88 \)
\( \Rightarrow n = 89 \)
Thus, the number of natural numbers between 101 and 999 which are divisible by both 2 and 5 is 89.

Question. The sum of the first \( n \) terms of an A.P. is \( 3n^2 + 6n \). Find the \( n^{th} \) term of this A.P.
Answer: We have,
\( S_n = 3n^2 + 6n \)
\( \therefore S_{n - 1} = 3(n - 1)^2 + 6(n - 1) \)
Now, \( t_n = S_n - S_{n - 1} \)
\( = 3n^2 + 6n - 3(n - 1)^2 - 6(n - 1) \)
\( = 3n^2 + 6n - 3n^2 + 6n - 3 - 6n + 6 \)
\( t_n = 6n + 3 \)
OR
We know that
\( S_n = \frac{n}{2} \{2a + (n - 1)d\} \)
Also, \( S_n = 3n^2 + 6n \)(Given)
\( \Rightarrow \frac{n}{2} [2a + (n - 1)d] = 3n^2 + 6n \)
\( \Rightarrow [2a + (n - 1)d] = 6(n + 2) \)
\( \Rightarrow 2a + dn - d = 6n + 12 \)
\( \Rightarrow dn + (2a - d) = 6n + 12 \)
On comparing we get
\( \therefore d = 6 \)
and \( 2a - d = 12 \)
\( \Rightarrow 2a - 6 = 12 \)
\( \Rightarrow 2a = 18 \)
\( \Rightarrow a = 9 \)
Thus, \( t_n = 9 + (n - 1)6 = 9 + 6n - 6 = 6n + 3 \).

Question. If the numbers \( 2n - 1, 3n + 2 \) and \( 6n - 1 \) are in A.P., find \( n \) and hence find the numbers.
Answer: The terms of an A.P. are \( (2n - 1), (3n + 2) \) and \( (6n - 1) \).
Thus, \( d = (3n + 2) - (2n - 1) = (6n - 1) - (3n + 2) \)
\( \Rightarrow n + 3 = 3n - 3 \)
\( \Rightarrow 2n = 6 \)
\( \Rightarrow n = 3 \)
Hence, the numbers are
\( = 2(3) - 1, 3(3) + 2 \) and \( 6(3) - 1 \)
\( = 5, 11 \) and \( 17 \).

Question. Find the sum of all three-digit natural numbers which are multiples of 11.
Answer: The given A.P. series is
110, 121, 132, 143, …, 990
Here, \( a = 110, d = 11 \) and \( t_n = 990 \)
We know that
\( t_n = a + (n - 1)d \)
\( \Rightarrow 990 = 110 + (n - 1)11 \)
\( \Rightarrow 90 = 10 + n - 1 \)
\( \Rightarrow n = 81 \)
Now \( S_n = \frac{n}{2} (a + l) \)
\( \Rightarrow S_{81} = \frac{81}{2} (110 + 990) \)
\( = \frac{81}{2} (1100) = 81 \times 550 \)
\( = 44550 \).

Question. Find an A.P. whose fourth term is 9 and the sum of its sixth term and thirteenth term is 40.
Answer: Let the first term of an A.P. be \( a \) and the common difference be \( d \).
Thus, \( t_4 = \{a + (4 - 1)d\} = 9 \)(Given)
\( \Rightarrow a + 3d = 9 \) …(i)
Also \( t_6 + t_{13} = 40 \)(Given)
\( \Rightarrow a + 5d + a + 12d = 40 \)
\( \Rightarrow 2a + 17d = 40 \) …(ii)
Multiplying (i) with 2 and then subtracting from (ii), we have
\( 11d = 22 \)
\( \Rightarrow d = 2 \)
Thus, \( a = 9 - 3(2) = 3 \)
Thus, the A.P. is 3, 5, 7, 9, 11, …

Question. Find whether \( -150 \) is a term of the A.P. 17, 12, 7, 2, …, ?
Answer: Given A.P. is 17, 12, 7, 2, ...
Here, \( a = 17, d = 12 - 17 = -5 \)
Let \( -150 \) be the \( n^{th} \) term.
Thus \( t_n = 17 + (n - 1)(-5) \)
\( \Rightarrow 17 - 5n + 5 = -150 \)
\( \Rightarrow -5n = -150 - 22 \)
\( \Rightarrow 5n = 172 \)
\( \Rightarrow n = \frac{172}{5} = 34.4 \)
Thus, \( -150 \) is not a term of this A.P. as \( n \) is not a natural number.

Question. Which term of the A.P. 5, 9, 13, 17, …is 81 ? Also, find the sum.
Answer: From the given A.P, we have \( a = 5 \) and \( d = 9 - 5 = 4 \)
Let 81 be the \( n^{th} \) term of the A.P.
Thus, \( t_n = 5 + (n - 1)(4) = 81 \)
\( \Rightarrow 5 + 4n - 4 = 81 \)
\( \Rightarrow 4n = 80 \)
\( \Rightarrow n = 20 \)
Thus, 81 is the \( 20^{th} \) term of the A.P.
Hence \( S_{20} = \frac{20}{2} \{2(5) + (20 - 1)4\} \)
\( = 10(10 + 76) = 10(86) = 860 \).

Question. Which term of the A.P. 3, 15, 27, 39,… will be 120 more than its \( 21^{st} \) term?
Answer: From the given A.P., we have \( a = 3, d = 15 - 3 = 12 \)
Thus, \( 21^{st} \) term, \( t_{21} = 3 + (21 - 1)(12) \)
\( = 3 + 240 = 243 \)
Thus, the number 120 more than its \( 21^{st} \) term is \( 243 + 120 = 363 \)
Let 363 be the \( m^{th} \) term.
Hence, \( t_m = 3 + (m - 1)(12) = 363 \)
\( \Rightarrow 3 + 12m - 12 = 363 \)
\( \Rightarrow 12m - 9 = 363 \)
\( \Rightarrow 12m = 372 \)
\( \Rightarrow m = 31 \)
Thus, the \( 31^{st} \) term will be 120 more than the \( 21^{st} \) term.

Question. Find the value of the middle term of the following A.P., 10, 7, 4, …, (– 62).
Answer: From the given A.P., we have \( a = 10, d = -3 \) and \( t_n = -62 \)
Thus \( t_n = 10 + (n - 1)(-3) = -62 \)
\( \Rightarrow 10 - 3n + 3 = -62 \)
\( \Rightarrow -3n = -75 \)
\( \Rightarrow n = 25 \)
Thus, the middle term would be 13.
Hence \( t_{13} = 10 + (13 - 1)(-3) \)
\( = 10 - 36 = -26 \).

Question. Find the middle term of the A.P. 6, 13, 20, ......... 216.
Answer: Given A.P. is 6, 13, 20, ........., 216
Here, \( a = 6, d = 13 - 6 = 20 - 13 = 7 \)
Let \( n \) be the number of terms.
Then, \( T_n = a + (n - 1)d \)
\( \Rightarrow 216 = 6 + (n - 1)7 \)
\( \Rightarrow 216 = 6 + 7n - 7 \)
\( \Rightarrow 217 = 7n \)
\( \therefore n = 31 \)
So middle term is \( (\frac{n+1}{2})^{th} \) term i.e., \( 16^{th} \) term
\( \therefore T_{16} = 6 + (16 - 1)7 \)
\( = 6 + 15 \times 7 = 6 + 105 = 111 \)
\( \therefore \) Middle term of the A.P. is 111.

Question. The \( 8^{th} \) term of an A.P. is zero. Prove that the \( 38^{th} \) term is triple its \( 18^{th} \) term.
Answer: Let the first term of the A.P. be \( a \) and the common difference be \( d \).
Thus, \( t_8 = a + (8 - 1)d = 0 \)[Given]
or \( a + 7d = 0 \) ... (i)
Also \( t_{18} = a + (18 - 1)d \)
\( = a + 17d \)
\( = (a + 7d ) + 10d = 10d \)[from (i)]
and \( t_{38} = a + (38 - 1)d \)
\( = a + 37d \)
\( = (a + 7d) + 30d \)
\( = 30d \)[from (i)]
\( = 3(10d) \)
\( = 3t_{18} \). Hence Proved.

Question. If \( m \) times the \( m^{th} \) term of an A.P. is equal to \( n \) times the \( n^{th} \) term and \( m \neq n \), then show that its \( (m + n)^{th} \) term is zero.
Answer: Let the first term be \( a \) and the common difference be \( d \).
Hence \( t_m = a + (m - 1)d \) and \( t_n = a + (n - 1)d \)
Now \( m\{a + (m - 1)d\} = n\{a + (n - 1)d\} \)
\( \Rightarrow ma + dm^2 - dm = na + dn^2 - dn \)
\( \Rightarrow (m - n)a + d(m^2 - n^2) - d(m - n) = 0 \)
\( \Rightarrow (m - n)a + (m - n)(m + n)d - (m - n)d = 0 \)
Divide by \( (m - n) \):
\( \Rightarrow a + (m + n)d - d = 0 \)
\( \Rightarrow a + \{(m + n) - 1\}d = 0 \)
\( \Rightarrow t_{m + n} = 0 \). Hence Proved.

Question. For what value of \( n \), are the \( n^{th} \) terms of two A.Ps. 63, 65, 67,....... and 3, 10, 17,...... equal ?
Answer: \( 1^{st} \) A.P. is 63, 65, 67, .....
Here, \( a = 63, d = 65 - 63 = 2 \)
\( \therefore a_n = a + (n - 1)d = 63 + (n - 1) (2) = 63 + 2n - 2 = 61 + 2n \)
\( 2^{nd} \) A.P. is 3, 10, 17, ....
Here, \( a = 3, d = 10 - 3 = 7 \)
\( \therefore a_n = a + (n - 1)d = 3 + (n - 1) (7) = 3 + 7n - 7 = 7n - 4 \)
According to question, \( 61 + 2n = 7n - 4 \)
\( \Rightarrow 61 + 4 = 7n - 2n \)
\( \Rightarrow 65 = 5n \)
\( \Rightarrow n = \frac{65}{5} = 13 \)
Hence, \( 13^{th} \) term of both A.P. is equal.

Question. Which term of the A.P. 3, 8, 13, 18, … will be 55 more than its \( 20^{th} \) term ?
Answer: Given, \( a = 3, d = 5 \)
Thus, \( t_{20} = 3 + (20 - 1)5 = 3 + 95 = 98 \)
So \( 98 + 55 = 153 \)
Let the required term be \( n \).
So \( t_n = 3 + (n - 1)5 = 153 \)
\( \Rightarrow (n - 1)5 = 150 \)
\( \Rightarrow n - 1 = 30 \)
\( \Rightarrow n = 31 \).

Question. Which term of the A.P. 5, 15, 25, … will be 130 more than its \( 31^{st} \) term ?
Answer: Given, \( a = 5, d = 10 \)
Thus, \( t_{31} = 5 + (31 - 1)10 = 5 + 300 = 305 \)
Then, \( 305 + 130 = 435 \)
Let 435 be the \( m^{th} \) term.
So \( t_n = 5 + (n - 1)10 = 435 \)
\( \Rightarrow (n - 1)10 = 430 \)
\( \Rightarrow n - 1 = 43 \)
\( \Rightarrow n = 44 \).

Question. Find the \( 6^{th} \) term from the end of the A.P. 17, 14, 11, …, – 40.
Answer: Given, \( a = 17, d = - 3 \) and \( t_n = - 40 \)
Now taking from the end \( a = - 40, d = -(-3) = 3 \) and \( t_n = 17 \)
Hence, the \( 6^{th} \) term from the end is \( t_6 = \{- 40 + (6 - 1)3\} \)
\( = - 40 + 15 = - 25 \)
Thus, the \( 6^{th} \) term from the end is \( - 25 \).

Question. Find the \( 21^{st} \) term of the A.P. \( -4\frac{1}{2}, -3, -1\frac{1}{2} \), …
Answer: Given \( -4\frac{1}{2}, -3, -1\frac{1}{2} \), … or \( -\frac{9}{2}, -3, -\frac{3}{2}, \dots \)
Here \( a = -\frac{9}{2}, d = -3 - (-\frac{9}{2}) = -3 + \frac{9}{2} = \frac{-6+9}{2} = \frac{3}{2} \)
\( T_n = a + (n - 1) d \)
\( T_{21} = -\frac{9}{2} + (21 - 1)\frac{3}{2} \)
\( T_{21} = -\frac{9}{2} + 20 \times \frac{3}{2} \)
\( T_{21} = -\frac{9}{2} + 30 \)
\( T_{21} = \frac{-9 + 60}{2} = \frac{51}{2} = 25\frac{1}{2} \).

Question. Determine \( k \) such that \( (3k - 2), (4k - 6) \) and \( (k + 2) \) are three consecutive terms of an A.P.
Answer: The terms of an A.P. are \( (3k - 2), (4k - 6) \) and \( (k + 2) \).
Thus, \( d = (4k - 6) - (3k - 2) = (k + 2) - (4k - 6) \)
\( \Rightarrow k - 4 = - 3k + 8 \)
\( \Rightarrow 4k = 12 \)
\( \Rightarrow k = 3 \).

Question. Write the value of \( x \) for which \( (x + 2), 2x \) and \( (2x + 3) \) are three consecutive terms of an A.P.
Answer: The terms of an A.P. are \( (x + 2), 2x \) and \( (2x + 3) \).
Thus, \( d = 2x - (x + 2) = (2x + 3) - 2x \)
\( \Rightarrow x - 2 = 3 \)
\( \Rightarrow x = 5 \).

Question. Find the sum of \( 25 + 28 + 31 + … + 100 \).
Answer: Here, \( a = 25, d = 3 \) and \( t_n = 100 \)
Now, \( t_n = 25 + (n - 1)3 = 100 \)
\( \Rightarrow 3(n - 1) = 75 \)
\( \Rightarrow n - 1 = 25 \)
\( \Rightarrow n = 26 \)
\( \therefore S_n = \frac{26}{2} \{2(25) + (26 - 1)3\} \)
\( = 13[50 + 75] = 13 \times 125 = 1625 \).

Question. If the \( n^{th} \) term of an A.P. is \( 2n + 1 \), find the sum of the first \( n \) terms of an A.P.
Answer: Given, \( t_n = 2n + 1 \)
\( \therefore t_1 = 2 \times 1 + 1 = 3 \)
and \( t_2 = 2 \times 2 + 1 = 5 \)
Thus, \( a = 3 \) and \( d = t_2 - t_1 = 2 \)
Hence, \( S_n = \frac{n}{2} \{2(3) + (n - 1)(2)\} \)
\( = n\{3 + n - 1\} = n(2 + n) \).

Question. How many terms of the A.P. 65, 60, 55,.... be taken so that their sum is zero ?
Answer: Given, A.P. is 65, 60, 55......
There, \( a = 65, d = 60 - 65 = -5 \)
Now, \( S_n = 0 \)
\( S_n = \frac{n}{2} [2a + (n - 1)d ] = 0 \)
\( \Rightarrow \frac{n}{2} [2(65) + (n - 1) (-5)] = 0 \)
\( \Rightarrow 130 - 5n + 5 = 0 \)
\( \Rightarrow 135 - 5n = 0 \)
\( \Rightarrow 5n = 135 \)
\( \therefore n = 27 \)
Hence, the number of terms are 27.

Question. Find the sum of the first hundred even natural numbers, which are divisible by 5.
Answer: First hundred even natural numbers divisible by 5 are 10, 20, 30, 40, …
This is an A.P. Here, \( a = 10, d = 10 \) and \( n = 100 \)
\( S_{100} = \frac{100}{2} \{2(10) + (100 - 1)(10)\} \)
\( = 50\{20 + 990\} = 50\{1010\} = 50500 \).

Question. The sum of the first \( n \) terms of an A.P. is \( 2n^2 + 5n \). Find the \( n^{th} \) term of this A.P.
Answer: Given, \( S_n = 2n^2 + 5n \)
\( \therefore S_{n - 1} = 2(n - 1)^2 + 5(n - 1) \)
Now \( t_n = S_n - S_{n - 1} \)
\( = 2n^2 + 5n - 2(n - 1)^2 - 5(n - 1) \)
\( = 2n^2 + 5n - 2n^2 + 4n - 2 - 5n + 5 \)
\( = 4n + 3 \).

Question. The sum of the first \( n \) terms of an A.P. is \( 3n^2 - n \). Find the \( n^{th} \) term, first term and the common difference of this A.P.
Answer: Given, \( S_n = 3n^2 - n \)
\( \therefore S_{n - 1} = 3(n - 1)^2 - (n - 1) \)
Now, \( t_n = S_n - S_{n - 1} \)
\( t_n = 3n^2 - n - 3 (n - 1)^2 + n - 1 \)
\( = 3n^2 - n - 3n^2 + 6n - 3 + n - 1 = 6n - 4 \)
\( \therefore a = t_1 = 6 - 4 = 2 \)
\( t_2 = 6 \times 2 - 4 = 12 - 4 = 8 \)
and \( d = t_2 - t_1 = 8 - 2 = 6 \)
Hence \( n^{th} \) term \( = 6n - 4, a = 2 \) and \( d = 6 \).

Question. The first and last terms of an A.P. are 4 and 81 respectively. If its common difference is 7, how many terms are there and what is their sum ?
Answer: Given, \( a = 4, d = 7 \) and \( t_n = 81 \)
We know \( t_n = a + (n - 1)d \)
\( \Rightarrow 4 + (n - 1)7 = 81 \)
\( \Rightarrow 7n - 3 = 81 \)
\( \Rightarrow 7n = 84 \)
\( \Rightarrow n = 12 \)
Thus, \( S_{12} = \frac{12}{2} \{2(4) + (12 - 1)7\} \)
\( = 6\{8 + 77\} = 6\{85\} = 510 \).

Question. Which term of the progression \( 20, 19\frac{1}{4}, 18\frac{1}{2}, 17\frac{3}{4} \dots \) is the first negative term ?
Answer: Given, A.P. is 20, \( 19\frac{1}{4}, 18\frac{1}{2}, 17\frac{3}{4}, \dots \) = \( 20, \frac{77}{4}, \frac{37}{2}, \frac{71}{4}, \dots \)
Here, \( a = 20, d = \frac{77}{4} - 20 = \frac{77-80}{4} = -\frac{3}{4} \)
Let \( a_n \) be its first negative term. Then \( a + (n - 1)d < 0 \).
\( \Rightarrow 20 + (n - 1) (-\frac{3}{4}) < 0 \)
\( \Rightarrow 20 - \frac{3}{4}n + \frac{3}{4} < 0 \)
\( \Rightarrow 20 + \frac{3}{4} < \frac{3}{4}n \)
\( \Rightarrow \frac{83}{4} < \frac{3}{4}n \)
\( \Rightarrow n > \frac{83}{4} \times \frac{4}{3} \)
\( \Rightarrow n > \frac{83}{3} = 27.66 \dots \)
\( 28^{th} \) term will be first negative term of given A.P.

Question. Which term of the A.P. 8, 14, 20, 26, ......, will be 72 more than its \( 41^{st} \) term ?
Answer: Given, AP is 8, 14, 20, 26,.....
Here, \( a = 8, d = 14 - 8 = 6 \)
Let \( a_n = a_{41} + 72 \)
\( \Rightarrow a + (n - 1)d = a + 40d + 72 \)
\( \Rightarrow (n - 1) 6 = 40 \times 6 + 72 \)
\( = 240 + 72 = 312 \)
\( \Rightarrow n - 1 = \frac{312}{6} = 52 \)
\( \Rightarrow n = 52 + 1 = 53^{rd} \) term.

Question. The \( n^{th} \) term of an A.P. is \( (7 - 4n) \). Find its common difference.
Answer: Given, \( t_n = 7 - 4n \)
\( \therefore t_{n-1} = 7 - 4(n - 1) = 7 - 4n + 4 = 11 - 4n \)
Now, \( d = t_n - t_{n-1} \)
\( \Rightarrow d = 7 - 4n - (11 - 4n) \)
\( \Rightarrow d = - 4 \).
OR
Given, \( t_n = (7 - 4n) \).
Let the first term be \( a \) and the common difference be \( d \).
Then, \( t_n = a + (n - 1)d = (7 - 4n) \)
\( \Rightarrow \{a + (n - 1)d\} = 7 - 4n \)
\( \Rightarrow a + dn - d = 7 - 4n \)
\( \Rightarrow (a - d) + dn = 7 - 4n \)
On comparing we get \( \Rightarrow d = - 4 \).

Question. The first term of an A.P. is \( p \) and its common difference is \( q \). Find its \( 10^{th} \) term.
Answer: Given, \( a = p \) and \( d = q \)
So \( t_n = \{p + (10 - 1)q\} \)
\( \Rightarrow t_{10} = p + 9q \).

Question. Write the next term of the A.P. \( \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots \)
Answer: The given A.P. is \( \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots \)
Thus, \( a = \sqrt{8} \)
and \( d = \sqrt{18} - \sqrt{8} \)
\( = 3\sqrt{2} - 2\sqrt{2} = \sqrt{2} \)
Hence, \( t_4 = \sqrt{8} + (4 - 1)\sqrt{2} \) (\( i.e. \), next term)
\( = 2\sqrt{2} + 3\sqrt{2} \)
\( = 5\sqrt{2} \)
\( = \sqrt{50} \)
Thus, the next term is \( \sqrt{50} \).

Question. Which term of the A.P. 21, 18, 15, … is zero ?
Answer: The given A.P. is 21, 18, 15, …
Thus, \( a = 21 \) and \( d = 18 - 21 = - 3 \)
Let the \( n^{th} \) term be zero.
Hence, \( t_n = 21 - (n - 1)3 = 0 \)
\( \Rightarrow 21 - 3n + 3 = 0 \)
\( \Rightarrow 24 - 3n = 0 \)
\( \Rightarrow 3n = 24 \)
\( \Rightarrow n = 8 \).

Question. Find the number of all three-digit natural number divisible by 9.
Answer: Three digit natural numbers divisible by 9 are 108, 117, 126, 135, …, 999.
The above series is an A.P.
with, \( a = 108, d = 9 \) and \( t_n = 999 \)
Thus, \( t_n = 108 + (n - 1)9 = 999 \)
\( \Rightarrow 108 + 9n - 9 = 999 \)
\( \Rightarrow 9n = 999 - 99 \)
\( \Rightarrow 9n = 900 \)
\( \Rightarrow n = 100 \)
Thus, the number of three-digit natural numbers divisible by 9 are 100.

Question. The \( 4^{th} \) term of an A.P. is zero. Prove that the \( 25^{th} \) term of the A.P. is three times its \( 11^{th} \) term.
Answer: We know that
\( T_n = a + (n - 1)d \)
Given, \( T_4 = a + (4 - 1)d = 0 \)
\( \Rightarrow a + 3d = 0 \)
\( \Rightarrow a = - 3d \) ...(i)
Now, \( T_{25} = a + (25 - 1)d \)
\( = a + 24d = (- 3d) + 24d \)
\( = 21d \)
and \( T_{11} = a + (11 - 1)d \)
\( = a + 10d \)
then, \( 3T_{11} = 3(a + 10d) \)
\( = 3a + 30d \)
\( = 3(-3d) + 30d \)[From (i)]
\( = 30d - 9d = 21d = T_{25} \)
\( \therefore 3T_{11} = T_{25} \)
Hence Proved.

Question. How many terms of the A.P. 18, 16, 14,... be taken so that their sum is zero ?
Answer: Given, A.P. is 18, 16, 14,...
There, \( a = 18, d = 16 - 18 = 14 - 16 = - 2 \)
Now, \( S_n = 0 \)
\( \therefore S_n = \frac{n}{2} [2a + (n - 1)d] = 0 \)
\( \Rightarrow \frac{n}{2} [2 \times 18 + (n - 1)(- 2)] = 0 \)
\( \Rightarrow 36 - 2n + 2 = 0 \)
\( \Rightarrow 2n = 38 \)
\( \Rightarrow n = 19 \)
Hence, the number of terms are 19.

Question. If in an A.P., \( S_5 + S_7 = 167 \) and \( S_{10} = 235 \), then find the A.P., where \( S_n \) denotes the sum of its first \( n \) terms.
Answer: Given, \( S_5 + S_7 = 167 \)
\( \Rightarrow \frac{5}{2} (2a + 4d) + \frac{7}{2} (2a + 6d) = 167 \)
\( \Rightarrow \frac{5}{2} \times 2(a + 2d) + \frac{7}{2} \times 2(a + 3d) = 167 \)
\( \Rightarrow 5a + 10d + 7a + 21d = 167 \)
\( \Rightarrow 12a + 31d = 167 \) ...(i)
Also, \( S_{10} = 235 \)
\( \Rightarrow \frac{10}{2} (2a + 9d) = 235 \)
\( \Rightarrow 10a + 45d = 235 \)
\( \Rightarrow 2a + 9d = 47 \) ...(ii)
On multiplying equation (ii) by 6, we get
\( 12a + 54d = 282 \) ...(iii)
On subtracting equation (i) from (iii), we get
\( 23d = 115 \)
\( d = 5 \)
Substituting value of \( d \) in equation (i), we get
\( 12a + 31 \times 5 = 167 \)
\( 12a + 155 = 167 \)
\( \Rightarrow 12a = 12 \)
\( \Rightarrow a = 1 \)
Hence, A.P. is 1, 6, 11....

Question. The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.
Answer: Let, 1st term = \( a \)
Common difference = \( d \)
Given, \( a_4 = a + 3d \)
\( 0 = a + 3d \)
\( \Rightarrow a = - 3d \) ...(i)
To prove : \( a_{25} = 3 \times a_{11} \)
\( a + 24d = 3(a + 10d) \) [From (i)]
\( \Rightarrow - 3d + 24d = 3(- 3d + 10d) \)
\( \Rightarrow 21d = 21d \)
From above,
\( a_{25} = 3(a_{11}) \)
Hence Proved.

Question. Find how many two-digit numbers are divisible by 6?
Answer: Two-digit numbers divisible by 6 are :
12, 18, 24, ...., 96
Here, \( a = 12, d = 18 - 12 = 6, a_n = 96 \)
Since, \( a + (n - 1)d = a_n \)
\( \therefore 12 + (n - 1)6 = 96 \)
\( \Rightarrow (n - 1)6 = 96 - 12 = 84 \)
\( \Rightarrow n - 1 = \frac{84}{6} \)
\( \Rightarrow n - 1 = 14 \)
\( \Rightarrow n = 14 + 1 = 15 \)
\( \therefore \) There are 15 two-digit numbers divisible by 6.

Question. Find the middle term of the A.P. 6, 13, 20, ...., 216.
Answer: The given A.P. is 6, 13, 20, ....., 216
Let \( n \) be the number of terms,
\( d = 13 - 6 = 7 \)
\( a_n = 216 \)
Since, \( a_n = a + (n - 1)d \)
\( \therefore 216 = 6 + (n - 1)7 \)
\( \Rightarrow 216 - 6 = (n - 1)7 \)
\( \Rightarrow \frac{210}{7} = n - 1 \)
\( \Rightarrow 30 + 1 = n \)
\( \Rightarrow n = 31 \)
Middle term = \( (\frac{n+1}{2})^{th} \) term
\( = (\frac{32}{2})^{th} \) term
\( = 16^{th} \) term
Now, \( a_{16} = a + 15d \)
\( = 6 + 15 \times 7 = 111 \).

Question. Find the A.P. whose \( n^{th} \) term is \( 7 - 3k \). Also find the \( 20^{th} \) term.
Answer: From the question it is given that, \( n^{th} \) term is \( 7 - 3k \)
So, \( T_n = 7 - 3n \)
Now, we start giving values, 1, 2, 3, ... in the place of \( n \), we get
\( T_1 = 7 - (3 \times 1) = 7 - 3 = 4 \)
\( T_2 = 7 - (3 \times 2) = 7 - 6 = 1 \)
\( T_3 = 7 - (3 \times 3) = 7 - 9 = - 2 \)
\( T_4 = 7 - (3 \times 4) = 7 - 12 = - 5 \)
\( T_{20} = 7 - (3 \times 20) = 7 - 60 = - 53 \)
Therefore, A.P. is 4, 1, – 2, – 5, ...
And \( 20^{th} \) term is – 53

Question. Which term of the Arithmetic Progression – 7, – 12, – 17, – 22, ...... will be – 82 ? Is – 100 any term of the A.P. ? Give reason for your answer.*
Answer: Given A.P. is
– 7, – 12, – 17, – 22, .....
Here \( a = - 7, d = - 12 - (- 7) \)
\( = - 12 + 7 \)
\( = - 5 \)
Let \( T_n = - 82 \)
\( T_n = a + (n - 1) d \Rightarrow - 82 = - 7 + (n - 1) (- 5) \)
\( \Rightarrow - 82 = - 7 - 5n + 5 \)
\( \Rightarrow - 82 = - 2 - 5n \Rightarrow - 82 + 2 = - 5n \)
\( \Rightarrow - 80 = - 5n \)
\( \Rightarrow n = 16 \)
Therefore, \( 16^{th} \) term will be – 82.
Let \( T_n = - 100 \)
Again, \( T_n = a + (n - 1) d \)
\( \Rightarrow - 100 = - 7 + (n - 1) (- 5) \)
\( \Rightarrow - 100 = - 7 - 5n + 5 \)
\( \Rightarrow - 100 = - 2 - 5n \)
\( \Rightarrow - 100 + 2 = - 5n \)
\( \Rightarrow - 98 = - 5n \)
\( \Rightarrow n = \frac{98}{5} \)
But the number of terms can not be in fraction.
So, – 100 can not be the term of this A.P.

Question. If the pth term of an A.P. is \( \frac{1}{q} \) and qth term is \( \frac{1}{p} \), prove that the sum of first pq terms of the A.P. is \( \frac{pq+1}{2} \).*
Answer: Let \( a \) be the first term and \( d \) is common difference.
Then, \( a_p = \frac{1}{q} \Rightarrow a + ( p - 1) d = \frac{1}{q} \)...(i)
and \( a_q = \frac{1}{p} \Rightarrow a + (q - 1) d = \frac{1}{p} \)...(ii)
Subtracting equation (ii) from equation (i)
\( pd - qd + d - d = \frac{1}{q} - \frac{1}{p} \)
\( \Rightarrow ( p - q) d = \frac{p - q}{pq} \)
\( \Rightarrow d = \frac{1}{pq} \)
Putting the value of \( d \) in equation (i), we get
\( a + ( p - 1) \frac{1}{pq} = \frac{1}{q} \)
\( \Rightarrow a = \frac{1}{q} - \frac{p}{pq} + \frac{1}{pq} \)
\( \Rightarrow a = \frac{1}{q} - \frac{1}{q} + \frac{1}{pq} \)
\( \Rightarrow a = \frac{1}{pq} \)
\( \therefore \) Sum of first \( pq \) terms
\( S_{pq} = \frac{pq}{2} [2a + ( pq - 1)d] \)
\( = \frac{pq}{2} \left[ \frac{2}{pq} + (pq - 1) \frac{1}{pq} \right] \)
\( = \frac{pq}{2} \left[ \frac{2}{pq} + \frac{pq}{pq} - \frac{1}{pq} \right] \)
\( = \frac{pq}{2} \left( \frac{1}{pq} + 1 \right) \)
\( S_{pq} = \frac{pq}{2} \left( \frac{1 + pq}{pq} \right) \)
\( = \frac{pq + 1}{2} \) Hence Proved.

Question. In an A.P., the sum of the first 10 terms is – 150 and the sum of the next ten terms is – 550. Find the A.P.*
Answer: Let the first term be \( a \) and the common difference be \( d \).
Thus, \( S_{10} = \frac{10}{2} \{2a + (10 - 1)d\} = – 150 \)
\( \Rightarrow 2a + 9d = – 30 \)...(i)
and \( S_{20} = \frac{20}{2} \{2a + (20 - 1)d\} \)
\( = 10\{2a + 19d\} \)
Sum of next ten terms
\( S_{10}' = S_{20} – S_{10} \)
\( \Rightarrow – 550 = S_{20} + 150 \)
\( \Rightarrow S_{20} = – 550 – 150 = – 700 \)
\( \Rightarrow 10\{2a + 19d\} = – 700 \)
\( \Rightarrow 2a + 19d = – 70 \)...(ii)
Subtracting (i) from (ii), we have
\( 10d = – 40 \)
\( \Rightarrow d = – 4 \)
Substituting \( d = – 4 \) in (i), we have
\( 2a + 9(– 4) = – 30 \)
\( \Rightarrow 2a – 36 = – 30 \)
\( \Rightarrow 2a = 6 \)
\( \Rightarrow a = 3 \)
Thus, the A.P. is 3, – 1, – 5, – 9, …

Question. The sum of first n terms of an A.P. is \( 5n^2 + 3n \). If \( m^{th} \) term is 168, find the value of \( m \) and also the \( 20^{th} \) term.*
Answer: Given, \( S_n = 5n^2 + 3n \) and \( t_m = 168 \)
Thus, \( S_n = \frac{n}{2} \{2a + (n - 1)d\} = 5n^2 + 3n \)
\( \Rightarrow n\{2a + (n - 1)d\} = 2n(5n + 3) \)
\( \Rightarrow 2a + dn - d = 10n + 6 \)
\( \Rightarrow (2a - d) + dn = 6 + 10n \)
On comparing both sides, we get
\( d = 10 \)
and \( 2a - d = 6 \)
\( \Rightarrow 2a = d + 6 \)
\( \Rightarrow 2a = 10 + 6 \)
\( \Rightarrow a = 8 \)
Now, \( t_m = a + (m - 1)d = 168 \)
\( \Rightarrow 8 + (m - 1)10 = 168 \)
\( \Rightarrow 10(m - 1) = 160 \)
\( \Rightarrow m - 1 = 16 \)
\( \Rightarrow m = 17 \)
Also, \( t_{20} = 8 + (20 - 1)10 = 198 \).

Question. If the sum of first \( m \) terms of an A.P. is the same as the sum of its first \( n \) terms, show that the sum of its first \( (m + n) \) terms is zero.*
Answer: Let \( a \) be first term and \( d \) be common difference of given A.P.
Then, \( S_m = S_n \)(Given)
\( \frac{m}{2} \{2a + (m - 1)d\} = \frac{n}{2} \{2a + (n - 1) d\} \)
\( \Rightarrow 2am + m(m - 1)d - 2an - n(n - 1) d = 0 \)
\( \Rightarrow 2am – 2an + \{m(m - 1) – n (n - 1)\} d = 0 \)
\( \Rightarrow 2a (m – n) + (m^2 – m – n^2 + n) d = 0 \)
\( \Rightarrow 2a (m – n) + \{m^2 – n^2 – (m – n)\} d = 0 \)
\( \Rightarrow 2a (m – n) + (m – n) (m + n – 1) d = 0 \)
\( \Rightarrow (m – n) \{2a + (m + n – 1) d\} = 0 \)
\( \Rightarrow 2a + (m + n – 1) d = 0 \)
Now, \( S_{m + n} = \frac{m + n}{2} \{2a + (m + n – 1) d \} \)
\( = \frac{m + n}{2} \times 0 = 0 \). Hence Proved.