CBSE Class 10 Mathematics Arithmetic Progressions Worksheet Set 13

Question. 200 logs are stacked in a such a way that 20 logs in the bottom row, 19 logs in the next row, 18 logs in the row next to it and so on. The total number of rows is 
(a) 16
(b) 12
(c) 15
(d) 10
Answer: (a) 16
Answer: The number of logs in the row from bottom to the top are 20, 19, 18, …. which form an AP with first term 20 and common difference \( 19 - 20 = -1 \).
Let the 200 logs be arranged in \( n \) rows.
Then \( S_n = 200 \)

\( \implies 200 = \frac{n}{2} [2a + (n - 1)d] \)

\( \implies 400 = n [2 \times 20 + (n - 1) (-1)] \)

\( \implies n^2 - 41n + 400 = 0 \)

\( \implies n^2 - 16n - 25n + 400 = 0 \)

\( \implies n(n - 16) - 25(n - 16) = 0 \)

\( \implies (n - 25)(n - 16) = 0 \)

\( \implies n - 25 = 0 \text{ or } n - 16 = 0 \)

\( \implies n = 25 \text{ or } n = 16 \)
\( n = 25 \) is not possible as on calculating number of logs in 25th row, there is negative number of logs, which is not possible.
Therefore, number of rows are 16.

Question. Sum of n terms of the series, \( \sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} + \dots \) is 
(a) \( \frac{n(n-1)}{2} \)
(b) \( \frac{n(n+1)}{\sqrt{2}} \)
(c) \( \frac{n(n+1)}{2} \)
(d) \( \frac{n(n-1)}{\sqrt{2}} \)
Answer: (b) \( \frac{n(n+1)}{\sqrt{2}} \)
Answer: Given: \( \sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} + \dots \)

\( \implies \sqrt{2} + 2\sqrt{2} + 3\sqrt{2} + 4\sqrt{2} + \dots \)
Here, \( a = \sqrt{2}, d = 2\sqrt{2} - \sqrt{2} = \sqrt{2} \)
\( \therefore S_n = \frac{n}{2} [2a + (n - 1)d] \)

\( \implies S_n = \frac{n}{2} [2 \times \sqrt{2} + (n - 1) \times \sqrt{2}] \)

\( \implies S_n = \frac{n}{2} [2\sqrt{2} + n\sqrt{2} - \sqrt{2}] \)

\( \implies S_n = \frac{n}{2} [\sqrt{2} + n\sqrt{2}] \)

\( \implies S_n = \frac{n\sqrt{2}}{2} (1 + n) = \frac{n(n+1)}{\sqrt{2}} \)

Question. A sum of Rs.700 is to be used to award 7 prizes. If each prize is Rs.20 less than its preceding prize, then the value of the first prize is 
(a) Rs.160
(b) Rs.100
(c) Rs.180
(d) Rs.200
Answer: (a) Rs.160
Answer: Let the first prize be \( a \).
The seven prizes form an AP with first term \( a \) and common difference, \( d = -20, n = 7 \).
Now the sum of all seven prizes = Rs. 700
\( \therefore S_n = 700 \)

\( \implies \frac{n}{2} [2a + (n - 1)d] = 700 \)

\( \implies \frac{7}{2} [2a + (7 - 1)(-20)] = 700 \)

\( \implies 2a - 120 = 200 \)

\( \implies 2a = 320 \)

\( \implies a = 160 \)
Therefore, the value of first prize is Rs. 160.

Question. The sum of odd numbers between 0 and 50 is (1)
(a) 625
(b) 600
(c) 500
(d) 2500
Answer: (a) 625
Answer: Odd numbers between 0 and 50 are 1, 3, 5, 7, ………, 49. Here \( a = 1, d = 3 - 1 = 2 \) and
\( n = 25 \)
\( S_{25} = \frac{25}{2} \times [1 + 49] \)
\( S_{25} = 25 \times 25 \)
= 625

Question. The common difference of an A.P. in which \( a_{18} - a_{14} = 32 \) is 
(a) – 8
(b) 6
(c) 8
(d) – 6
Answer: (c) 8
Answer: Given: \( a_{18} - a_{14} = 32 \)

\( \implies a + (18 - 1)d - [a + (14 - 1)d] = 32 \)

\( \implies a + 17d - a - 13d = 32 \)

\( \implies 4d = 32 \)

\( \implies d = 8 \)

Question. If \( S_n \) denotes the sum of first n terms of an AP, prove that \( S_{12} = 3(S_8 - S_4) \). 
Answer: Let \( a \) be the first term and \( d \) be the common difference of the given AP. Then,
\( S_n = \frac{n}{2} [2a + (n - 1)d] \),
\( \therefore 3(S_8 - S_4) = 3[ \frac{8}{2}(2a + 7d) - \frac{4}{2}(2a + 3d) ] \)
\( = 3[4(2a + 7d) - 2(2a + 3d)] = 3[8a + 28d - 4a - 6d] \)
\( = 3[4a + 22d] = 12a + 66d \)
\( = 6(2a + 11d) \)
\( = \frac{12}{2} (2a + 11d) = S_{12} \).
Hence, \( S_{12} = 3(S_8 - S_4) \).

Question. For an AP, if \( a_{18} - a_{14} = 32 \) then find the common difference d. 
Answer: We know, \( n^{th} \) term of an AP is given by \( a_n = a + (n - 1)d \), where \( a \) is the first term and \( d \) is the common difference.
Given, \( a_{18} - a_{14} = 32 \)

\( \implies (a + (18 - 1)d) - (a + (14 - 1)d) = 32 \)

\( \implies (a + 17d) - (a + 13d) = 32 \)

\( \implies 4d = 32 \)

\( \implies d = 8 \)

Question. Find the first four terms of an A.P. whose first term is - 2 and common difference is - 2. 
Answer: given \( a_1 = -2 \), common difference \( d = -2 \)
\( a_1 = -2 \),
\( a_2 = a_1 + d = -2 + (-2) = -4 \)
\( a_3 = a_2 + d = -4 + (-2) = -6 \)
\( a_4 = a_3 + d = -6 + (-2) = -8 \)
\( \therefore \) First four terms are - 2, - 4, - 6, - 8

Question. If the sum of n terms of an A.P. is \( 2n^2 + 5n \), then find the 4th term. 
Answer: Let the sum of n terms of A.P. = \( S_n \).
Given, \( S_n = 2n^2 + 5n \)
Now, \( n^{th} \) term of A.P, \( a_n = S_n - S_{n-1} \)
or, \( a_n = (2n^2 + 5n) - [2(n - 1)^2 + 5(n - 1)] \)
\( a_n = (2n^2 + 5n) - [2(n^2 - 2n + 1) + 5n - 5] \)
\( a_n = 2n^2 + 5n - [2n^2 - 4n + 2 + 5n - 5] \)
\( a_n = 2n^2 + 5n - [2n^2 + n - 3] \)
\( a_n = 2n^2 + 5n - 2n^2 - n + 3 \)
\( a_n = 4n + 3 \)
\( 4^{th} \) term \( a_4 = 4 \times 4 + 3 = 16 + 3 = 19 \)

Question. What is the common difference of the A.P. \( \frac{1}{p}, \frac{1-p}{p}, \frac{1-2p}{p}, \dots \)? 
Answer: Common difference(d) = \( a_2 - a_1 = \frac{1-p}{p} - \frac{1}{p} = \frac{1-p-1}{p} = \frac{-p}{p} = -1 \)
therefore, \( d = -1 \)

Question. In the following AP's find the missing terms: 2, __, 26. 
Answer: We know that the difference between consecutive terms is equal in any A.P.
Let the missing term be \( x \).
\( x - 2 = 26 - x \)

\( \implies 2x = 28 \)

\( \implies x = 14 \)
Therefore, missing term is 14.

Question. Is 68 a term of the AP : 7, 10, 13, ....?
Answer: Given AP 7, 10, 13...
Here, first term \( a = 7 \) and common difference \( d = a_2 - a_1 = 10 - 7 = 3 \),
Assume that 68 is \( n^{th} \) term of given AP.
We know that \( n^{th} \) term is given by \( a_n = a + (n - 1)d \)

\( \implies 68 = 7 + (n - 1) \times 3 \)

\( \implies 61 = (n - 1) \times 3 \)

\( \implies \frac{61}{3} + 1 = n \)

\( \implies n = \frac{64}{3} = 21 \frac{1}{3} \)
Since, \( n \) cannot be fraction, but a whole number. Therefore, 68 is not a term of given AP.

Question. Find 51 is a term of given A.P. or not where the A.P. is 5, 8, 11, 14, ......
Answer: Given, A.P. is 5, 8, 11, 14, ......
Here \( a = 5 \) and \( d = (8 - 5) = 3 \).
Let the \( n^{th} \) term of the given AP be 51. Then,
\( T_n = 51 \)
We know that \( T_n = a + (n - 1)d \).

\( \implies a + (n - 1)d = 51 \)

\( \implies 5 + (n - 1) \times 3 = 51 \) [Because, \( a = 5 \) and \( d = 3 \)]
\( 5 + 3n - 3 = 51 \)

\( \implies 3n = 51 - 2 \)

\( \implies 3n = 49 \)

\( \implies n = 16 \frac{1}{3} \).
But, the number of terms cannot be a fraction.
Therefore, 51 is not a term of the given Arithmetic progression.

Question. In the following situation, does the list of numbers involved make an arithmetic progression, and why? The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum. 
Answer: Amount of money after 1 year \( = Rs 10000 (1 + \frac{8}{100}) = a_1 \)
Amount of money after 2 year \( = Rs 10000 (1 + \frac{8}{100})^2 = a_2 \)
Amount of money after 3 year \( = Rs 10000 (1 + \frac{8}{100})^3 = a_3 \)
Amount of money after 4 year \( = Rs 10000 (1 + \frac{8}{100})^4 = a_4 \)
\( a_2 - a_1 = Rs 10000 (1 + \frac{8}{100})^2 - Rs 10000 (1 + \frac{8}{100}) \)
\( = Rs 10000 (1 + \frac{8}{100}) (1 + \frac{8}{100} - 1) \)
\( = 10000 (1 + \frac{8}{100}) (\frac{8}{100}) \)
\( a_3 - a_2 = Rs 10000 (1 + \frac{8}{100})^3 - Rs 10000 (1 + \frac{8}{100})^2 \)
\( = 10000 (1 + \frac{8}{100})^2 (1 + \frac{8}{100} - 1) \)
\( = 10000 (1 + \frac{8}{100})^2 (\frac{8}{100}) \)
Since \( a_3 - a_2 \neq a_2 - a_1 \). It does not form AP.

Question. If \( 7^{th} \) term of an A.P. is \( \frac{1}{9} \) and \( 9^{th} \) term is \( \frac{1}{7} \), find \( 63^{rd} \) term. 
Answer: Let the first term be \( a \) and the common difference be \( d \).
\( t_n = a + (n - 1)d \)
Given, \( t_7 = \frac{1}{9} \)
\( t_9 = \frac{1}{7} \)
\( a + 6d = \frac{1}{9} \) .....(i)
and \( a + 8d = \frac{1}{7} \) ....(ii)
On subtracting eqn.(i) from (ii)
\( a + 8d - a - 6d = \frac{1}{7} - \frac{1}{9} \)
or, \( 2d = \frac{2}{63} \)
or, \( d = \frac{1}{63} \)
Substituting the value of \( d \) in (ii) we get
\( a + 8 \times \frac{1}{63} = \frac{1}{7} \)
or, \( a = \frac{1}{7} - \frac{8}{63} \)
or, \( a = \frac{9-8}{63} = \frac{1}{63} \)
\( t_{63} = a + 62d \)
\( \therefore t_{63} = \frac{1}{63} + 62 \times \frac{1}{63} = \frac{1+62}{63} = \frac{63}{63} = 1 \)
Hence, \( t_{63} = 1 \).

Question. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149. 
Answer: Here, \( d = 7, a_{22} = 149 \)
Let the first term of the AP be \( a \).
We know that \( a_n = a + (n - 1)d \)

\( \implies a_{22} = a + (22 - 1)d \)

\( \implies a_{22} = a + 21d \)

\( \implies 149 = a + (21)(7) \)

\( \implies 149 = a + 147 \implies a = 2 \)
Again, we know that
\( S_n = \frac{n}{2} [2a + (n - 1)d] \)

\( \implies S_{22} = \frac{22}{2} [2(2) + (22 - 1)7] \)

\( \implies S_{22} = (11) [4 + 147] \)

\( \implies S_{22} = (11) (151) \implies S_{22} = 1661 \)
Hence, the sum of the first 22 terms of the AP is 1661.

Question. In an A.P. the first term is 8, nth term is 33 and the sum to first n terms is 123. Find n and the common difference(d). 
Answer: Given First term \( (a) = 8 \) and, \( n^{th} \) term \( (a_n) = 33 \)

\( \implies a + (n - 1)d = 33 \)

\( \implies 8 + (n - 1)d = 33 \)

\( \implies (n - 1)d = 33 - 8 \)

\( \implies (n - 1)d = 25 \) .....(i)
and, Sum of first n terms = 123

\( \implies \frac{n}{2} [a + a_n] = 123 \)

\( \implies \frac{n}{2} [8 + 33] = 123 \)

\( \implies \frac{n}{2} \times 41 = 123 \)

\( \implies n = \frac{123 \times 2}{41} \implies n = 6 \)
Put value of \( n \) in equation (i)
\( (6 - 1)d = 25 \)

\( \implies 5d = 25 \)

\( \implies d = \frac{25}{5} = 5 \)

Question. Each year, a tree grows 5 cm less than it did the preceding year. If it grew by 1 m in the first year, then in how many years will it have ceased growing? 
Answer: Given that, tree grows 5 cm less than preceding year, and grew by 1m (100 cm) in the first year.
means, 95cm in the 2nd year, 90 in the 3rd year, 85 in the fourth year and so on.
growth in the year in which it will stop growing will be 0cm.
Therefore, The following sequence can be formed.
100, 95, 90, …., 0 which is an AP.
Here, \( a = 100, d = 95 - 100 = -5 \) and \( l = 0 \)
Let \( l = a_n = a + (n - 1)d \)
Then, \( 0 = 100 + (n - 1)(-5) \)
\( 0 = 100 - 5n + 5 \)
\( 0 = 105 - 5n \)
\( 5n = 105 \implies n = 21 \)
Hence, Tree will ceased growing in 21 years.

Question. If the sum of the first m terms of an AP be n and the sum of its first n terms be m then show that the sum of its first (m + n) terms is -(m + n). 
Answer: Let \( a \) be the first term and \( d \) be the common difference of the given AP. Then,
\( S_m = n \implies \frac{m}{2} [2a + (m-1)d] = n \)

\( \implies 2am + m(m-1)d = 2n \) ...... (i)
And, \( S_n = m \implies \frac{n}{2} [2a + (n - 1)d] = m \)

\( \implies 2an + n(n-1)d = 2m \) ...... (ii)
On subtracting (ii) from (i), we get
\( 2a(m-n) + [ (m^2 - n^2) - (m - n) ]d = 2(n - m) \)

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

\( \implies 2a + (m + n - 1)d = -2 \) ..... (iii)
Sum of the first (m + n) terms of the given AP
\( = \frac{(m+n)}{2} \{2a + (m + n - 1)d\} \)
\( = \frac{(m+n)}{2} \cdot (-2) = -(m + n) \) [using (iii)].
Hence, the sum of first (m + n) terms of the given AP is -(m + n).

Question. Find the sum of all integers between 100 and 550, which are divisible by 9. 
Answer: According to the question,
All integers between 100 and 550, which are divisible by 9
= 108, 117, 126, ………., 549
First term (a) = 108
Common difference(d) = 117 - 108 = 9
Last term(\( a_n \)) = 549

\( \implies a + (n - 1)d = 549 \)

\( \implies 108 + (n - 1)(9) = 549 \)

\( \implies 108 + 9n - 9 = 549 \)

\( \implies 9n = 549 + 9 - 108 \)

\( \implies 9n = 450 \)

\( \implies n = \frac{450}{9} = 50 \)
Sum of 50 terms = \( \frac{n}{2} [a + a_n] \)
\( = \frac{50}{2} [108 + 549] \)
= 25 \(\times\) 657 = 16425

Question. Which term of the A.P. 121, 117, 113, ……… is its first negative term? 
(a) 32
(b) 33
(c) 30
(d) 31
Answer: (a) 32
Explanation: Here \( a = 121, d = 117 - 121 = -4 \)
For the first negative term, \( a_n < 0 \)

\( \implies 121 + (n - 1)(-4) < 0 \)

\( \implies 121 - 4n + 4 < 0 \)

\( \implies 125 - 4n < 0 \)

\( \implies n > \frac{125}{4} \)

\( \implies n > 31 \frac{1}{4} \)
Therefore 32nd term is the first negative term.

Question. The A.P. whose third term is 16 and the difference of 5th term from the 7th term is 12, then the A.P. is 
(a) 4, 11, 18, 25, ......
(b) 4, 14, 24, 34, .......
(c) 4, 10, 16, 22, .....
(d) 4, 6, 8, 10, ........
Answer: (c) 4, 10, 16, 22, .....
Explanation: Given: \( a_3 = 16 \implies a + 2d = 16 \dots \dots \dots \dots \)(i)
And \( a_7 - a_5 = 12 \implies a + 6d - a - 4d = 12 \)

\( \implies 2d = 12 \implies d = 6 \)
Putting value of \( d \) in eq. (i), we get
\( a + 2 \times 6 = 16 \)

\( \implies a = 4 \)
Therefore, A.P. is 4, 10, 16, 22, …………

Question. If \( a_1 = 4 \) and \( a_n = 4a_{n-1} + 3, n > 1 \), then the value of \( a_4 \) is 
(a) 320
(b) 329
(c) 319
(d) 300
Answer: (c) 319
Explanation: Given: \( a_1 = 4 \) and \( a_n = 4a_{n-1} + 3, n > 1 \),
\( \therefore a_2 = 4a_{2-1} + 3 = 4a_1 + 3 = 4 \times 4 + 3 = 16 + 3 = 19 \)
and \( a_3 = 4a_{3-1} + 3 = 4a_2 + 3 = 4 \times 19 + 3 = 76 + 3 = 79 \)
and \( a_4 = 4a_{4-1} + 3 = 4a_3 + 3 = 4 \times 79 + 3 = 316 + 3 = 319 \)

Question. In an A.P. it is given that a = 5, d = 3 and \( a_n = 50 \), then the value of ‘n’ is 
(a) 16
(b) 18
(c) 20
(d) 15
Answer: (a) 16
Explanation: Given: a = 5, d = 3 and \( a_n = 50 \)
\( \therefore a_n = a + (n - 1)d \)

\( \implies 50 = 5 + (n - 1) \times 3 \)

\( \implies 45 = (n - 1) \times 3 \)

\( \implies \frac{45}{3} = n - 1 \)

\( \implies n - 1 = 15 \)

\( \implies n = 16 \)

Question. If the second term of an AP is 13 and its fifth term is 25, then its 7th term is 
(a) 37
(b) 33
(c) 39
(d) 35
Answer: (b) 33
Explanation: Given: \( a_2 = 13 \)

\( \implies a + (2 - 1)d = 13 \)

\( \implies a + d = 13 \dots \dots \dots \dots \)(i)
And \( a_5 = 25 \)

\( \implies a + (5 - 1)d = 25 \)

\( \implies a + 4d = 25 \dots \dots \dots \dots \)(ii)
Solving eq. (i) and (ii), we get a = 9 and d = 4
\( \therefore a_7 = a + (7 - 1)d \)
= 9 + (7 – 1) \( \times \) 4
= 9 + 6 \( \times \) 4
= 9 + 24 = 33

Question. In the A.P. 2, x, 26 find the value of x. 
Answer: 2, x and 26 are in A.P.
\( \therefore x - 2 = 26 - x \)
or, \( x + x = 26 + 2 \)
or, 2x = 28
or, \( x = \frac{28}{2} = 14 \)

Question. Find \( 9^{th} \) term of the A.P. \( \frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \frac{9}{4}, \dots \) 
Answer: Given, A.P = \( \frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \frac{9}{4}, \dots \)
First term(a) = \( \frac{3}{4} \)
Common difference (d) = \( \frac{5}{4} - \frac{3}{4} = \frac{2}{4} \)
As we know,
\( a_n = a + (n - 1)d \)

\( \implies a_9 = \frac{3}{4} + (9 - 1) \times \frac{2}{4} \)
= \( \frac{3}{4} + 8 \times \frac{2}{4} \)
= \( \frac{3}{4} + \frac{16}{4} = \frac{19}{4} \)

Question. If they form an AP, find the common difference d and write three more terms. 0.2, 0.22, 0.222, 0.2222, .... 
Answer: \( a_2 - a_1 = 0.22 - 0.2 = 0.02 \)
\( a_3 - a_2 = 0.222 - 0.22 = 0.002 \)
As \( a_2 - a_1 \neq a_3 - a_2 \), the given list of numbers does not form an AP.

Question. If sum of first n terms of an AP is \( 2n^2 + 5n \). Then find \( S_{20} \). (1)
Answer: \( S_n = 2n^2 + 5n \)
\( S_{20} = 2(20)^2 + 5(20) \)
= 2(400) + 100 = 900.

Question. Find the sum of each of the following APs: 0.6, 1.7, 2.8,... to 100 terms. (1)
Answer: Here, a = 0.6, d = 1.7 – 0.6 = 1.1 and n = 100
Now we know that, \( S_n = \frac{n}{2} [2a + (n - 1)d] \)
Therefore, \( S_n = \frac{100}{2} [2 \times 0.6 + (100 - 1)(1.1)] \)
= 50[1.2 + 108.9]
= 50 \( \times \) 110.1
= 5505

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: We have,
\( a_4 = 0 \)
a + 3d = 0
3d = –a
or –3d = a \(\dots\dots\dots\dots\)(i)
Now,
\( a_{25} = a + 24d \)
= –3d + 24d [Putting value of a from eq(i)]
= 21d \(\dots\dots\dots\dots\dots\)(ii)
\( a_{11} = a + 10d \)
= –3d + 10d
= 7d \(\dots\dots\dots\dots\)(iii)
From eq(ii) and (iii), we get
\( a_{25} = 21d \)
\( a_{25} = 3(7d) \)
\( a_{25} = 3a_{11} \)
Hence Proved

Question. In a certain A.P. \( 32^{th} \) term is twice the \( 12^{th} \) term. Prove that \( 70^{th} \) term is twice the \( 31^{st} \) term. 
Answer: Let the 1st term be a and common difference be d.
\( a_n = a + (n-1)d \)
According to the question, \( a_{32} = 2a_{12} \)
a + 31d = 2 (a + 11d)
a + 31d = 2a + 22d
a - 2a = 22d - 31d
a = 9d
\( a_{70} = a + 69d = 9d + 69d = 78d \)
\( a_{31} = a + 30d = 9d + 30d = 39d \)
\( a_{70} = 78d \)
\( a_{70} = 2(39d) \)
\( a_{70} = 2a_{31} \)
\( a_{70} = 2a_{31} \) Hence Proved.

Question. The sum of first six terms of an arithmetic progression is 42. The ratio of its 10th term to its 30th term is 1 : 3. Calculate the first and the thirteenth term of the A.P. 
Answer: Let a be the first term and d be the common difference of the given A.P. Then,
\( \therefore S_6 = 42 \)
\( \frac{6}{2} (2a + 5d) = 42 \)
2a + 5d = 14 \(\dots\dots\dots\dots\)(i)
It is given that
\( a_{10} : a_{30} = 1 : 3 \)

\( \implies \frac{a+9d}{a+29d} = \frac{1}{3} \)
3a + 27d = a + 29d
2a = 2d
a = d \(\dots\dots\dots\dots\dots\) (ii)
From (i) and (ii) we get
a = d = 2
Hence \( a_{13} = 2 + 12 \times 2 = 26 \) and \( a_1 = 2 \)

Question. Find the sum of the first 25 terms of an A.P. whose \( n^{th} \) term is given by \( a_n = 7 - 3n \). 
Answer: Given, \( a_n = 7 - 3n \)
Put n = 1, \( a_1 = 7 - 3 \times 1 = 7 - 3 = 4 \)
Put n = 2, \( a_2 = 7 - 3 \times 2 = 7 - 6 = 1 \)
Common difference(d) = 1 - 4 = -3
\( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( S_{25} = \frac{25}{2} [2 \times 4 + (25 - 1)(-3)] \)
= \( \frac{25}{2} [8 - 72] \)
= \( \frac{25}{2} \times -64 \)
= -800

Question. The 12th term of an AP is -13 and the sum of its first four terms is 24. Find the sum of its first 10 terms. 
Answer: 12th term = \( T_{12} \) = -13
a + 11d = -13 \dots (i)
\( S_4 = 24 \)

\( \implies \frac{4}{2} [2a + 3d] = 24 \)

\( \implies 2[2a + 3d] = 24 \)

\( \implies 2a + 3d = 12 \dots \dots \)(ii)
Multiplying equation (i) by 2, we get
2a + 22d = -26 \dots (iii)
Subtracting (ii) from (iii), we get
19d = -38
d = -2
a + 11(-2) = -13 \dots [from (i)]
a = 9
\( \therefore \) Sum of first 10 terms, \( S_{10} = \frac{10}{2} [2(9) + 9(-2)] = 5[18 - 18] = 5 \times 0 = 0 \)

Question. Prove that the 11th term of an A.P. cannot be \( n^2 + 1 \). Justify your answer. 
Answer: Let \( n^{th} \) term of A.P.
\( a_n = n^2 + 1 \)
Putting the value of n = 1, 2 ,3 ,..... we get
\( a_1 = 1^2 + 1 = 2 \)
\( a_2 = 2^2 + 1 = 5 \)
\( a_3 = 3^2 + 1 = 10 \)
The obtained sequence = 2, 5, 10, 17……..
Its common difference
\( a_2 - a_1 \neq a_3 - a_2 \neq a_4 - a_3 \)
or, 5 - 2 \( \neq \) 10 - 5 \( \neq \) 17 - 10
\( \therefore 3 \neq 5 \neq 7 \)
Since the sequence has no common difference
Hence, \( n^2 + 1 \) is not a form of \( n^{th} \) term of an A.P.

Question. Let there be an A.P. with first term 'a', common difference 'd'. If \( a_n \) denotes its \( n^{th} \) term and \( S_n \) the sum of first n terms, find n and \( a_n \), if a = 2, d = 8 and \( S_n = 90 \). 
Answer: Given that, a = 2, d = 8 and \( S_n = 90 \).
As, \( S_n = \frac{n}{2} [2a + (n - 1)d] \)
90 = \( \frac{n}{2} [4 + (n - 1)8] \)
90 = \( n[2 + (n - 1)4] \)
90 = n[2 + 4n - 4]
90 = n[2 + 4n - 4]
90 = n(4n - 2) = \( 4n^2 - 2n \)
\( 4n^2 - 2n - 90 = 0 \)
\( 4n^2 - 20n + 18n - 90 = 0 \)
4n(n - 5) + 18(n - 5) = 0
(n - 5)(4n + 18) = 0
Either n = 5 or \( n = \frac{18}{4} = \frac{-9}{2} \)
However, n can neither be negative nor fractional.
Therefore, n = 5
\( a_n = a + (n - 1)d \)
\( a_5 = 2 + (5 - 1)8 \)
= 2 + 4(8)
= 2 + 32 = 34

Question. Find the sum of all natural numbers between 200 and 300 which are divisible by 4. 
Answer: All the numbers between 200 and 300 which are divisible by 4 are 204, 208, 212, 216, ....., 296
Here, \( a_1 = 204, a_2 = 208, a_3 = 212, a_4 = 216 \)
\( \therefore a_2 - a_1 = 208 - 204 = 4 \)
\( a_3 - a_2 = 212 - 208 = 4 \)
\( a_4 - a_3 = 216 - 212 = 4 \)
\( \because a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = \dots (= 4 \text{ each}) \)
\( \therefore \) This sequence is an arithmetic progression whose common difference is 4.
Here, a = 204, d = 4, l = 296
Let the number of terms be n. Then,
l = a + (n - 1)d

\( \implies 296 = 204 + (n - 1)4 \)

\( \implies 92 = (n - 1)4 \)

\( \implies n - 1 = 23 \)

\( \implies n = 23 + 1 \)

\( \implies n = 24 \)
\( \therefore S_n = \frac{n}{2} (a + l) = \left( \frac{24}{2} \right) (204 + 296) \)
= (12) (500)
= 6000.
Hence, the sum of all the natural numbers between 200 and 300 which are divisible by 4 is 6000.

Question. In a garden bed, there are 23 rose plants in the first row, 21 are in the \( 2^{nd} \), 19 in \( 3^{rd} \) row and so on. There are 5 plants in the last row. How many rows are there of rose plants? Also find the total number of rose plants in the garden. 
Answer: The number of rose plants in the \( 1^{st}, 2^{nd}, \dots \) are 23, 21, 19, ……….. 5
a = 23, d = 21 - 23 = - 2 , \( a_n = 5 \)
\( \therefore a_n = a + (n - 1 )d \)
or, 5 = 23 + (n - 1)(-2)
or, 5 = 23 - 2n + 2
or, 5 = 25 - 2n
or, 2n = 20
or, n = 10
Total number of rose plants in the flower bed,
\( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( S_{10} = \frac{10}{2} [2(23) + (10 - 1)(-2)] \)
= 5[46 – 20 + 2]
\( S_{10} = 5(46 - 18) \)
= 5 (28)
\( S_{10} = 140 \)