CBSE Class 10 Mathematics Arithmetic Progressions VBQs Set 13

Short Answer Type Questions

Question. Justify whether it is true to say that the following are the \( n^{th} \) terms of an AP: (i) \( 2n – 3 \), (ii) \( 3n^2 + 5 \), (iii) \( 1 + n + n^2 \)
Answer: (i) Yes, \( (2n – 3) \) is the \( n^{th} \) term of an A.P.
It is given that \( a_n = 2n – 3 \)
Put \( n = 1, a_1 = 2(1) – 3 = 2 – 3 = –1 \)
\( n = 2, a_2 = 2(2) – 3 = 4 – 3 = 1 \)
\( n = 3, a_3 = 2(3) – 3 = 6 – 3 = 3 \)
\( n = 4, a_4 = 2(4) – 3 = 8 – 3 = 5 \)
List of numbers becomes –1, 1, 3, 5...
Here, \( a_2 – a_1 = 1 – (–1) = 2 \)
\( a_3 – a_2 = 3 – 1 = 2 \)
\( a_4 – a_3 = 5 – 3 = 2 \)
Clearly, \( a_2 – a_1 = a_3 – a_2 = a_4 – a_3 \)
Hence, \( 2n – 3 \) is the \( n^{th} \) term of an AP.
(ii) No, \( (3n^2 + 5) \) is not the \( n^{th} \) term of an AP.
It is given that \( a_n = 3n^2 + 5 \)
Put \( n = 1, a_1 = 3(1)^2 + 5 = 3 + 5 = 8 \)
\( n = 2, a_2 = 3(2)^2 + 5 = 12 + 5 = 17 \)
\( n = 3, a_3 = 3(3)^2 + 5 = 27 + 5 = 32 \)
List of number becomes 8, 17, 32 ...
Here, \( a_2 – a_1 = 17 – 8 = 9 \), \( a_3 – a_2 = 32 – 17 = 15 \)
Clearly, \( a_2 – a_1 \neq a_3 – a_2 \)
Hence, \( (3n^2 + 5) \) is not the \( n^{th} \) term of an AP.
(iii) No, \( (1 + n + n^2) \) is not the \( n^{th} \) term of an AP.
It is given that \( a_n = 1 + n + n^2 \)
Put \( n = 1, a_1 = 1 + (1) + (1)^2 = 3 \)
\( n = 2, a_2 = 1 + (2) + (2)^2 = 7 \)
\( n = 3, a_3 = 1 + (3) + (3)^2 = 13 \)
List of number becomes 3, 7, 13 ...
Here, \( a_2 – a_1 = 7 – 3 = 4 \), \( a_3 – a_2 = 13 – 7 = 6 \)
Clearly, \( a_2 – a_1 \neq a_3 – a_2 \)
Hence, \( (1 + n + n^2) \) is not the \( n^{th} \) term of an AP.

Question. Find a, b and c such that the following numbers in AP: a, 7, b, 23, c. 
Answer: It is given that a, 7, b, 23, c are in AP
They have a common difference
i.e., \( 7 – a = b – 7 = 23 – b = c – 23 \)
Taking second and third terms, we get
\( b – 7 = 23 – b \) \( \implies 2b = 30 \) \( \implies b = 15 \)
Taking first and second terms, we get
\( 7 – a = b – 7 \)
\( \implies 7 – a = 15 – 7 \) [As \( b = 15 \)]
\( \implies 7 – a = 8 \) \( \implies a = –1 \)
Taking third and fourth terms, we get
\( 23 – b = c – 23 \)
\( \implies 23 – 15 = c – 23 \) [As \( b = 15 \)]
\( \implies 8 = c – 23 \) \( \implies c = 31 \)
Hence, \( a = –1, b = 15 \) and \( c = 31 \).

Question. Determine the AP whose 5th term is 19 and the difference of the 8th term from the 13th term is 20.
Answer: Let the first term of an AP be a and common difference be d.
It is given that \( a_5 = 19 \) and \( a_{13} – a_8 = 20 \)
We know that \( a_n = a + (n – 1) d \)
\( a_5 = a + (5 – 1) d = a + 4d = 19 \) ...(i)
Also, \( a_{13} – a_8 = 20 \)
\( \implies (a + 12d) – (a + 7d) = 20 \)
\( \implies a + 12d – a – 7d = 20 \)
\( \implies 5d = 20 \) \( \implies d = 4 \)
Putting the value of \( d = 4 \) in equation (i), we get
\( a + 4d = 19 \)
\( \implies a + 4(4) = 19 \)
\( \implies a + 16 = 19 \)
\( \implies a = 3 \)
So, the required AP will be:
a, a + d, a + 2d, a + 3d ...
i.e., 3, 3 + 4, 3 + 2(4), 3 + 3(4) ...
i.e., 3, 7, 11, 15 ...

Question. The sum of the first 30 terms of an A.P. is 1920. If the fourth term is 18, find its 11th term. 
Answer: Let a be the first term and d, the common difference of the AP
Here, number of terms, \( n = 30 \)
Then, \( S_{30} = \frac{30}{2} [2a + 29d] = 1920 \)
or \( 2a + 29d = 128 \) ...(i)
Also, \( a_4 = a + 3d = 18 \) ...(ii)
On solving equations (i) and (ii), we get
\( d = 4 \) and \( a = 6 \)
Thus, \( a_{11} = a + 10d = 6 + 40 = 46 \).
Hence, 11th term of the AP is 46.

Question. Which term of the A.P. 20, \( 19\frac{1}{4} \), \( 18\frac{1}{2} \), \( 17\frac{3}{4} \), .... is the first negative term.
Answer: Here, first term \( a = 20 \) and common difference \( d = -\frac{3}{4} \)
Let \( a_n \) be the first negative term, i.e.,
\( a + (n – 1)d < 0 \)
\( \implies 20 - \frac{3}{4}(n - 1) < 0 \)
or \( 20 - \frac{3n}{4} + \frac{3}{4} < 0 \)
or \( \frac{3n}{4} > \frac{83}{4} \)
or \( n > \frac{83}{3} \)
So, the nth term is 28.

Question. Find the sum of the two middle most terms of the AP: \( -\frac{4}{3}, -1, -\frac{2}{3}, \dots, 4\frac{1}{3} \).
Answer: The given AP is \( -\frac{4}{3}, -1, -\frac{2}{3}, \dots, \frac{13}{3} \)
Here, first term, \( a = -\frac{4}{3} \)
Common difference, \( d = -1 - (-\frac{4}{3}) = -1 + \frac{4}{3} = \frac{1}{3} \)
Last term, \( L = 4\frac{1}{3} = \frac{13}{3} \)
We know that \( a_n = a + (n – 1)d \)
If \( a_n \) is the last term, then
\( L = a + (n – 1)d \)
\( \frac{13}{3} = -\frac{4}{3} + (n-1)(\frac{1}{3}) \)
\( \implies 13 = –4 + (n – 1) \)
\( \implies (n – 1) = 17 \)
\( \implies n = 18 \)
So, the two middle most terms are \( (\frac{n}{2})^{th} \) and \( (\frac{n}{2}+1)^{th} \) as numbers of terms are even.
The two middle most terms are \( (\frac{18}{2})^{th} \) and \( (\frac{18}{2}+1)^{th} \) i.e., 9th and 10th term
\( a_9 = a + (9 – 1)d = a + 8d \)
\( = -\frac{4}{3} + 8(\frac{1}{3}) = \frac{-4 + 8}{3} = \frac{4}{3} \)
\( a_{10} = a + (10 – 1)d = a + 9d \)
\( = -\frac{4}{3} + 9(\frac{1}{3}) = \frac{-4 + 9}{3} = \frac{5}{3} \)
So, the sum of the two middle most terms \( = a_9 + a_{10} \)
\( = \frac{4}{3} + \frac{5}{3} = \frac{9}{3} = 3 \)

Question. Show that the sum of all terms of an A.P. whose first term is a, the second term is b and the last term is c is equal to \( \frac{(a + c)(b + c - 2a)}{2(b - a)} \). 
Answer: Here, first term is a, and, common difference \( = b – a = d \) and last term is c.
Let A.P. contains ‘n’ terms. Then, \( a_n = c \), i.e., \( a + (n – 1)(b – a) = c \) or \( n = \frac{b + c - 2a}{b - a} \)
Now \( S_n = \frac{n}{2}[a + c] \) [Since c is last term]
\( = \frac{b + c - 2a}{2(b - a)} [a + c] \)
\( = \frac{(a + c)(b + c - 2a)}{2(b - a)} \) Hence, proved.

Question. The first term of an AP is –5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference. 
Answer: Let a be the first term, d be the common difference and n be the number of terms.
It is given that:
first term, \( a = –5 \)
Last term, \( L = 45 \)
We know that, if the last term of an AP is known, then the sum of n terms of an AP is
\( S_n = \frac{n}{2}(a + l) \)
\( 120 = \frac{n}{2}(-5 + 45) \)
\( 120 \times 2 = 40 \times n \)
\( \implies n = 6 \)
\( L = a + (n – 1)d \)
\( 45 = –5 + (6 – 1)d \)
\( \implies 50 = 5d \implies d = 10 \)
Hence, number of terms = 6
and common difference = 10

Question. If \( S_n \) denotes the sum of first n terms of an AP, prove that \( S_{12} = 3(S_8 – S_4) \) 
Answer: We know that sum of n terms of an AP, \( S_n = \frac{n}{2}[2a + ( n – 1) d] \)
\( S_4 = \frac{4}{2} [2a + (4 – 1)d] = 2 [2a + 3d] = 4a + 6d \)
\( S_8 = \frac{8}{2} [2a + (8 – 1)d] = 4 [2a + 7d] = 8a + 28d \)
\( (S_8 – S_4) = (8a + 28d) – (4a + 6d) \)
\( = 8a + 28d – 4a – 6d \)
\( = 4a + 22d \) ...(i)
\( S_{12} = \frac{12}{2} [2a + (12 – 1)d] \)
\( = 6 [2a + 11d] = 12a + 66d \)
\( = 3 [4a + 22d] \)
\( \implies S_{12} = 3 [S_8 – S_4] \) [using equation (i)]
Hence proved.

Question. If the sum of the first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of the first 10 terms. 
Answer: It is given that \( S_6 = 36 \) and \( S_{16} = 256 \)
Let a be the first term and d be the common difference of an AP.
We know that sum of n terms of an AP, \( S_n = \frac{n}{2} [2a + (n – 1)d] \)
Now \( S_6 = 36 \) (Given)
\( \implies \frac{6}{2} [2a + (6 – 1)d] = 36 \)
\( \implies 3 [2a + 5d] = 36 \)
\( \implies 2a + 5d = 12 \) ...(i)
Also, \( S_{16} = 256 \)
\( \implies \frac{16}{2} [2a + (16 – 1)d] = 256 \)
\( \implies 8[2a + 15d] = 256 \)
\( \implies 2a + 15d = 32 \) ...(ii)
Subtracting equation (i) from equation (ii), we get
\( (2a + 15d) – (2a + 5d) = 32 – 12 \)
\( \implies 2a + 15d – 2a – 5d = 20 \)
\( \implies 10d = 20 \)
\( \implies d = 2 \)
Putting the value of d in equation (i), we get
\( 2a + 5d = 12 \)
\( 2a + 5(2) = 12 \)
\( \implies 2a + 10 = 12 \)
\( \implies 2a = 2 \implies a = 1 \)
We have to find sum of first 10 terms, \( S_{10} = ? \)
\( S_{10} = \frac{10}{2} [2a + (10 – 1)d] \)
\( = 5 [2(1) + 9(2)] \)
\( = 5 [2 + 18] = 5 \times 20 = 100 \)
Hence, the required sum of the first 10 terms is 100.

Question. The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is –30 and the common difference is 8. Find n.
Answer: It is given that
first term of first AP, \( a = 8 \)
and common difference of first AP, \( d = 20 \)
Let n be the number of terms in first AP.
We know that sum of first n terms of an AP,
\( S_n = \frac{n}{2} [2a + (n – 1)d] \)
\( = \frac{n}{2} [2 \times 8 + (n – 1)20] \)
\( = \frac{n}{2} [16 + 20n – 20] \)
\( = \frac{n}{2} [20n – 4] = n[10n – 2] \)
\( \implies S_n = n[10n – 2] \) ...(i)
Now, first term of second AP \( (a’) = –30 \)
Common difference of second \( (d’) = 8 \)
\( \therefore \) Sum of first 2n terms of second AP
\( S_{2n} = \frac{2n}{2} [2a’ + (2n – 1)d’] \)
\( = n[2(–30) + (2n –1)8] \)
\( = n[–60 + 16n – 8] \)
\( S_{2n} = n[16n – 68] \) ...(ii)
By given condition,
Sum of first n terms of first AP = sum of first 2n terms of second AP
\( \implies S_n = S_{2n} \)
Using equation (i) and equation (ii), we get
\( n[10n – 2] = n[16n – 68] \)
\( \implies 10n – 2 – 16n + 68 = 0 \)
\( \implies –6n + 66 = 0 \)
\( \implies –6(n – 11) = 0 \)
\( \implies n = 11 \)
Hence, the required value of n is 11.

Question. If \( m^{th} \) term of an A.P. is \( \frac{1}{n} \) and \( n^{th} \) term is \( \frac{1}{m} \), find the sum of its first mn terms. 
Answer: Let, the first term of an A. P. be ‘a’ and its common difference be ‘d’.
Now, \( a_m = a + (m – 1)d \)
\( \implies \frac{1}{n} = a + (m – 1)d \) ...(i) (given)
and \( a_n = a + (n – 1)d \)
\( \implies \frac{1}{m} = a + (n – 1)d \) ...(ii) (given)
On subtracting equation (ii) from (i), we get:
\( \frac{1}{n} - \frac{1}{m} = [(m – 1) – (n – 1)]d \)
\( \implies \frac{m - n}{mn} = (m – n)d \)
\( \implies d = \frac{1}{mn} \) ...(iii)
If we put the value of d in equation (i), we get
\( \frac{1}{n} = a + (m – 1) \times \frac{1}{mn} \)
\( \implies a = \frac{1}{n} - (\frac{m - 1}{mn}) = \frac{m - m + 1}{mn} = \frac{1}{mn} \) ...(iv)
Sum of mn terms,
\( S_{mn} = \frac{mn}{2} [2 \times \frac{1}{mn} + (mn – 1) \times \frac{1}{mn}] \)
[ \( \because S_n = \frac{n}{2} (2a + (n – 1)d] \)
\( S_{mn} = \frac{1}{2} [2 + mn – 1] = \frac{mn + 1}{2} \)
Hence, the sum of (mn) terms is \( \frac{mn + 1}{2} \).

Question. Find the sum of n terms of the series \( (4 - \frac{1}{n}) + (4 - \frac{2}{n}) + (4 - \frac{3}{n}) + \dots \) 
Answer: Given series is \( (4 - \frac{1}{n}) + (4 - \frac{2}{n}) + (4 - \frac{3}{n}) + \dots n \) terms.
\( = (4 + 4 + 4 + \dots n \) terms) \( – (\frac{1}{n} + \frac{2}{n} + \frac{3}{n} + \dots n \) terms)
\( = 4(1 + 1 + 1 + \dots \) upto n terms) \( – \frac{1}{n} (1 + 2 + 3 + \dots \) upto n terms)
\( = 4n - \frac{1}{n} \times \frac{n(n + 1)}{2} \)
\( = 4n - \frac{n + 1}{2} \)
\( = \frac{8n - n - 1}{2} = \frac{7n - 1}{2} \)
Hence, the sum of the series is \( \frac{7n - 1}{2} \).

Question. For what value of n are the nth terms of two A.P.’s 63, 65, 67, .... and 3, 10, 17, ..... equal ? 
Answer: Given, APs are 63, 65, 67....... and 3, 10, 17........
For first A.P., first term, \( a = 63 \) and common difference, \( d = 2 \)
Then, its \( n^{th} \) term = \( 63 + (n – 1) \times 2 \) ...(i)
For second A.P., first term, \( a’ = 3 \) and common difference, \( d’ = 7 \)
Then, its \( n^{th} \) term = \( 3 + (n – 1) \times 7 \)
According to the question:
\( 63 + (n – 1) \times 2 = 3 + (n – 1) \times 7 \)
\( (n – 1) \times 5 = 60 \)
\( \implies n – 1 = 12 \)
\( \implies n = 13 \)
The 13th term of both given APs are equal.

Question. Find the sum of the first 40 positive integers which give a remainder 1 when divided by 6. 
Answer: The numbers which when divided by 6 gives 1 as remainder are 7, 13, 19, 25, 31, 37, ...
They form an A.P., as their common difference is the same, \( d = 6 \)
Here first term, \( a = 7 \), common difference, \( d = 6 \)
Sum of first 40 positive integers:
\( \therefore S_{40} = \frac{n}{2} [2a + (n – 1)d] \)
\( = \frac{40}{2} [2 \times 7 + (40 – 1) \times 6] \)
\( = 20[14 + 39 \times 6] \)
\( = 20 \times 248 = 4,960 \)
Hence, the sum of the first 40 positive integers is 4,960.

Question. Divide 56 in four parts in A.P such that the ratio of the product of their extremes (1st and 4th) to the product of means (2nd and 3rd) is 5 : 6. 
Answer: Let the four parts of the A.P. are \( (a – 3d) \), \( (a – d) \), \( (a + d) \), \( (a + 3d) \)
Now, \( (a – 3d) + (a – d) + (a + d) + (a + 3d) = 56 \)
\( \implies 4a = 56 \)
\( \implies a = 14 \)
According to question,
\( \frac{(a-3d)(a+3d)}{(a-d)(a+d)} = \frac{5}{6} \)
\( \implies \frac{(14-3d)(14+3d)}{(14-d)(14+d)} = \frac{5}{6} \) [Putting \( a = 14 \)]
\( \implies \frac{196-9d^2}{196-d^2} = \frac{5}{6} \)
\( \implies 1176 – 54d^2 = 980 – 5d^2 \)
\( \implies 49d^2 = 196 \)
\( \implies d^2 = 4 \implies d = \pm 2 \)
when, \( a = 14 \) and \( d = 2 \)
The 4 parts are \( (a – 3d), (a – d), (a + d), (a + 3d) \), i.e., 8, 12, 16, 20.
when, \( a = 14 \) and \( d = –2 \)
The 4 parts are 20, 16, 12 and 8.

Question. If the sum of the first 7 terms of an A.P. is 49 and that of its first 17 terms is 289, find the sum of first n terms of the A.P. 
Answer: Let the first term of an A.P be ‘a’ and its common difference be ‘d’.
Given \( S_7 = 49 \) and \( S_{17} = 289 \)
Then, \( S_7 = \frac{7}{2} [2a + (7 – 1) \times d] \)
\( \implies 49 = \frac{7}{2} [2a + 6d] \)
\( \implies 7 = a + 3d \) ...(i)
and \( S_{17} = \frac{17}{2} [2a + (17 – 1) \times d] \)
\( \implies 289 = \frac{17}{2} \times [2a + 16 \times d] \)
\( \implies 17 = a + 8d \) ...(ii)
On subtracting equation (i) from equation (ii), we get
\( (a + 8d) - (a + 3d) = 17 - 7 \)
\( \implies 5d = 10 \implies d = 2 \)
If we put the value of ‘d’ in equation (i), we get
\( a + 3 \times (2) = 7 \)
\( \implies a = 7 – 6 = 1 \)
\( \therefore \) Sum of the first ‘n’ terms,
\( S_n = \frac{n}{2} [2a + (n – 1)d] \)
\( = \frac{n}{2} [2 + (n – 1) \times 2] \)
\( = \frac{n}{2} [2 + 2n – 2] \)
\( = n^2 \)
Hence, the sum of the first ‘n’ terms is \( n^2 \).

Question. If the sum of first m terms of an AP is the same as the sum of its first n terms, show that the sum of its first (m + n) terms is zero. 
Answer: \( S_m = S_n \)
\( \implies \frac{m}{2} [2a + (m - 1)d] = \frac{n}{2} [2a + (n - 1)d] \)
\( \implies 2a(m - n) + d(m^2 - m - n^2 + n) = 0 \)
\( \implies (m - n)[2a + (m + n - 1)d] = 0 \)
\( \implies 2a + (m + n - 1)d = 0 \)
\( \implies S_{m+n} = \frac{m+n}{2} [2a + (m+n-1)d] = 0 \)
or \( S_{m+n} = 0 \)

Question. Among the natural numbers 1 to 49, find a number x, such that the sum of numbers preceding it is equal to sum of numbers succeeding it. 
Answer: Let, the number be x.
1, 2, 3, 4, ..., x – 1, x, x + 1, ... 49
1 + 2 + 3 + 4 + ... + x – 1 = (x + 1) + ... + 49
\( S_{x – 1} = S_{49} – S_x \)
\( \implies \frac{(x – 1)x}{2} = \frac{49 \times 50}{2} - \frac{x(x+1)}{2} \)
\( \implies x^2 – x = 2450 – x^2 – x \)
\( \implies x^2 = 1225 \)
\( \implies x = \pm 35 \) (–35 is not between 1 to 49 therefore rejected)
\( x = 35 \).

Question. The 14th term of an AP is twice its 8th term. If its 6th term is –8, then find the sum of its first 20 terms. 
Answer: \( a_{14} = 2a_8 \)
\( \implies a + 13d = 2(a + 7d) \implies a = –d \)
\( a_6 = –8 \)
\( \implies a + 5d = –8 \)
solving to get \( a = 2, d = –2 \)
\( S_{20} = 10(2a + 19d) \)
\( = 10(4 – 38) \)
\( = –340 \)

Question. The digits of a positive number of three digits are in A.P. and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number. 
Answer: Let the required numbers in A.P. are \( (a – d) \), a, \( (a + d) \) respectively.
Now, \( (a – d) + a + (a + d) = 15 \)
\( 3a = 15 \implies a = 5 \)
According to question, number is
\( 100(a – d) + 10a + a + d = 111a – 99d \)
Number on reversing the digits is
\( 100(a + d) + 10a + a – d = 111a + 99d \)
Now, as per given condition in question,
\( (111a – 99d) – (111a + 99d) = 594 \)
\( \implies –198d = 594 \implies d = –3 \)
So, digits of number are \( [5 – (–3)], 5, [5 + (–3)] = 8, 5, 2 \)
Required number is \( 111 \times (5) – 99 \times (– 3) \)
\( = 555 + 297 = 852 \)
The number is 852.
 

Long Answer Type Questions

Question. The \( 26^{th} \), \( 11^{th} \) and the last term of an AP are 0, 3 and \( -\frac{1}{5} \) respectively. Find the common difference and the number of terms. 
Answer: It is given that:
\( 26^{th} \) term of AP, \( a_{26} = 0 \)
\( 11^{th} \) term of AP, \( a_{11} = 3 \)
Last term, \( L = -\frac{1}{5} \)
Let the AP contain \( n \) term, last term (\( L \)) is the \( n^{th} \) term.
Let first term be \( a \) and common difference be \( d \) of an AP.
We know that \( a_n = a + (n - 1)d \)
Also, \( a_{26} = 0 \) [Given]
\( \implies a_{26} = a + (26 - 1)d \)
\( \implies 0 = a + 25d \)
\( \implies a + 25d = 0 \) ...(i)
Also, \( a_{11} = 3 \)
\( \implies a + (11 - 1) d = 3 \)
\( \implies a + 10d = 3 \) ...(ii)
Last term \( L = -\frac{1}{5} \)
\( \implies a + (n - 1) d = -\frac{1}{5} \) ...(iii)
Subtracting equation (ii) from equation (i), we get
\( (a + 25d) - (a + 10d) = 0 - 3 \)
\( \implies a + 25d - a - 10d = 0 - 3 \)
\( \implies 15d = -3 \)
\( \implies d = \frac{-3}{15} = \frac{-1}{5} \)
Putting the value of \( d \) in equation (i)
\( \implies a + 25 \left( \frac{-1}{5} \right) = 0 \)
\( \implies a - 5 = 0 \)
\( \implies a = 5 \)
Putting the value of \( a \) and \( d \) in equation (iii)
\( \implies a + (n - 1) d = -\frac{1}{5} \)
\( \implies 5 + (n - 1) \left( -\frac{1}{5} \right) = -\frac{1}{5} \)
\( \implies 25 - (n - 1) = -1 \)
\( \implies 25 + 1 = (n - 1) \)
\( \implies n = 27 \)
Hence, the common difference = \( -\frac{1}{5} \) and number of terms = 27.

Question. Find the sum of the following series : \( 5 + (-41) + 9 + (-39) + 13 + (-37) + 17 + .... + (-5) + 81 + (-3) \) 
Answer: Given: series is
\( 5 + (-41) + 9 + (-39) + 13 + (-37) + 17 + ... + (-5) + 81 + (-3) \)
\( = [5 + 9 + 13 + ... + 81] + [(-41) + (-39) + (-37) + ... + (-5) + (-3)] \)
Here are the two series:
1. \( 5 + 9 + 13 + ... + 81 \)
2. \( (-41) + (-39) + (-37) + ... + (-3) \)
For the first series:
first term \( a = 5 \), common difference, \( d = 9 - 5 = 4 \)
last term, \( l(a_n) = 81 \)
Then, \( a_n = a + (n - 1)d \) [where '\( n \)' is the number of terms]
\( 81 = 5 + (n - 1) \times 4 \)
\( \implies (n - 1) = \frac{76}{4} = 19 \)
\( \implies n = 20 \)
Sum of the first series, \( S_{20} \)
\( S_{20} = \frac{20}{2} [2 \times 5 + (20 - 1) \times 4] \)
\( = 10(10 + 19 \times 4) \)
\( = 10(10 + 76) \)
\( = 860 \)
For the second series:
first term, \( a' = -41 \),
common difference, \( d' = (-39) - (-41) = 2 \)
last term, \( l'(a_{n'}) = -3 \)
Then, \( a_{n'} = a' + (n' - 1)d' \) [where \( n' \) are the number of terms]
\( \implies -3 = -41 + (n' - 1) \times 2 \)
\( \implies (n' - 1) = 19 \)
\( \implies n' = 20 \)
Sum of the second series, \( S'_{20} = \frac{n'}{2} [2a' + (n' - 1)d'] \)
\( = \frac{20}{2} [2 \times (-41) + (19) \times 2] \)
\( = 10[-82 + 38] \)
\( = 10 \times (-44) \)
\( = -440 \)
\( \therefore \text{Total sum} = S_{20} + S'_{20} \)
\( = 860 - 440 = 420 \)
Hence, the total sum of series is 420.

Question. The sum of four consecutive number in AP is 32 and the ratio of the product of the first and last terms to the product of two middle term is 7 : 15. Find the numbers. 
Answer: Let '\( a \)' be the first term and '\( d \)' be the common difference to the AP. Then,
\( a - 3d, a - d, a + d, a + 3d \),
are four consecutive terms of the AP.
As per the question,
\( (a - 3d) + (a - d) + (a + d) + (a + 3d) = 32 \)
\( \implies 4a = 32 \), or \( a = 8 \) ...(i)
and \( \frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{7}{15} \)
\( \implies \frac{a^2 - 9d^2}{a^2 - d^2} = \frac{7}{15} \)
\( \implies 15a^2 - 135d^2 = 7a^2 - 7d^2 \)
\( \implies 8a^2 = 128 d^2 \)
Using (i), we have:
\( 8 \times 8^2 = 128 d^2 \)
\( \implies d^2 = 4 \), or \( d = \pm 2 \)
Thus, the four numbers are 2, 6, 10 and 14.

Question. Solve : \( 1 + 4 + 7 + 10 + .... + x = 287 \). 
Answer: In the given AP, \( a = 1 \) and \( d = 3 \).
Let AP contains '\( n \)' terms. Then, \( a_n = x \)
\( \implies a + (n - 1)d = x \)
\( \implies 1 + 3(n - 1) = x \)
\( \implies n = \frac{x + 2}{3} \)
Further, \( S_n = \frac{n}{2} [\text{first term} + \text{last term}] \)
\( \implies 287 = \frac{x + 2}{3 \times 2} [1 + x] \)
\( \implies (x + 1) (x + 2) = 1722 \)
or \( x^2 + 3x - 1720 = 0 \)
\( \implies x^2 + 43x - 40x - 1720 = 0 \)
\( \implies x(x + 43) - 40(x + 43) = 0 \)
\( \implies (x + 43) (x - 40) = 0 \)
\( \implies x - 40 = 0 \) (\( \because x + 43 \neq 0 \))
\( \implies x = 40 \).
Thus, \( x = 40 \).

Question. The sum of the first 5 terms of an AP and the sum of the first 7 terms of the same AP is 167. If the sum of the first 10 terms of this AP is 235, find the sum of its first 20 terms. 
Answer: Let \( a \) be the first term, \( d \) be the common difference and \( n \) be the number of terms of an AP.
We know that \( S_n = \frac{n}{2} [2a + (n - 1)d] \)
Sum of first five terms, \( S_5 \)
\( S_5 = \frac{5}{2} [2a + (5 - 1)d] = \frac{5}{2} [2a + 4d] \)
\( S_5 = 5[a + 2d] \)
\( S_5 = 5a + 10d \) ...(i)
Sum of first seven terms, \( S_7 \)
\( S_7 = \frac{7}{2} [2a + (7 - 1)d] = \frac{7}{2} [2a + 6d] \)
\( S_7 = 7[a + 3d] \)
\( S_7 = 7a + 21d \) ...(ii)
Now, by given condition
\( S_5 + S_7 = 167 \)
\( \implies 5a + 10d + 7a + 21d = 167 \) [using equation (i) & equation (ii)]
\( \implies 12a + 31d = 167 \) ...(iii)
Also, it is given that sum of first 10 terms of this AP is 235.
\( \implies S_{10} = 235 \)
\( \implies \frac{10}{2} [2a + (10 - 1)d] = 235 \)
\( \implies 5[2a + 9d] = 235 \)
\( \implies 2a + 9d = \frac{235}{5} = 47 \)
\( \implies 2a + 9d = 47 \) ...(iv)
Multiplying equation (iv) by 6 and subtracting it from equation (iii), we get
\( (12a + 31d) - (12a + 54d) = 167 - 282 \)
\( \implies -23d = -115 \implies d = 5 \)
Putting the value of \( d \) in equation (iv), we get
\( 2a + 9(5) = 47 \)
\( \implies 2a + 45 = 47 \)
\( \implies 2a = 2 \implies a = 1 \)
Sum of first 20 terms of this AP,
\( S_{20} = \frac{20}{2} [2a + (20 - 1)d] \)
\( = 10[2(1) + 19(5)] \)
\( = 10[2 + 95] \)
\( = 10 \times 97 = 970 \)
Hence, the required sum of first 20 terms is 970.

Question. Find the:
(A) sum of those integers between 1 and 500 which are multiples of 2 as well as of 5. 
(B) sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.
(C) sum of those integers from 1 to 500 which are multiples of 2 or 5.
[Hint (iii): These numbers will be: multiple of 2 + multiple of 5 – multiples of 2 as well as of 5] 
Answer: (A) Multiples of 2 and 5 will be multiples of LCM of 2 and 5. LCM of (2, 5) = 10
Multiples of 2 as well as 5 between 1 and 500 is 10, 20, 30, 40, ...490
This forms an AP with first term, \( a = 10 \)
Common difference, \( d = 20 - 10 = 10 \)
Last term, \( L = 490 \)
We know that sum of \( n \) terms between 1 and 500 is \( S_n = \frac{n}{2} [a + L] \) ...(i)
Also, \( L = a + (n - 1) d \)
\( \implies 490 = 10 + (n - 1)10 \)
\( \implies 480 = (n - 1)10 \)
\( \implies (n - 1) = 48 \)
\( \implies n = 49 \)
Putting this value in equation (i), we get
\( S_{49} = \frac{49}{2} [10 + 490] \)
\( = \frac{49}{2} \times 500 = 49 \times 250 = 12250 \)
\( \implies S_{49} = 12250 \)
(B) Here, multiples of 2 as well as 5 from 1 to 500 are 10, 20, 30, ...500
Here, first term, \( a = 10 \)
common difference, \( d = 20 - 10 = 10 \)
Last term, \( L = 500 \)
We know that \( a_n = a + (n - 1)d \)
\( L = a + (n - 1)d \) [Where, \( n \) is total no. of terms]
\( 500 = 10 + (n - 1)10 \)
\( 490 = (n - 1)10 \)
\( \implies (n - 1) = 49 \implies n = 50 \)
Also we know that \( S_n = \frac{n}{2} (a + L) \)
\( \implies S_{50} = \frac{50}{2} (10 + 500) = 25 \times 510 = 12750 \)
Hence, \( S_{50} = 12750 \)
(C) Multiples of 2 or 5 = Multiples of 2 + multiples of 5 – [Multiples of 2 and 5] ...(i)
Multiples of 2 [2, 4, 6, ...500]
Multiples of 5 [5, 10, 15, ...500]
Multiples 2 and of 5 [10, 20, 30, ...500]
1st list of multiples of 2 [2, 4, 6, ...500]
Here, first term, \( a_1 = 2 \) and common difference \( d_1 = 2 \)
Let number of terms be \( n_1 \)
Then, last term, \( L = a + (n_1 - 1)d \)
\( 500 = 2 + (n_1 - 1)(2) \)
\( 498 = (n_1 - 1)2 \)
\( \implies (n_1 - 1) = 249 \implies n_1 = 250 \)
Sum of [2, 4, 6, ... 500]
\( S_{n1} = \frac{n_1}{2} [a + l] = \frac{250}{2} [2 + 500] = 125 \times 502 = 62750 \)
2nd list of multiples of 5 [5, 10, 15, ...500]
Here, first term, \( a' = 5 \), common difference, \( d' = 5 \)
Last term, \( L' = 500 \)
Let \( n_2 \) be the number of terms of second list. We know that \( a_n = a + (n - 1)d \)
\( L' = a' + (n_2 - 1) d' \)
\( 500 = 5 + (n_2 - 1)5 \)
\( \implies 495 = (n_2 - 1)5 \)
\( \implies n_2 - 1 = 99 \implies n_2 = 100 \)
Sum of 2nd List, \( S_{n2} = \frac{n_2}{2} (a + L) = \frac{100}{2} (5 + 500) = 50 \times 505 = 25250 \)
3rd list of multiples of 2 as well as 5 [10, 20, 30, ...500]
Here, first term, \( a'' = 10 \), common difference, \( d'' = 10 \)
Last term, \( L'' = 500 \)
Let \( n_3 \) be the number of terms, then \( L'' = a'' + (n_3 - 1) d'' \)
\( 500 = 10 + (n_3 - 1)(10) \)
\( 490 = (n_3 - 1)10 \implies n_3 - 1 = 49 \implies n_3 = 50 \)
Sum of 3rd List, \( S_{n3} = \frac{n_3}{2} (10 + 500) = \frac{50}{2} \times 510 = 12750 \)
\( \implies \text{Sum of multiples of 2 or 5} = S_{n1} + S_{n2} - S_{n3} \)
\( = 62750 + 25250 - 12750 = 88000 - 12750 = 75250 \)

Question. An AP consists of 37 terms. The sum of the 3 middle most terms is 225 and the sum of the last 3 is 429. Find the AP.
Answer: It is given that Total number of terms, \( n = 37 \)
Since \( n \) is odd, therefore middle most term = \( \left(\frac{n+1}{2}\right)^{th} \) term = \( 19^{th} \) term
3 middle most terms = \( 18^{th} \), \( 19^{th} \) and \( 20^{th} \) term by given condition
\( a_{18} + a_{19} + a_{20} = 225 \)
We know that \( a_n = a + (n - 1) d \)
\( \implies (a + 17d) + (a + 18d) + (a + 19d) = 225 \)
\( \implies 3a + 54d = 225 \)
\( \implies a + 18d = 75 \) ...(i)
Also, it is given that sum of last 3 terms = 429
\( \implies a_{35} + a_{36} + a_{37} = 429 \)
\( \implies (a + 34d) + (a + 35d) + (a + 36d) = 429 \)
\( \implies 3a + 105d = 429 \)
\( \implies a + 35d = 143 \) ...(ii)
Subtracting equation (i) from equation (ii), we get
\( (a + 35d) - (a + 18d) = 143 - 75 \)
\( \implies 17d = 68 \implies d = 4 \)
Putting value of \( d \) in equation (i), we get
\( a + 18(4) = 75 \)
\( \implies a + 72 = 75 \)
\( \implies a = 3, d = 4 \)
Required AP is \( a, a + d, a + 2d, a + 3d, ... \)
\( 3, 3 + 4, 3 + 2(4), 3 + 3(4), ... \)
3, 7, 11, 15, ...

Question. If the sum of the first 'p' terms of an A.P. is 'q' and sum of the first 'q' terms is 'p'; then show that the sum of the first (p + q) terms is {– (p + q)}. 
Answer: Consider an A.P., whose first term be '\( a \)' and common difference be '\( d \)'.
Given, \( S_p = q \)
\( S_q = p \)
To prove : \( S_{p+q} = -(p + q) \)
Proof : \( \therefore S_p = \frac{p}{2} [2a + (p - 1)d] = q \)
or \( 2a + (p - 1)d = \frac{2q}{p} \) ...(i)
and \( S_q = \frac{q}{2} [2a + (q - 1)d] = p \)
or \( 2a + (q - 1)d = \frac{2p}{q} \) ...(ii)
On subtracting equation (ii) from (i), we get
\( [(p - 1) - (q - 1)]d = \frac{2q}{p} - \frac{2p}{q} \)
\( \implies (p - q)d = \frac{2q^2 - 2p^2}{pq} \)
\( \implies d = \frac{2(q^2 - p^2)}{pq} \times \frac{1}{p-q} = \frac{-2(p^2 - q^2)}{pq(p-q)} \)
\( \implies d = \frac{-2(p + q)}{pq} \) ...(iii)
and from equation (i), \( 2a = \frac{2q}{p} - (p - 1)d \)
Now, \( S_{p+q} = \frac{p + q}{2} [2a + (p + q - 1)d] \)
\( = \frac{p + q}{2} [\frac{2q}{p} - (p - 1)d + (p + q - 1)d] \)
\( = \frac{p + q}{2} [\frac{2q}{p} + qd] \)
\( = \frac{p + q}{2} [\frac{2q}{p} + q(\frac{-2(p + q)}{pq})] \) [From (iii)]
\( = \frac{p + q}{2} [\frac{2q}{p} - \frac{2(p + q)}{p}] \)
\( = \frac{p + q}{2} [\frac{2q - 2p - 2q}{p}] \)
\( = \frac{p + q}{2} [\frac{-2p}{p}] = -(p + q) \)
Hence proved.

Question. Find the sum of the integers between 100 and 200 that are:
(A) divisible by 9
(B) not divisible by 9 
Answer: (A) Integers between 100 and 200 that are divisible by 9 are 108, 117, 126, ...198
Let \( n \) be the number of terms between 100 and 200
Here, first term, \( a = 108 \), Common difference, \( d = 117 - 108 = 9 \), Last term, \( L = 198 \)
We know that \( a_n = a + (n - 1)d \)
\( L = a + (n - 1)d \implies 198 = 108 + (n - 1)9 \)
\( \implies 198 - 108 = (n - 1)9 \implies 90 = (n - 1)9 \implies (n - 1) = 10 \implies n = 11 \)
Sum of \( n \) terms, \( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( \therefore \) Sum of 11 terms between 100 and 200 which is divisible by 9 is
\( S_{11} = \frac{11}{2} [2(108) + (11 - 1)(9)] = \frac{11}{2} [216 + 90] = \frac{11}{2} [306] = 11 \times 153 = 1683 \)
Hence, the required sum of integers is 1683.
(B) Sum of integers between 100 and 200 which is not divisible by 9
= sum of total no. between 100 and 200 – sum of no. between 100 and 200 which are divisible by 9 ...(i)
From part (A), we know that sum of no. between 100 and 200 divisible by 9 = 1683
Total numbers between 100 and 200 is 101, 102, 103, ... 199.
Here, first term, \( a = 101 \), Common difference, \( d = 102 - 101 = 1 \), Last term, \( L = 199 \)
Let \( n \) be total no. of terms. We know that \( a_n = a + (n - 1)d \)
\( \implies L = a + (n - 1)d \implies 199 = 101 + (n - 1)(1) \implies 199 - 101 = (n - 1) \implies n = 99 \)
\( \therefore \) Sum of 99 terms between 100 and 200
\( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( S_{99} = \frac{99}{2} [2(101) + (99 - 1)(1)] = \frac{99}{2} [202 + 98] = \frac{99}{2} \times 300 = 99 \times 150 = 14850 \)
Putting this value in equation (i), we get sum of no. between 100 and 200 not divisible by 9
\( = 14850 - 1683 = 13167 \)
Hence, the required sum is 13167.

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: The first term of the A.P., \( a = -7 \)
Common difference, \( d = (-12) - (-7) = -5 \)
Let the \( n^{th} \) term of A.P. be – 82.
Then, \( a_n = -82 \implies a + (n - 1) d = -82 \)
\( \implies -7 + (n - 1)(-5) = -82 \)
\( \implies -7 - 5n + 5 = -82 \)
\( \implies -5n = -80 \implies n = 16 \)
Let, \( m^{th} \) term of the given A.P. be – 100
\( \therefore a_m = -100 \implies a + (m - 1)d = -100 \)
\( \implies -7 + (m - 1)(-5) = -100 \)
\( \implies (m - 1)5 = 93 \implies m = \frac{93}{5} + 1 = \frac{98}{5} \notin N \)
Therefore, – 100 is not any term of the given A.P.
Hence, – 82 is the \( 16^{th} \) term of the A.P. and – 100 is not a term of the given A.P.

Question. How many terms of the arithmetic progression 45, 39, 33, ... must be taken so that their sum is 180? Explain the double answer.
Answer: Given: The arithmetic progression is 45, 39, 33, ......
Here, \( a = 45 \), \( d = 39 - 45 = -6 \), and the sum of \( n^{th} \) term, \( S_n = 180 \)
\( \because S_n = \frac{n}{2} [2a + (n - 1)d] \) where, \( n \) is the number of terms.
\( 180 = \frac{n}{2} [2 \times 45 + (n - 1) (-6)] \)
\( \implies 360 = 96n - 6n^2 \)
\( \implies 6n^2 - 96n + 360 = 0 \implies 6[n^2 - 16n + 60] = 0 \)
\( \implies n^2 - 10n - 6n + 60 = 0 \implies n(n - 10) - 6(n - 10) = 0 \implies (n - 6)(n - 10) = 0 \)
\( \implies n = 6, 10 \)
\( \therefore \) Sum of \( a_7, a_8, a_9 \) and \( a_{10} \) terms = 0
Hence, on either adding 6 terms or 10 terms we get a total of 180.

Question. Show that the sum of an AP whose \( 1^{st} \) term is \( a \), the \( 2^{nd} \) turn is \( b \) and the last term \( c \), is equal to \( \frac{(a + c)(b + c - 2a)}{2(b - a)} \) 
Answer: It is given that First term, \( a_1 = a \), Second term, \( a_2 = b \), Last term, \( L = c \)
Common difference = \( b - a \). AP is \( a, b, ...c \)
Let \( n \) be the number of terms of the given AP. We know that \( a_n = a + (n - 1)d \)
\( \implies L = a + (n - 1)(b - a) \implies c = a + (n - 1)(b - a) \)
\( \implies (c - a) = (n - 1)(b - a) \)
\( \implies (n - 1) = \frac{c - a}{b - a} \) ...(i)
\( \implies n = \frac{c - a}{b - a} + 1 = \frac{(c - a) + (b - a)}{b - a} = \frac{c + b - 2a}{b - a} \) ...(ii)
Now, we know that sum of an AP, \( S_n = \frac{n}{2} [a + L] \)
\( = \frac{b + c - 2a}{2(b - a)} [a + c] \) [using equation (ii)]
\( = \frac{(a + c)(b + c - 2a)}{2(b - a)} \)
Hence, proved.

Question. If the sum of the first four terms of an AP is 40 and that of the first 14 terms is 280, find the sum of its first 'n' terms.
Answer: Let the first term of the A.P. be '\( a \)' and common difference be '\( d \)'
Then, sum of '\( n \)' terms, \( S_n = \frac{n}{2} [2a + (n - 1)d] \). According to the question
Given, \( S_4 = 40 \implies \frac{4}{2} [2a + (4 - 1) \times d] = 40 \implies 2a + 3d = 20 \) ...(i)
and \( S_{14} = 280 \implies \frac{14}{2} [2a + 13d] = 280 \implies 2a + 13d = 40 \) ...(ii)
On subtracting equation (i) from equation (ii), we get \( 10d = 20 \implies d = 2 \)
Put the value of '\( d \)' in equation (i), \( \implies a = \frac{20 - 3 \times 2}{2} = 7 \)
\( \therefore \) Sum of \( n \) terms = \( \frac{n}{2} [2 \times 7 + (n - 1) \times 2] = \frac{n}{2} [14 + 2(n - 1)] = n[7 + n - 1] = n(n + 6) \text{ or } n^2 + 6n \)
Hence, the sum of first '\( n \)' terms is \( n(n + 6) \text{ or } n^2 + 6n \).

Question. The sum of the \( 4^{th} \) and the \( 8^{th} \) terms of an AP is 24 and the sum of the \( 6^{th} \) and the \( 10^{th} \) terms is 44. Find the sum of the first 10 terms of the AP. 
Answer: Let the first term of A.P. be '\( a \)' and its common difference be '\( d \)'.
Given, \( a_4 + a_8 = 24 \)
\( \implies a + 3d + a + 7d = 24 \) [\( \because a_n = a + (n - 1)d \)]
\( \implies 2a + 10d = 24 \) or \( a + 5d = 12 \) ...(i)
and \( a_6 + a_{10} = 44 \) (given)
\( \implies a + 5d + a + 9d = 44 \implies 2a + 14d = 44 \) or \( a + 7d = 22 \) ...(ii)
On subtracting equation (i) from (ii), we get:
\( a + 7d = 22 \)
\( a + 5d = 12 \)
\( \implies 2d = 10 \implies d = 5 \)
If we put the value of '\( d \)' in equation (i), we get \( a = 12 - 25 = -13 \)
Sum of first \( n \) terms, \( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( \therefore S_{10} = \frac{10}{2} [2 \times (-13) + (10 - 1) \times 5] = 5(-26 + 45) = 95 \)
Hence, the sum of the first 10 terms of A.P. is 95.

Question. If the ratio of the \( 11^{th} \) term of an AP to its \( 18^{th} \) term is 2 : 3, find the ratio of the sum of the first five terms to the sum of its first 10 terms. 
Answer: Let, the first term of A.P. be '\( a \)' and its common difference be '\( d \).
Then, \( 11^{th} \) term of A.P., \( a_{11} = a + 10d \); \( 18^{th} \) term of A.P., \( a_{18} = a + 17d \)
\( \frac{a + 10d}{a + 17d} = \frac{2}{3} \) (given)
\( \implies 3(a + 10d) = 2(a + 17d) \implies 3a + 30d = 2a + 34d \implies a = 4d \) ...(i)
Now, sum of first five terms, \( S_5 = \frac{5}{2} [2a + (5 - 1)d] = \frac{5}{2} [2 \times 4d + 4d] = \frac{5}{2} \times 12d = 30d \) [from (i)]
Sum of first 10 terms, \( S_{10} = \frac{10}{2} [2a + (10 - 1)d] = 5[2 \times 4d + 9d] = 5(8d + 9d) = 85d \)
\( \therefore \frac{S_5}{S_{10}} = \frac{30d}{85d} = \frac{6}{17} \)
Hence, the required ratio is 6 : 17.

Question. The ratio of the sums of first m and first n terms of an A.P. is \( m^2 : n^2 \). Show that the ratio of its \( m^{th} \) and \( n^{th} \) terms is \( (2m - 1):(2n - 1) \). 
Answer: Let, '\( a \)' be the first term of an A.P. and '\( d \)' be the common difference.
Let, \( S_m \) and \( S_n \) be the sum of the first '\( m \)' and first '\( n \)' terms of the A.P. respectively.
Then, \( S_m = \frac{m}{2} [2a + (m - 1)d] \) and \( S_n = \frac{n}{2} [2a + (n - 1)d] \)
But \( \frac{S_m}{S_n} = \frac{m^2}{n^2} \) (given)
\( \therefore \frac{\frac{m}{2}[2a + (m - 1)d]}{\frac{n}{2}[2a + (n - 1)d]} = \frac{m^2}{n^2} \implies \frac{2a + (m - 1)d}{2a + (n - 1)d} = \frac{m}{n} \)
\( \implies n[2a + (m - 1)d] = m[2a + (n - 1)d] \implies 2an + mnd - nd = 2am + mnd - md \)
\( \implies md - nd = 2a(m - n) \implies d = 2a \)
Now, the ratio of the \( m^{th} \) and \( n^{th} \) terms is:
\( \frac{a_m}{a_n} = \frac{a + (m - 1)d}{a + (n - 1)d} = \frac{a + (m - 1)2a}{a + (n - 1)2a} \) [\( \because d = 2a \)]
\( = \frac{a[1 + 2m - 2]}{a[1 + 2n - 2]} = \frac{2m - 1}{2n - 1} \)
Hence, the ratio of \( m^{th} \) to \( n^{th} \) term is \( (2m - 1):(2n - 1) \).

Question. Solve the equation \( -4 + (-1) + 2 + ... + x = 437 \). 
Answer: The given equation is \( -4 + (-1) + 2 + ... + x = 437 \). This is an A.P. with first term, \( a = -4 \) and common difference, \( d = (-1) - (-4) = 3 \). Last term, \( L = x \).
We know that \( a_n = a + (n - 1)d \implies x = -4 + (n - 1)3 \implies n = \frac{x+7}{3} \)
Also, \( S_n = \frac{n}{2} [2a + (n - 1)d] \implies 437 = \frac{x+7}{2 \times 3} [2(-4) + (\frac{x+7}{3}-1)3] \)
\( = \frac{x+7}{6} [-8 + x + 4] = \frac{(x+7)(x-4)}{6} \)
\( \implies (x+7)(x-4) = 437 \times 6 \implies x^2 + 3x - 28 = 2622 \implies x^2 + 3x - 2650 = 0 \)
By quadratic formula, \( x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-2650)}}{2(1)} = \frac{-3 \pm \sqrt{10609}}{2} = \frac{-3 \pm 103}{2} \)
\( \implies x = 50, -53 \). But \( x \) cannot be negative, so \( x = 50 \).

Question. A thief runs with a uniform speed of 100 m/minute. After one minute a policeman after the thief to catch him. He goes with a speed of 100 m/minute in the first minute and increases his speed by 10 m/minute every succeeding minute. After how many minutes the policeman will catch the thief? 
Answer: Let, the total time to catch the thief be '\( n \)' minutes. Speed of thief = 100 m/min. Total distance covered by the thief = \( (100 n) \) meters.
The speed of the policeman in the first minute is 100 m/min, in the \( 2^{nd} \) minute is 110 m/min, and so on. This forms an A.P. with \( a = 100, d = 10 \). As the policeman starts after a minute, he runs for \( (n - 1) \) minutes.
Total distance by policeman \( = 100 + 110 + 120 + ... (n - 1) \) terms
\( \therefore 100n = \frac{n - 1}{2} [200 + (n - 2) \times 10] \)
\( \implies 200n = (n - 1) [200 + 10n - 20] \implies 200n = (n - 1) (180 + 10n) \)
\( \implies 200n = 180n - 180 + 10n^2 - 10n \implies 10n^2 - 30n - 180 = 0 \implies n^2 - 3n - 18 = 0 \)
\( \implies (n - 6) (n + 3) = 0 \implies n = 6 \) (since \( n = -3 \) is not possible).
Hence, the policeman takes \( (n - 1) = 5 \) minutes to catch the thief.