CBSE Class 10 Mathematics Arithmetic Progressions VBQs Set D

Question. How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3? 
Answer: Numbers that lie between 10 and 300 are 11, 12, ... 299.
Numbers between 10 and 300 which when divided by 4 leave a remainder 3 and are 11, 15, 19, ...299
Here, first number, \( a = 11 \)
and common difference, \( d = 4 \)
number of terms, \( n = ? \)
We know that \( a_n = a + (n - 1)d \)
\( a + (n - 1)d = 299 \)
\( \Rightarrow 11 + (n - 1)(4) = 299 \)
\( \Rightarrow 11 + 4n - 4 = 299 \)
\( \Rightarrow 4n = 292 \Rightarrow n = 73 \)

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 \( -\frac{4}{3}, -1, -\frac{2}{3}, \dots, 4\frac{1}{3} \)
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
\( L = a + (n - 1)d \)
\( \frac{13}{3} = -\frac{4}{3} + (n - 1)(\frac{1}{3}) \)
\( \Rightarrow 13 = -4 + (n - 1) \)
\( \Rightarrow (n - 1) = 17 \)
\( \Rightarrow n = 18 \)
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 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
\( S_n = \frac{n}{2}(a + l) \)
\( 120 = \frac{n}{2}(-5 + 45) \)
\( 120 \times 2 = 40 \times n \)
\( \Rightarrow n = 6 \)
\( L = a + (n - 1)d \)
\( 45 = -5 + (6 - 1)d \)
\( 50 = 5d \Rightarrow d = 10 \)
Hence, number of terms = 6
and common difference = 10

Question. If Sn 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)
\( S_{12} = \frac{12}{2} [2a + (12 - 1)d] = 6 [2a + 11d] = 12a + 66d = 3 [4a + 22d] \)
\( S_{12} = 3 [S_8 - S_4] \) [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)
\( \Rightarrow \frac{6}{2} [2a + (6 - 1)d] = 36 \)
\( \Rightarrow 3 [2a + 5d] = 36 \)
\( \Rightarrow 2a + 5d = 12 \) ...(i)
Also, \( S_{16} = 256 \)
\( \Rightarrow \frac{16}{2} [2a + (16 - 1)d] = 256 \)
\( \Rightarrow 8 [2a + 15d] = 256 \)
\( \Rightarrow 2a + 15d = 32 \) ...(ii)
Subtracting equation (i) from equation (ii), we get
\( (2a + 15d) - (2a + 5d) = 32 - 12 \)
\( \Rightarrow 10d = 20 \Rightarrow d = 2 \)
Putting the value of d in equation (i), we get
\( 2a + 5(2) = 12 \Rightarrow 2a + 10 = 12 \Rightarrow 2a = 2 \Rightarrow a = 1 \)
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,
\( S_n = \frac{n}{2} [2a + (n - 1)d] = \frac{n}{2} [2 \times 8 + (n - 1)20] = \frac{n}{2} [16 + 20n - 20] = \frac{n}{2} [20n - 4] = n[10n - 2] \)
\( \Rightarrow S_n = n[10n - 2] \) ...(i)
Now, first term of second AP (\( a' \)) = –30, Common difference of second (\( d' \)) = 8
\( \therefore \) Sum of first 2n terms of second AP,
\( S_{2n} = \frac{2n}{2} [2a' + (2n - 1)d'] = n[2(-30) + (2n - 1)8] = n[-60 + 16n - 8] = n[16n - 68] \) ...(ii)
By given condition, Sum of first n terms of first AP = sum of first 2n terms of second AP
\( S_n = S_{2n} \)
Using equation (i) and equation (ii), we get \( n(10n - 2) = n(16n - 68) \)
\( \Rightarrow 10n - 2 = 16n - 68 \Rightarrow 6n = 66 \Rightarrow n = 11 \)
Hence, the required value of n is 11.

Question. If mth term of an A.P. is \( \frac{1}{n} \) and nth 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, \( a_m = a + (m - 1)d \)
\( \Rightarrow \frac{1}{n} = a + (m - 1)d \) ...(i) (given)
and \( a_n = a + (n - 1)d \)
\( \Rightarrow \frac{1}{m} = a + (n - 1)d \) ...(ii) (given)
On subtracting equation (ii) from (i), we get:
\( \frac{1}{n} - \frac{1}{m} = [(m - 1) - (n - 1)]d \)
\( \Rightarrow \frac{m - n}{mn} = (m - n)d \Rightarrow d = \frac{1}{mn} \) ...(iii)
If we put the value of d in equation (i), we get
\( \frac{1}{n} = a + (m - 1) \times \frac{1}{mn} \Rightarrow a = \frac{1}{n} - (\frac{m - 1}{mn}) = \frac{m - m + 1}{mn} = \frac{1}{mn} \) ...(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 nth term = \( 63 + (n - 1) \times 2 \) ...(i)
For second A.P., first term, \( a' = 3 \) and common difference, \( d' = 7 \)
Then, its nth term = \( 3 + (n - 1) \times 7 \)
According to the question
\( 63 + (n - 1) \times 2 = 3 + (n - 1) \times 7 \)
\( \Rightarrow 60 = (n - 1) \times 5 \Rightarrow n - 1 = 12 \Rightarrow n = 13 \)
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:
\( S_{40} = \frac{40}{2} [2a + (n - 1)d] = 20 [2 \times 7 + (40 - 1) \times 6] = 20 [14 + 39 \times 6] = 20 \times 248 = 4,960 \)
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, \( (a - 3d) + (a - d) + (a + d) + (a + 3d) = 56 \Rightarrow 4a = 56 \Rightarrow a = 14 \)
According to question, \( \frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{5}{6} \)
\( \Rightarrow \frac{(14 - 3d)(14 + 3d)}{(14 - d)(14 + d)} = \frac{5}{6} \) [Putting \( a = 14 \)]
\( \Rightarrow \frac{196 - 9d^2}{196 - d^2} = \frac{5}{6} \Rightarrow 1176 - 54d^2 = 980 - 5d^2 \Rightarrow 49d^2 = 196 \Rightarrow d^2 = 4 \Rightarrow d = \pm 2 \)
when, \( a = 14 \) and \( d = 2 \), the parts are 8, 12, 16, 20.
when, \( a = 14 \) and \( d = -2 \), the 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, \( S_7 = \frac{7}{2} [2a + (7 - 1) \times d] \Rightarrow 49 = \frac{7}{2} [2a + 6d] \Rightarrow 7 = a + 3d \) ...(i)
And \( S_{17} = \frac{17}{2} [2a + (17 - 1) \times d] \Rightarrow 289 = \frac{17}{2} [2a + 16d] \Rightarrow 17 = a + 8d \) ...(ii)
On subtracting equation (i) from equation (ii), we get \( 5d = 10 \Rightarrow d = 2 \).
If we put the value of ‘d’ in equation (i), we get \( a + 3 \times (2) = 7 \Rightarrow a = 7 - 6 = 1 \).
\( \therefore \) 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: \( S_m = S_n \)
\( \Rightarrow \frac{m}{2} [2a + (m - 1)d] = \frac{n}{2} [2a + (n - 1)d] \)
\( \Rightarrow 2a(m - n) + d(m^2 - m - n^2 + n) = 0 \)
\( \Rightarrow (m - n)[2a + (m + n - 1)d] = 0 \)
or \( S_{m+n} = \frac{m + n}{2} [2a + (m + n - 1)d] = 0 \).

Question. How many terms of an A.P. 9, 17, 25,... must be taken to give a sum of 636? 
Answer: \( a = 9, d = 8, S_n = 636 \).
\( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( 636 = \frac{n}{2} [18 + (n - 1)8] \)
\( 636 = n(9 + 4n - 4) \)
\( 636 = n(5 + 4n) \Rightarrow 4n^2 + 5n - 636 = 0 \).
Solving the quadratic equation: \( 4n^2 + 53n - 48n - 636 = 0 \)
\( n(4n + 53) - 12(4n + 53) = 0 \)
\( (4n + 53)(n - 12) = 0 \)
\( \therefore n = -\frac{53}{4} \) or \( 12 \).
As n is a natural number, \( n = 12 \).
\( \therefore 12 \) terms are required to give sum 636.

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
\( S_{x – 1} = S_{49} – S_x \)
\( \Rightarrow \frac{(x – 1)x}{2} = \frac{49 \times 50}{2} – \frac{(x + 1)x}{2} \)
\( \Rightarrow x^2 – x = 2450 – x^2 – x \)
\( \Rightarrow x^2 = 1225 \)
\( \Rightarrow x = \pm 35 \) (–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: \( a_{14} = 2a_8 \)
\( \Rightarrow a + 13d = 2(a + 7d) \Rightarrow a = –d \)
\( a_6 = –8 \)
\( \Rightarrow a + 5d = –8 \)
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, \( a – d + a + a + d = 15 \)
\( 3a = 15 \Rightarrow a = 5 \)
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,
\( (111a – 99d) – (111a + 99d) = 594 \)
\( \Rightarrow –198d = 594 \Rightarrow d = –3 \)
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.

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 three digits of 3-digit number be \( a-d, a, a+d \).
Their sum = 15
\( a-d + a + a+d = 15 \Rightarrow 3a = 15 \Rightarrow a = 5 \)
Required 3-digit no = \( 100(a-d) + 10a + a+d \)
\( = 100a - 100d + 10a + a+d \)
\( = 111a - 99d \)
No obtained by reversing digits = \( 100(a+d) + 10a + a-d \)
\( = 100a + 100d + 10a + a-d \)
\( = 111a + 99d \)
According to question,
\( 111a + 99d = 111a - 99d - 594 \)
\( 594 = 111a - 99d - 111a - 99d \)
\( 594 = -198d \)
\( d = \frac{594}{-198} = -3 \)
The no. = \( 111a - 99d \)
\( = 111 \times 5 - 99 \times (-3) \)
\( = 555 + 297 = 852 \)
No = 852 or 258.

Question. Find the sum of first 24 terms of an A.P. whose \( n^{th} \) term is given by \( a_n = 3 + 2n \). 
Answer: Given, \( n^{th} \) term of an A.P., \( a_n = 3 + 2n \)
First term of A.P., \( a_1 = 3 + 2 \times 1 = 5 \)
Second term of A.P., \( a_2 = 3 + 2 \times 2 = 7 \)
third term of A.P., \( a_3 = 3 + 2 \times 3 = 9 \)
and \( 24^{th} \) term of A.P., \( a_{24} = 3 + 2 \times 24 = 51 \)
common difference of A.P., \( d = a_2 – a_1 = a_3 – a_2 \)
\( = 7 – 5 = 9 – 7 = 2 \)
Sum of 24 terms, \( S_{24} = \frac{n}{2}[2a + (n – 1)d] \)
Here, \( a = 5, n = 24, d = 2 \)
\( \therefore S_{24} = \frac{24}{2}[2 \times 5 + (24 – 1) \times 2] \)
\( = 12 [10 + 46] \)
\( = 12 \times 56 = 672 \)
Hence the sum of first 24 terms of an A.P. is 672.

Question. Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12. 
Answer: Let, the first term of A.P. be ‘a’ and its common difference be ‘d’.
Given, \( a_3 = 16 \)
i.e. \( a + (3 – 1)d = 16 \) [\( \because a_n = a + (n – 1)d \)]
\( \Rightarrow a + 2d = 16 \) ...(i)
And \( a_7 = a_5 + 12 \)
\( \Rightarrow a + (7 – 1)d = a + (5 – 1)d + 12 \)
\( \Rightarrow a + 6d = a + 4d + 12 \)
\( \Rightarrow 2d = 12 \)
\( \Rightarrow d = 6 \)
If we put the value of ‘d’ in equation (i), we get:
\( a + 2 \times 6 = 16 \)
\( \Rightarrow a = 16 – 12 = 4 \)
First term of AP = 4
Second term of AP = 4 + 6 = 10
Third term of AP = 10 + 6 = 16 and so on.
Hence, the required A.P is 4, 10, 16, 22, ....

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 terms, 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]
\( \Rightarrow a_{26} = a + (26 – 1)d \)
\( \Rightarrow 0 = a + 25d \)
\( \Rightarrow a + 25d = 0 \) ...(i)
Also, \( a_{11} = 3 \)
\( \Rightarrow a + (11 – 1)d = 3 \)
\( \Rightarrow a + 10d = 3 \) ...(ii)
Last term \( L = -\frac{1}{5} \)
\( a + (n – 1)d = -\frac{1}{5} \) ...(iii)
Subtracting equation (ii) from equation (i), we get
\( (a + 25d) – (a + 10d) = 0 – 3 \)
\( \Rightarrow a + 25d – a – 10d = -3 \)
\( \Rightarrow 15d = -3 \)
\( \Rightarrow d = -\frac{3}{15} = -\frac{1}{5} \)
Putting the value of d in equation (i)
\( a + 25(-\frac{1}{5}) = 0 \)
\( \Rightarrow a – 5 = 0 \)
\( \Rightarrow a = 5 \)
Putting the value of a and d in equation (iii)
\( 5 + (n – 1)(-\frac{1}{5}) = -\frac{1}{5} \)
\( \Rightarrow 25 – (n – 1) = -1 \)
\( \Rightarrow 25 + 1 = (n – 1) \)
\( \Rightarrow n = 27 \)
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]
\( 81 = 5 + (n – 1) \times 4 \)
\( \Rightarrow (n – 1) = \frac{76}{4} = 19 \)
\( \Rightarrow n = 20 \)
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, \( a_{n'} = a' + (n' – 1)d' \)
\( \Rightarrow – 3 = – 41 + (n' – 1) \times 2 \)
\( \Rightarrow (n' – 1) = 19 \)
\( \Rightarrow n' = 20 \)
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
\( \therefore \) Total sum = \( S_{20} + S'_{20} = 860 – 440 = 420 \).
Hence, the total sum of series is 420.

Question. The sum of four consecutive numbers in AP is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7 : 15. Find the numbers. 
Answer: Let ‘a - 3d’, ‘a - d’, ‘a + d’, ‘a + 3d’ be the common difference to the AP. Then, are four consecutive terms of the AP.
As per the question,
\( (a – 3d) + (a – d) + (a + d) + (a + 3d) = 32 \)
\( \Rightarrow 4a = 32 \), or \( a = 8 \) ...(i)
and \( \frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{7}{15} \)
\( \Rightarrow \frac{a^2 - 9d^2}{a^2 - d^2} = \frac{7}{15} \)
\( \Rightarrow 15a^2 – 135d^2 = 7a^2 – 7d^2 \)
\( \Rightarrow 8a^2 = 128 d^2 \)
Using (i), we have:
\( 8 \times 8^2 = 128 d^2 \)
\( \Rightarrow d^2 = 4 \), 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, \( a_n = x \)
\( \Rightarrow a + (n – 1)d = x \)
\( \Rightarrow 1 + 3(n – 1) = x \Rightarrow n = \frac{x + 2}{3} \)
Further, \( S_n = \frac{n}{2} \) [first term + last term]
\( \Rightarrow 287 = \frac{x + 2}{3 \times 2} [1 + x] \)
\( \Rightarrow (x + 1) (x + 2) = 1722 \)
or \( x^2 + 3x – 1720 = 0 \)
\( \Rightarrow x^2 + 43x – 40x – 1720 = 0 \)
\( \Rightarrow x(x + 43) – 40(x + 43) = 0 \)
\( \Rightarrow (x + 43) (x – 40) = 0 \)
\( \Rightarrow x – 40 = 0 \) (\( \because x + 43 \neq 0 \))
\( x = 40 \).
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, \( S_5 \)
\( S_5 = \frac{5}{2} [2a + (5 – 1)d] = \frac{5}{2} [2a + 4d] = 5[a + 2d] = 5a + 10d \) ...(i)
Sum of first seven terms, \( S_7 \)
\( S_7 = \frac{7}{2} [2a + (7 – 1)d] = \frac{7}{2} [2a + 6d] = 7[a + 3d] = 7a + 21d \) ...(ii)
Now, by given condition: \( S_5 + S_7 = 167 \)
\( \Rightarrow 5a + 10d + 7a + 21d = 167 \)
\( \Rightarrow 12a + 31d = 167 \) ...(iii)
Also, it is given that sum of first 10 terms is 235.
\( S_{10} = 235 \Rightarrow \frac{10}{2} [2a + 9d] = 235 \Rightarrow 5[2a + 9d] = 235 \Rightarrow 2a + 9d = 47 \) ...(iv)
Multiplying equation (iv) by 6 and subtracting it from equation (iii), we get:
\( (12a + 31d) – (12a + 54d) = 167 – 282 \)
\( \Rightarrow -23d = -115 \Rightarrow d = 5 \)
Putting value of d in eq (iv): \( 2a + 45 = 47 \Rightarrow 2a = 2 \Rightarrow a = 1 \)
Sum of first 20 terms: \( S_{20} = \frac{20}{2} [2a + 19d] = 10[2(1) + 19(5)] = 10[2 + 95] = 970 \).

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. 
Answer: (A) Multiples of 2 and 5 are multiples of 10. Between 1 and 500: 10, 20...490. \( a=10, d=10, L=490 \). \( 490 = 10 + (n-1)10 \Rightarrow n=49 \). \( S_{49} = \frac{49}{2}(10 + 490) = 12250 \).
(B) Multiples from 1 to 500: 10, 20...500. \( a=10, L=500, n=50 \). \( S_{50} = \frac{50}{2}(10 + 500) = 12750 \).
(C) Multiples of 2 or 5 = Sum(2) + Sum(5) – Sum(10).
Sum(2) [2, 4...500]: \( n_1=250 \), \( S_1 = \frac{250}{2}(2+500) = 62750 \).
Sum(5) [5, 10...500]: \( n_2=100 \), \( S_2 = \frac{100}{2}(5+500) = 25250 \).
Sum(10) [10, 20...500]: \( n_3=50 \), \( S_3 = \frac{50}{2}(10+500) = 12750 \).
Total \( = 62750 + 25250 – 12750 = 75250 \).

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: \( n = 37 \). Middle term = \( (\frac{37+1}{2})^{th} = 19^{th} \). Middle most terms are \( 18^{th}, 19^{th}, 20^{th} \).
\( a_{18} + a_{19} + a_{20} = 225 \Rightarrow (a+17d) + (a+18d) + (a+19d) = 225 \Rightarrow 3a + 54d = 225 \Rightarrow a + 18d = 75 \) ...(i)
Sum of last 3: \( a_{35} + a_{36} + a_{37} = 429 \Rightarrow 3a + 105d = 429 \Rightarrow a + 35d = 143 \) ...(ii)
Subtracting (i) from (ii): \( 17d = 68 \Rightarrow d = 4 \).
From (i): \( a + 72 = 75 \Rightarrow a = 3 \).
AP is 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: \( S_p = q \Rightarrow \frac{p}{2}[2a + (p-1)d] = q \Rightarrow 2a + (p-1)d = \frac{2q}{p} \) ...(i)
\( S_q = p \Rightarrow \frac{q}{2}[2a + (q-1)d] = p \Rightarrow 2a + (q-1)d = \frac{2p}{q} \) ...(ii)
Subtracting: \( (p-q)d = \frac{2q}{p} - \frac{2p}{q} = \frac{2(q^2 - p^2)}{pq} \Rightarrow d = \frac{-2(p+q)}{pq} \) ...(iii)
From (i) and (iii), \( 2a = \frac{2q}{p} - (p-1)d \).
Now, \( S_{p+q} = \frac{p+q}{2} [2a + (p+q-1)d] \).
Substituting 2a and d leads to \( S_{p+q} = \frac{p+q}{2} [\frac{2q}{p} - (p-1)d + (p+q-1)d] \)
\( = \frac{p+q}{2} [\frac{2q}{p} + qd] = \frac{p+q}{2} [\frac{2q}{p} + q(\frac{-2(p+q)}{pq})] = -(p+q) \). 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 divisible by 9: 108, 117... 198. \( a=108, d=9, L=198 \). \( 198 = 108 + (n-1)9 \Rightarrow n=11 \). \( S_{11} = \frac{11}{2}(108 + 198) = 1683 \).
(B) Total integers 101 to 199: \( n=99 \). Sum \( S_{99} = \frac{99}{2}(101 + 199) = 14850 \).
Sum not divisible by 9 = \( 14850 – 1683 = 13167 \).

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, \( d = (– 12) – (– 7) = (–12 + 7) = – 5 \)
Let the \( n^{th} \) term of A.P. be – 82.
Then, \( a_n = – 82 \)
\( a + (n – 1) d = – 82 \)
\( \Rightarrow – 7 + (n – 1) (– 5) = – 82 \)
\( \Rightarrow – 7 – 5n + 5 = – 82 \)
\( \Rightarrow – 5n = – 80 \)
\( \Rightarrow n = 16 \)
Let, \( m^{th} \) term of the given A.P. be – 100
\( \therefore a_m = – 100 \)
\( a + (m – 1) d = – 100 \)
\( \Rightarrow – 7 + (m – 1) (– 5) = – 100 \)
\( \Rightarrow (m – 1) 5 = 93 \)
\( \Rightarrow m = \frac{93}{5} + 1 = \frac{98}{5} \notin N \)
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, \( a = 45 \)
\( d = 39 – 45 = 33 – 39 = – 6 \)
and the sum the of \( n^{th} \) term, \( S_n = 180 \)
\( \because S_n = \frac{n}{2} [2a + (n – 1)d] \)
where, \( n \) is the number of terms.
\( 180 = \frac{n}{2} [2 \times 45 + (n – 1) (– 6)] \)
\( \Rightarrow 360 = 96n – 6n^2 \)
\( \Rightarrow 6n^2 – 96n + 360 = 0 \)
\( \Rightarrow 6[n^2 - 16n + 60] = 0 \)
\( \Rightarrow 6[n^2 - 10n - 6n + 60] = 0 \)
\( \Rightarrow 6[n(n - 10) - 6(n - 10)] = 0 \)
\( \Rightarrow 6[(n – 6) (n – 10)] = 0 \)
\( \Rightarrow n = 6, 10 \)
\( \therefore \) 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 \( d = b – a \)
AP is \( a, b, \dots c \)
Let \( n \) be the number of terms of the given AP.
We know that \( a_n = a + (n – 1)d \)
\( \Rightarrow L = a + (n – 1)(b – a) \)
\( \Rightarrow c = a + (n – 1)(b – a) \)
\( \Rightarrow (c – a) = (n – 1)(b – a) \)
\( \Rightarrow (n – 1) = \frac{(c – a)}{(b – a)} \) ...(i)
\( \Rightarrow n = \frac{c – a}{b – a} + 1 = \frac{(c – a) + (b – a)}{b – a} \)
\( = \frac{c + b – 2a}{b – a} \) ...(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)]
\( = \frac{(b + c – 2a)(a + c)}{2(b – a)} \)
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, \( S_4 = 40 \)
\( \Rightarrow \frac{4}{2} [2a + (4 – 1) \times d] = 40 \)
\( \Rightarrow 2a + 3d = 20 \) ...(i)
and \( S_{14} = 280 \)
\( \Rightarrow \frac{14}{2} [2a + 13d] = 280 \)
\( \Rightarrow 2a + 13d = 40 \) ...(ii)
On subtracting equation (i) from equation (ii), we get:
\( 2a + 13d = 40 \)
\( 2a + 3d = 20 \)
\( 10d = 20 \)
\( d = 2 \)
Put the value of \( d \) in equation (i),
\( \Rightarrow a = \frac{20 - 3 \times 2}{2} = 7 \)
\( \therefore \) Sum of n terms = \( \frac{n}{2} [2 \times 7 + (n – 1) \times 2] \)
\( = \frac{n}{2} [14 + 2n - 2] \)
\( = n[7 + n – 1] \)
\( = n(n + 6) \) or \( n^2 + 6n \)
Hence, the sum of first 'n' terms is \( n(n + 6) \) 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, \( a_4 + a_8 = 24 \)
\( \Rightarrow a + 3d + a + 7d = 24 \) [\( \because a_n = a + (n – 1)d \)]
\( \Rightarrow 2a + 10d = 24 \)
or \( a + 5d = 12 \) ...(i)
and \( a_6 + a_{10} = 44 \) (given)
\( \Rightarrow a + 5d + a + 9d = 44 \)
\( \Rightarrow 2a + 14d = 44 \)
or \( a + 7d = 22 \) ...(ii)
On subtracting equation (i) from (ii), we get:
\( a + 7d = 22 \)
\( a + 5d = 12 \)
\( 2d = 10 \)
\( d = 5 \)
If we put the value of \( d \) in equation (i), we get
\( a = 12 – 25 = – 13 \)
Then, first term of A.P., \( a = –13 \)
Common difference of A.P. \( d = 5 \)
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)
\( \Rightarrow 3(a + 10d) = 2(a + 17d) \)
\( \Rightarrow 3a + 30d = 2a + 34d \)
\( \Rightarrow a = 4d \) ...(i)
Now, sum of first five terms,
\( S_5 = \frac{5}{2} [2a + (5 – 1)d] \)
\( = \frac{5}{2} [2 \times 4d + 4 \times d] \) [from (i)]
\( = \frac{5}{2} \times 12d = 30d \)
Sum of first 10 terms,
\( S_{10} = \frac{10}{2} [2a + (10 – 1)d] \)
\( = \frac{10}{2} [2 \times 4d + 9 \times d] \)
\( = 5(8d + 9d) \)
\( = 85d \)
\( \therefore \frac{S_5}{S_{10}} = \frac{30d}{85d} = \frac{6}{17} \)
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 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' of 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)
\( \therefore \frac{\frac{m}{2} [2a + (m – 1)d]}{\frac{n}{2} [2a + (n – 1)d]} = \frac{m^2}{n^2} \)
\( \Rightarrow \frac{2a + (m – 1)d}{2a + (n – 1)d} = \frac{m}{n} \)
\( \Rightarrow n[2a + (m – 1)d] = m[2a + (n – 1)d] \)
\( \Rightarrow 2an + mnd – nd = 2am + mnd – md \)
\( \Rightarrow md – nd = 2am – 2an \)
\( \Rightarrow d(m – n) = 2a(m – n) \)
\( \Rightarrow d = 2a \)
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
The given equation is A.P. with first term, \( a = –4 \)
and common difference, \( d = (–1) – (–4) = –1 + 4 = 3 \)
Last term, \( L = x \)
We know that \( a_n = a + (n – 1)d \)
\( L = x = a + (n – 1)(d) \)
\( x = –4 + (n – 1)(3) \)
\( x = –4 + 3n – 3 \)
\( n = \frac{x + 7}{3} \)
Also, we know that \( S_n = \frac{n}{2} [2a + (n – 1)d] \)
\( S_n = \frac{\frac{x + 7}{3}}{2} [2(-4) + (\frac{x + 7}{3} - 1)3] \)
\( = \frac{x + 7}{6} [-8 + x + 4] = \frac{(x + 7)(x - 4)}{6} \)
The given equation is (–4) + (–1) + 2... + x = 437
\( \therefore S_n = 437 \)
\( \frac{(x + 7)(x - 4)}{6} = 437 \)
\( x^2 – 4x + 7x – 28 = 437 \times 6 \)
\( x^2 + 3x – 28 = 2622 \)
\( x^2 + 3x – 2650 = 0 \)
By quadratic formula \( x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-2650)}}{2(1)} \)
\( = \frac{-3 \pm \sqrt{10609}}{2} \)
\( = \frac{-3 \pm 103}{2} \)
\( \Rightarrow x = \frac{-3 + 103}{2}, \frac{-3 - 103}{2} = \frac{100}{2}, \frac{-106}{2} = 50, –53 \)
But \( x \) can not 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 \( n \))
Here, the speed of the policeman in the first minute is 100 m/min, in the \( 2^{nd} \) minute is 110 m/min, in the \( 3^{rd} \) minute is 120 m/min and so on.
Then, the speed forms an A.P. with a constant increasing speed of 10 m/min. thus, the series is 100, 110, 120, 130
As the policeman starts after a minute (n – 1)
Total distance covered by the policeman = 100 + 110 + 120 + ... (n – 1) terms
\( \therefore 100n = \frac{n - 1}{2} [200 + (n – 2) \times 10] \)
[\( \because S_n = \frac{n}{2} [2a + (n – 1)d] \)]
\( \Rightarrow 200n = (n – 1) [200 + 10n – 20] \)
\( \Rightarrow 200n = (n – 1) (180 + 10n) \)
\( \Rightarrow 200n = 180n – 180 + 10n^2 – 10n \)
\( \Rightarrow 10n^2 – 30n – 180 = 0 \)
\( \Rightarrow n^2 – 3n – 18 = 0 \)
\( \Rightarrow (n – 6) (n + 3) = 0 \)
\( \Rightarrow n = 6 \) [\( \because n = – 3 \), is not possible]
Hence, the policeman takes \( n – 1 = 5 \) minutes to catch the thief.

Question. If the ratio of the sum of the first n terms of two A.Ps is (7n + 1) : (4n + 27), then find the ratio of their \( 9^{th} \) terms. 
Answer: Let \( a, d \) and \( A, D \) be the \( 1^{st} \) term and common difference of the 2 APs respectively.
Then, \( \frac{\frac{n}{2} [2a + (n-1)d]}{\frac{n}{2} [2A + (n-1)D]} = \frac{7n+1}{4n+27} \)
\( \frac{2a + (n-1)d}{2A + (n-1)D} = \frac{7n+1}{4n+27} \)
Replacing \( n \) by 17 in both LHS and RHS,
\( \frac{2a + (17-1)d}{2A + (17-1)D} = \frac{7(17)+1}{4(17)+27} \)
\( \frac{2a + 16d}{2A + 16D} = \frac{119+1}{68+27} \)
\( \frac{2(a+8d)}{2(A+8D)} = \frac{120}{95} \)
as \( a + (n-1)d = a_n \),
\( \frac{a_9}{A_9} = \frac{24}{19} \)
\( \therefore \) ratio of \( 9^{th} \) terms is 24 : 19.