CBSE Class 10 Maths HOTs Arithmetic Progressions Set L

Very Short Answer Type Questions

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 \( S_{m + n} = 0 \)

Question. If 6 times the \( 6^{th} \) term of an A.P. is equal to 9 times the \( 9^{th} \) term, show that its \( 15^{th} \) term is zero.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the AP.
Given, \( 6 a_6 = 9 \times a_9 \)
Thus, \( 6(a + 5d) = 9(a + 8d) \)
\( \Rightarrow 3a = – 42d \)
or \( a + 14d = 0 \) ...(i)
Thus, \( a_{15} = a + 14d = 0 \) (by (i))
Hence, \( 15^{th} \) term of the AP is zero.

Question. Find the sum of all 11 terms of an A.P., whose middle term is 30.
Answer: Let ‘a’ be the first term and d, the common difference of the given A.P.
Then, the middle term = \( a_6 = a + 5d = 30 \).
Now, \( S_{11} = \frac{11}{2} [2a + 10d] \)
\( = \frac{11}{2} \times 2(a + 5d) \)
\( = 11 \times 30 \) [\( \because a + 5d = 30 \)]
\( = 330 \)

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. Find the number of natural numbers between 102 and 998 which are divisible by 2 and 5 both.
Answer: Numbers divisible by both 2 and 5 are divisible by 10.
The series is 110, 120, 130, … , 990
\( a_n = 990 \)
\( \Rightarrow 110 + (n - 1) \times 10 = 990 \)
\( \therefore n = 89 \)

Question. Show that \( (a - b)^2, (a^2 + b^2) \) and \( (a + b)^2 \) are in AP.
Answer: \( (a - b)^2, (a^2 + b^2) \) and \( (a + b)^2 \) will be in AP, if
\( 2(a^2 + b^2) = (a - b)^2 + (a + b)^2 \)
L.H.S. = \( 2(a^2 + b^2) = 2a^2 + 2b^2 \)
R.H.S. = \( (a^2 + b^2 - 2ab) + (a^2 + b^2 + 2ab) \)
\( = 2(a^2 + b^2) \)
Since L.H.S. = R.H.S.
Hence, \( (a - b)^2, (a^2 + b^2) \) and \( (a + b)^2 \) are in A.P.

Question. If the 17th term of an A.P. exceeds its 10th term by 7, find the common difference.
Answer: Given : \( a_{17} = a_{10} + 7 \) ...(i)
Let ‘\( a \)’ be the first term of A.P. and ‘\( d \)’ be its common difference.
Then \( a_{10} = a + (10 - 1)d = a + 9d \)
and \( a_{17} = a + (17 - 1)d = a + 16d \)
put the values in (i), we get
\( a + 16d = a + 9d + 7 \)
\( \Rightarrow 7d = 7 \Rightarrow d = 1 \)
Hence, the common difference of given A.P. is 1.

Question. The common difference between the terms of two APs is same. If the difference between their 50th terms is 100, what is the difference between their 100th terms?
Answer: Let \( a_1 \) and \( a_2 \) be the first terms of two APs and \( d \) be their common difference.
\( (a_1 + 49d) - (a_2 + 49d) = 100 \)
\( a_1 - a_2 = 100 \) ...(i)
Then, difference between their 100th terms is
\( (a_1 + 99d) - (a_2 + 99d) = a_1 - a_2 = 100 \) [Using (i)]
The difference between their 100th terms is 100 i.e., same as difference in 50th terms.

Question. Find the mean of first eleven natural numbers.
Answer: Mean = \( \frac{1 + 2 + 3 + ... + 11}{11} = \frac{\frac{11(11 + 1)}{2}}{11} = \frac{11 \times 12}{2 \times 11} = 6 \)
[Sum of first \( n \) numbers \( \frac{n(n + 1)}{2} \)]

Short Answer Type Questions

Question. Find the number of terms in the A.P. : 18, \( 15\frac{1}{2} \), 13, ..., – 47.
Answer: Given A.P. is 18, \( 15\frac{1}{2} \), 13, ....., – 47
Here, first term, \( a = 18 \)
Common difference, \( d = \frac{31}{2} - 18 = 15.5 - 18 = -2.5 = \frac{-5}{2} \)
last term, \( a_n = - 47 \)
Now, \( a_n = a + (n - 1)d \)
\( - 47 = 18 + (n - 1) \times \frac{-5}{2} \)
\( (n - 1) \times \frac{-5}{2} = - 65 \)
\( (n - 1) = 26 \Rightarrow n = 27 \)
Hence, the number of terms in the given A.P. is 27.

Question. If the \( n \)th term of the A.P. –1, 4, 9, 14, .... is 129, find the value of \( n \).
Answer: Given, A.P. is –1, 4, 9, 14, ......
\( n \)th term, \( a_n = 129 \)
Here, first term, \( a = -1 \)
Common difference, \( d = 4 - (-1) = 5 \)
\( \therefore a + (n - 1)d = 129 \)
\( \Rightarrow -1 + (n - 1)5 = 129 \)
\( \Rightarrow -1 + 5n - 5 = 129 \)
\( \Rightarrow 5n - 6 = 129 \)
\( \Rightarrow 5n = 135 \)
\( \Rightarrow n = 27 \)
Hence, the value of \( n \) is 27.

Question. Find the common difference of the Arithmetic Progression (A.P.) \( \frac{1}{a}, \frac{3-a}{3a}, \frac{3-2a}{3a}, ... (a \neq 0) \)
Answer: Given: Arithmetic progression (AP) is \( \frac{1}{a}, \frac{3-a}{3a}, \frac{3-2a}{3a}, ... (a \neq 0) \)
In the given progression,
\( a_1 = \frac{1}{a}, a_2 = \frac{3-a}{3a}, a_3 = \frac{3-2a}{3a} \)
Common difference,
\( d = a_2 - a_1 = a_3 - a_2 \)
\( = \frac{3-a}{3a} - \frac{1}{a} \)
\( = \frac{3 - a - 3}{3a} \)
\( = \frac{-a}{3a} \)
\( = \frac{-1}{3} \)
Hence, the common difference of the A.P. is \( \frac{-1}{3} \).

Question. Justify whether it is true to say that –1, \( \frac{-3}{2} \), –2, \( \frac{5}{2} \), ... form an AP as \( a_2 – a_1 = a_3 – a_2 \).
Answer: False
Explanation: The given series of numbers is –1, \( \frac{-3}{2} \), –2, \( \frac{5}{2} \), ...
Here, \( a_1 = -1, a_2 = \frac{-3}{2}, a_3 = -2, a_4 = \frac{5}{2}, ... \)
Difference between two successive terms
\( a_2 - a_1 = \frac{-3}{2} - (-1) = \frac{-3}{2} + 1 = \frac{-1}{2} \)
\( a_3 - a_2 = -2 - (\frac{-3}{2}) = -2 + \frac{3}{2} = \frac{-1}{2} \)
\( a_4 - a_3 = \frac{5}{2} - (-2) = \frac{5}{2} + 2 = \frac{9}{2} \)
We can see that \( a_2 - a_1 = a_3 - a_2 = \frac{-1}{2} \), but \( a_4 - a_3 = \frac{9}{2} \).
Clearly, the difference of two successive terms is not the same, hence we can say that the given list of numbers does not form an AP.

Question. How many 2-digit numbers are divisible by 3?
Answer: Two digits numbers divisible by 3 are 12, 15, 18 ..., 99
Here, first term, \( a = 12 \)
Common difference, \( d = 15 - 12 = 3 \)
Last term, \( a_n = 99 \)
Now, \( n \)th term, \( a_n = a + (n - 1)d \), where, ‘\( n \)’ is the number of the terms
\( \Rightarrow 99 = 12 + (n - 1)3 \)
\( \Rightarrow 99 = 12 + 3n - 3 \)
\( \Rightarrow 3n = 90 \)
\( \Rightarrow n = 30 \)
Hence, the number of terms are 30.

Question. For the AP: -3, -7, -11, ..., can we directly find \( a_{30} - a_{20} \) without actually finding \( a_{30} \) and \( a_{20} \)? Give reasons for your answer.
Answer: Yes, we can find.
The given list of numbers of an AP –3, –7, –11 ...
We know that \( a_n = a + (n - 1)d \)
\( \Rightarrow a_{30} = a + (30 - 1)d = a + 29d \)
\( \Rightarrow a_{20} = a + (20 - 1)d = a + 19d \)
Now, \( a_{30} - a_{20} = (a + 29d) - (a + 19d) \)
\( = a + 29d - a - 19d = 10d \) ...(i)
For the given series, common difference,
\( d = -7 - (-3) = -7 + 3 = -4 \)
Hence, \( a_{30} - a_{20} = 10d = 10(-4) = -40 \) [Using eqn. (i)]

Question. How many multiples of 4 lie between 10 and 205?
Answer: Multiples of 4 between 10 and 205 are 12, 16, 20, 24, 28, ......, 204.
Let, the number of multiples be ‘\( n \)’.
Here, first term, \( a = 12 \)
Common difference, \( d = 4 \)
Last term, \( a_n = 204 \)
Since, \( n \)th term of an A.P. is \( a_n = a + (n - 1)d \)
\( \Rightarrow 204 = 12 + (n - 1) \times 4 \)
\( \Rightarrow 204 = 12 + 4n - 4 \)
\( \Rightarrow 4n = 204 - 8 \)
\( \Rightarrow n = \frac{196}{4} = 49 \)
Hence, the number of multiples of 4 are 49.

Question. Determine the A.P. whose third term is 16 and 7th term exceeds the 5th term by 12.
Answer: Let, the first term of an A.P. be ‘\( a \)’ and common difference be ‘\( d \)’.
Given, \( a_3 = 16 \)
i.e., \( a + (3 - 1)d = 16 \)
\( \Rightarrow a + 2d = 16 \) ...(i)
and \( a_7 = a_5 + 12 \) (given)
\( a + (7 - 1)d = a + (5 - 1)d + 12 \)
\( \Rightarrow a + 6d = a + 4d + 12 \)
\( \Rightarrow 2d = 12 \)
\( \Rightarrow d = 6 \)
If we put the value of \( d \) in equation (i), we get
\( a + 2 \times 6 = 16 \)
\( \Rightarrow a = 4 \)
Hence, the required A.P. is 4, 10, 16, 20........

Question. Which term of the AP 3, 15, 27, 39, .... will be 120 more than its 21st term?
Answer: Given: A.P. is 3, 15, 27, 39, .......
Here, the first term, \( a = 3 \)
Common difference, \( d = 15 - 3 = 27 - 15 = 12 \).
Now, 21st term
\( a_{21} = 3 + (21 - 1) \times 12 \)
\( = 3 + 20 \times 12 \)
\( = 3 + 240 = 243 \)
According to the given condition :
\( a_n = a_{21} + 120 \)
where, \( a_n \) is the \( n \)th term
\( a_n = 243 + 120 = 363 \)
\( \Rightarrow a + (n - 1)d = 363 \)
\( \Rightarrow 3 + (n - 1) \times 12 = 363 \)
\( \Rightarrow (n - 1) = \frac{360}{12} \)
\( \Rightarrow n = 31 \)
Hence, the 31st term is 120 more than the 21st term.

Question. If \( S_n \), the sum of first \( n \) terms of an AP is given by \( S_n = 3n^2 - 4n \), find the \( n \)th term.
Answer: Here, \( S_n = 3n^2 - 4n \)
\( \therefore S_{n - 1} = 3 (n - 1)^2 - 4(n - 1) \)
\( = 3(n^2 - 2n + 1) - 4n + 4 \)
\( = 3n^2 - 10n + 7 \)
Hence, the \( n \)th term, \( a_n = S_n - S_{n - 1} \)
\( = (3n^2 - 4n) - (3n^2 - 10n + 7) \)
\( = 3n^2 - 4n - 3n^2 + 10n - 7 \)
\( a_n = 6n - 7 \)
Hence, the \( n \)th term is \( 6n - 7 \).

Question. Find the sum of first 8 multiples of 3.
Answer: First 8 multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, which forms an A.P.
Here, first term, \( a = 3 \)
Common difference, \( d = 3 \)
Number of terms, \( n = 8 \)
\( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( S_8 = \frac{8}{2} [2 \times 3 + (8 - 1)3] \)
\( = 4[6 + 21] = 4 \times 27 = 108 \)
Thus, the sum of the first 8 multiples is 108.

Question. If seven times the 7th term of an A.P. is equal to eleven times the 11th term, then what will be its 18th term?
Answer: Let the first term of the A.P. be ‘\( a \)’ and its common difference be ‘\( d \)’.
Given, \( 7a_7 = 11a_{11} \)
Then, \( 7(a + 6d) = 11(a + 10d) \)
\( \Rightarrow 7a + 42d = 11a + 110d \)
\( \Rightarrow 7a - 11a = 110d - 42d \)
\( \Rightarrow - 4a = 68d \)
\( \Rightarrow a = - 17d \)
Now, 18th term of A.P.
\( a_{18} = a + (18 - 1)d \)
\( = a + 17d \)
putting \( a = - 17d \)
\( = -17d + 17d = 0 \)
Hence, the 18th term of A.P. is 0.

Question. The 10th term of an A.P. is –4 and its 22nd term is (– 16). Find its 38th term.
Answer: Let, the first term of A.P. be ‘\( a \)’ and its common difference be ‘\( d \)’.
\( a_{10} = - 4 \) (given)
i.e. \( a + 9d = - 4 \) ...(i)
\( a_{22} = - 16 \) (given)
\( a + 21d = - 16 \) ...(ii)
On solving (i) and (ii), we get
\( 12d = -12 \Rightarrow d = -1 \)
If we put the value of \( d = - 1 \) in equation (i), we get
\( a = - 4 + 9 = 5 \)
\( \therefore \) first term, \( a = 5 \)
and common difference, \( d = - 1 \)
38th term, \( a_{38} = a + (38 - 1)d \)
\( = 5 + 37(- 1) = - 32 \)
Hence, the 38th term of the A.P. is – 32.

Question. Find how many integers between 200 and 500 are divisible by 8.
Answer: Integers between 200 and 500 divisible by 8 are 208, 216, 224, ..., 496.
This series forms an A.P., where first term, \( a = 208 \), common difference, \( d = 8 \), and last term, \( l = 496 \)
Let, the number of integers be ‘\( n \)’
Then, \( n \)th term = \( a_n \) (last term \( l \)) = \( a + (n - 1)d \)
\( 496 = 208 + (n - 1) \times 8 \)
\( \Rightarrow (n - 1) \times 8 = 496 - 208 \)
\( \Rightarrow (n - 1) = \frac{288}{8} \)
\( \Rightarrow n = 37 \)
Hence, the number of terms are 37.

Question. Determine the AP whose third term is 5 and the seventh term is 9.
Answer: Given, third term of A.P., \( a_3 = 5 \)
Seventh term of A.P., \( a_7 = 9 \)
Let the first term of A.P. be ‘\( a \)’ and its common difference be ‘\( d \)’.
Now, \( a_3 = a + (3 - 1)d \)
or \( 5 = a + 2d \) ...(i)
and \( a_7 = a + (7 - 1)d \)
or \( 9 = a + 6d \) ...(ii)
On solving (i) and (ii), we get:
\( 4d = 4 \Rightarrow d = 1 \)
If we put the value of ‘\( d \)’ in equation (i), we get:
\( 5 = a + 2 \times 1 \Rightarrow a = 3 \)
Hence, the A.P. is 3, 4, 5, 6.

Long Short Answer Type Questions

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, \( S_5 \)
\( S_5 = \frac{5}{2} [2a + (5 - 1)d] = \frac{5}{2} [2a + 4d] \)
\( S_5 = 5[a + 2d] \)
\( S_5 = 5a + 10d \quad ...(i) \)
Sum of first seven terms, \( S_7 \)
\( S_7 = \frac{7}{2} [2a + (7 - 1)d] \)
\( = \frac{7}{2} [2a + 6d] \)
\( S_7 = 7[a + 3d] \)
\( S_7 = 7a + 21d \quad ...(ii) \)
Now, by given condition
\( S_5 + S_7 = 167 \)
\( \Rightarrow 5a + 10d + 7a + 21d = 167 \)
[using equation (i) & equation (ii)]
\( \Rightarrow 12a + 31d = 167 \quad ...(iii) \)
Also, it is given that sum of first 10 terms of this AP is 235.
\( \Rightarrow S_{10} = 235 \)
\( \Rightarrow \frac{10}{2} [2a + (10 - 1)d] = 235 \)
\( \Rightarrow 5[2a + 9d] = 235 \)
\( \Rightarrow 2a + 9d = \frac{235}{5} = 47 \)
\( \Rightarrow 2a + 9d = 47 \quad ...(iv) \)
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 the value of \( d \) in equation (4), we get
\( 2a + 9d = 47 \)
\( \Rightarrow 2a + 9(5) = 47 \)
\( \Rightarrow 2a + 45 = 47 \)
\( \Rightarrow 2a = 2 \Rightarrow a = 1 \)
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
\( S_n = \frac{n}{2} [a + L] \quad ...(i) \)
Also, \( L = a + (n - 1) d \)
\( \Rightarrow 490 = 10 + (n - 1)10 \)
\( \Rightarrow 480 = (n - 1)10 \)
\( \Rightarrow (n - 1) = 48 \)
\( \Rightarrow n = 49 \)
Putting this value in equation (i), we get
\( S_{49} = \frac{49}{2} [10 + 490] \)
\( = \frac{49}{2} \times 500 = \frac{49}{2} \times 250 = 12250 \)
\( \Rightarrow S_{49} = 12250 \)
(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]
\( 500 = 10 + (n - 1)d \)
\( 490 = (n - 1)10 \)
\( \Rightarrow (n - 1) = 49 \Rightarrow n = 50 \)
Also we know that \( S_n = \frac{n}{2} (a + L) \)
\( \Rightarrow S_{50} = \frac{50}{2} (10 + 500) \)
\( = 25 \times 510 = 12750 \)
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]
\( 1^{st} \) 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, \( L = a + (n_1 - 1)d \)
\( 500 = 2 + (n_1 - 1)(2) \)
\( 498 = (n_1 - 1)2 \)
\( \Rightarrow (n_1 - 1) = 249 \Rightarrow n_1 = 250 \)
Sum of [2, 4, 6, ... 500]
\( S_{n_1} = \frac{n_1}{2} [a + l] \)
\( = \frac{250}{2} [2 + 500] \)
\( S_{n_1} = 225 \times 502 = 62750 \) (Note: 250/2 is 125, but OCR says 225. Let's recalculate: 125 * 502 = 62750. So 225 is typo, 125 is correct. Wait, \( S_{n_1} \) calculation: \( \frac{250}{2} = 125 \). \( 125 \times 502 = 62750 \). Correct. The text has a typo '225' but the result '62750' is correct.)
\( 2^{nd} \) 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 \( a_n = a + (n - 1)d \)
\( L' = a' + (n_2 - 1) d' \)
\( 500 = 5 + (n_2 - 1)5 \)
\( \Rightarrow 495 = (n_2 - 1)5 \)
\( \Rightarrow n_2 - 1 = 99 \)
\( \Rightarrow n_2 = 100 \)
Sum of \( 2^{nd} \) List,
\( S_n = \frac{n}{2} (a + L) \)
\( S_{n_2} = \frac{n_2}{2} (5 + 500) \)
\( = \frac{100}{2} \times 505 = 25250 \)
\( 3^{rd} \) 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
\( L'' = a'' + (n_3 - 1) d'' \)
\( 500 = 10 + (n_3 - 1)(10) \)
\( 490 = (n_3 - 1)10 \)
\( \Rightarrow n_3 - 1 = 49 \)
\( \Rightarrow n_3 = 50 \)
Sum of 3rd List,
\( S_{n_3} = \frac{n_3}{2} (10 + 500) \)
\( = \frac{50}{2} \times 510 = 12750 \)
\( \Rightarrow \) Sum of multiples of 2 or 5
\( = S_{n_1} + S_{n_2} - S_{n_3} \)
\( = 62750 + 25250 - 12750 \)
\( = 88000 - 12750 = 75250 \)

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 \( a_n = a + (n - 1) d \)
\( \Rightarrow (a + 17d) + (a + 18d) + (a + 19d) = 225 \)
\( \Rightarrow 3a + 54d = 225 \)
\( \Rightarrow a + 18d = 75 \quad ...(i) \)
Also, it is given that sum of last 3 terms = 429
\( \Rightarrow a_{35} + a_{36} + a_{37} = 429 \)
\( \Rightarrow (a + 34d) + (a + 35d) + (a + 36d) = 429 \)
\( \Rightarrow 3a + 105d = 429 \)
\( \Rightarrow a + 35d = 143 \quad ...(ii) \)
Subtracting equation (i) from equation (ii), we get
\( (a + 35d) - (a + 18d) = 143 - 75 \)
\( a + 35d - a - 18d = 68 \)
\( \Rightarrow 17d = 68 \Rightarrow d = 4 \)
Putting value of \( d \) in equation (i), we get
\( a + 18(4) = 75 \)
\( \Rightarrow a + 72 = 75 \)
\( \Rightarrow a = 3, d = 4 \)
Required AP is
\( a, a + d, a + 2d, a + 3d, ... \)
\( 3, 3 + 4, 3 + 2(4), 3 + 3(4), ... \)
\( 3, 7, 11, 15, ... \)

Question. Given \( S_p = q \) and \( S_q = p \). Prove that \( S_{p+q} = -(p + q) \).
Answer: Proof:
\( S_p = \frac{p}{2} [2a + (p – 1)d] = q \)
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
\( [(p – 1) – (q – 1)]d = \frac{2q}{p} - \frac{2p}{q} \)
\( \Rightarrow (p - q)d = \frac{2q^2 - 2p^2}{pq} = \frac{2(q^2 - p^2)}{pq} \)
\( \Rightarrow d = \frac{2(q - p)(q + p)}{pq(p - q)} \)
\( d = \frac{-2(p + q)}{pq} \) ...(iii)
and substituting d in (i):
\( 2a = \frac{2q}{p} - (p-1)d \)
\( a = \frac{q}{p} - \frac{(p-1)}{2}d \)
Using the value of d, we can find \(S_{p+q}\):
Now, \( S_{p + q} = \frac{p + q}{2} [2a + (p + q – 1)d] \)
\( = \frac{p + q}{2} [2a + (p - 1)d + qd] \)
From (i), \( 2a + (p-1)d = \frac{2q}{p} \)
\( = \frac{p + q}{2} [\frac{2q}{p} + qd] \)
\( = \frac{p + q}{2} [\frac{2q}{p} + q(\frac{-2(p+q)}{pq})] \)
\( = \frac{p + q}{2} [\frac{2q}{p} - \frac{2(p+q)}{p}] \)
\( = \frac{p + q}{2} [\frac{2q - 2p - 2q}{p}] \)
\( = \frac{p + q}{2} [\frac{-2p}{p}] \)
\( = \frac{p + q}{2} [-2] = -(p + q) \)
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 \( a_n = a + (n – 1)d \)
\( \Rightarrow L = a + (n – 1)d \)
\( \Rightarrow 198 = 108 + (n – 1)9 \)
\( \Rightarrow 198 – 108 = (n – 1)9 \)
\( \Rightarrow 90 = (n – 1)9 \)
\( \Rightarrow (n – 1) = 10 \Rightarrow n = 11 \)
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
\( S_{11} = \frac{11}{2} [2(108) + (11 – 1)(9)] \)
\( = \frac{11}{2} [2(108) + 10 \times 9] \)
\( = 11 [108 + 5 \times 9] \)
\( = 11 [108 + 45] \)
\( = 11 \times 153 = 1683 \)
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 (i), 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 \( a_n = a + (n – 1)d \)
\( \Rightarrow L = a + (n – 1)d \)
\( 199 = 101 + (n – 1)(1) \)
\( 199 – 101 = (n – 1) \)
\( n = 99 \)
\( \therefore \) Sum of 99 terms between 100 and 200
\( S_{99} = \frac{99}{2} [2(101) + (99 – 1)(1)] \)
\( = \frac{99}{2} [202 + 98] \)
\( = \frac{99}{2} \times 300 \)
\( S_{99} = 99 \times 150 = 14850 \)
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, \( d = (– 12) – (– 7) = (–12 + 7) = – 5 \)
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
\( \therefore a_m = – 100 \)
\( a + (m – 1) d = – 100 \)
\( \Rightarrow – 7 + (m – 1) (– 5) = – 100 \)
\( \Rightarrow (m – 1) 5 = 93 \)
\( \Rightarrow m = \frac{93}{5} + 1 = \frac{98}{5} \notin \mathbb{N} \)
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 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 \( 2^{nd} \) term 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 \( a_n = a + (n – 1)d \)
\( \Rightarrow L = a + (n – 1)(b – a) \)
\( \Rightarrow c = a + (n – 1)(b – a) \)
\( \Rightarrow (c – a) = (n – 1)(b – a) \)
\( \Rightarrow (n – 1) = \frac{c - a}{b - a} \) ...(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
\( S_n = \frac{n}{2} [2a + (n – 1)d] \)
\( = \frac{b + c - 2a}{2(b - a)} \left[ 2a + \frac{c - a}{b - a}(b - a) \right] \) [using equation (i) and equation (ii)]
\( = \frac{(b + c - 2a)}{2(b - a)} [2a + c - a] \)
\( = \frac{(b + c - 2a)(a + c)}{2(b - a)} \)
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, \( S_n = \frac{n}{2} [2a + (n – 1)d] \)
where, ‘n’ is the number of terms.
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 = \( \frac{n}{2} [2 \times 7 + (n – 1) \times 2] \)
\( = \frac{n}{2} [14 + 2 (n – 1)] \)
\( = n[7 + n – 1] \)
\( = n(n + 6) \) 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:
\( a + 7d = 22 \)
\( a + 5d = 12 \)
(-) (-)
-----------
\( 2d = 10 \Rightarrow d = 5 \)
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)
\( \Rightarrow 3(a + 10d) = 2(a + 7d) \) Note: OCR error says 2(a+7d), but mathematically \( a_{18} \) is \( a+17d \). Let's recompute based on correct math from screenshot equation logic.
Wait, screenshot shows \( 3(a+10d) = 2(a+7d) \). This might be a typo in the book or OCR. Let's trace back. \( 3a + 30d = 2a + 34d \) implies the RHS was \( 2(a+17d) \).
\( \Rightarrow 3a + 30d = 2a + 34d \)
\( \Rightarrow a = 4d \) ...(i)
Now, sum of first five terms,
\( S_5 = \frac{n}{2} [2a + (n – 1)d] \)
\( = \frac{5}{2} [2 \times 4d + 4 \times d] \) [from (i)]
\( = \frac{5}{2} \times 12d = 30d \)
Sum of first 10 terms,
\( S_{10} = \frac{n}{2} [2a + (n – 1)d] \)
\( = \frac{10}{2} [2 \times 4d + 9 \times d] \)
\( = 5(8d + 9d) \)
\( = 85d \)
\( \therefore \frac{S_5}{S_{10}} = \frac{30d}{85d} = \frac{6}{17} \)
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’ 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)
\( \therefore \frac{\frac{m}{2}[2a + (m - 1)d]}{\frac{n}{2}[2a + (n - 1)d]} = \frac{m^2}{n^2} \)
\( \Rightarrow \frac{2a + (m - 1)d}{2a + (n - 1)d} = \frac{m}{n} \)
\( \Rightarrow n[2a + (m – 1)d] = m[2a + (n – 1)d] \)
\( \Rightarrow 2an + mnd – nd = 2am + mnd – md \)
\( \Rightarrow md – nd = 2am - 2an = 2a(m – n) \)
\( \Rightarrow d(m - n) = 2a(m - n) \)
\( \Rightarrow d = 2a \)
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 \( a_n = a + (n – 1)d \)
\( \Rightarrow L = x = a + (n – 1)(d) \)
\( \Rightarrow x = –4 + (n – 1)(3) \)
\( \Rightarrow x = –4 + 3n – 3 \)
\( \Rightarrow n = \frac{x + 7}{3} \)
Also, we know that
\( S_n = \frac{n}{2} [2a + (n – 1)d] \)
\( \Rightarrow S_n = \frac{x + 7}{2 \times 3} [2(-4) + (\frac{x + 7}{3} - 1)(3)] \)
\( = \frac{x + 7}{6} [-8 + x + 4] = \frac{(x + 7)(x - 4)}{6} \)
The given equation is (–4) + (–1) + 2... + x = 437
\( \Rightarrow S_n = 437 \)
\( \Rightarrow \frac{(x + 7)(x - 4)}{6} = 437 \)
\( \Rightarrow x^2 – 4x + 7x – 28 = 437 \times 6 \)
\( \Rightarrow x^2 + 3x – 28 = 2622 \)
\( \Rightarrow x^2 + 3x – 2650 = 0 \)
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 runs 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 = \( 100n \)
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, he runs for \( (n – 1) \) minutes.
Total distance covered by the policeman
= 100 + 110 + 120 + ... ((n – 1) terms)
\( \therefore 100n = \frac{n - 1}{2} [2(100) + ((n - 1) - 1) \times 10] \)
\( \left[ \because S_n = \frac{n}{2}[2a + (n-1)d] \right] \)
\( \Rightarrow 100n = \frac{n - 1}{2} [200 + (n - 2) \times 10] \)
\( \Rightarrow 200n = (n - 1) [200 + 10n - 20] \)
\( \Rightarrow 200n = (n - 1) (180 + 10n) \)
\( \Rightarrow 200n = 180n – 180 + 10n^2 – 10n \)
\( \Rightarrow 10n^2 – 30n – 180 = 0 \)
\( \Rightarrow n^2 – 3n – 18 = 0 \)
\( \Rightarrow (n – 6) (n + 3) = 0 \)
\( \Rightarrow n = 6 \) [\( \because n = – 3 \), is not possible]
Hence, the policeman takes \( (n – 1) = 5 \) minutes to catch the thief.

Question. 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
\( L = a + (n – 1)d \)
\( \frac{13}{3} = -\frac{4}{3} + (n - 1) \left(\frac{1}{3}\right) \)
\( \Rightarrow 13 = –4 + (n – 1) \)
\( \Rightarrow (n – 1) = 17 \)
\( \Rightarrow n = 18 \)
So, the two middle most terms are \( \left(\frac{n}{2}\right)^{th} \) and \( \left(\frac{n}{2} + 1\right)^{th} \) as numbers of terms are even.
The two middle most terms are \( \left(\frac{18}{2}\right)^{th} \) and \( \left(\frac{18}{2} + 1\right)^{th} \)
i.e., \( 9^{th} \) and \( 10^{th} \) term
\( a_9 = a + (9 – 1)d = a + 8d \)
\( = -\frac{4}{3} + 8\left(\frac{1}{3}\right) = \frac{-4 + 8}{3} = \frac{4}{3} \)
\( a_{10} = a + (10 – 1)d = a + 9d \)
\( = -\frac{4}{3} + 9\left(\frac{1}{3}\right) = \frac{-4 + 9}{3} = \frac{5}{3} \)
So, the sum of the two middle most terms = \( a_9 + a_{10} \)
\( = \frac{4}{3} + \frac{5}{3} = \frac{9}{3} = 3 \)

Question. 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, \( S_7 = \frac{7}{2} [2a + (7 - 1) \times d] \)
\[ [\because S_n = \frac{n}{2} [2a + (n - 1)d] \] \( \Rightarrow 49 = \frac{7}{2} [2a + 6d] \)
\( \Rightarrow 7 = a + 3d \quad ...(i) \)
and \( S_{17} = \frac{17}{2} [2a + (17 - 1) \times d] \)
\( \Rightarrow 289 = \frac{17}{2} \times [2a + 16 \times d] \)
\( \Rightarrow 17 = a + 8d \quad ...(ii) \)
On subtracting equation (i) from equation (ii), we get
\( a + 8d = 17 \)
\( a + 3d = 7 \)
(-) (-) (-)
----------------
\( 5d = 10 \)
\( d = 2 \)
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 \( n^2 \).