CBSE Class 10 Arithmetic Progressions Sure Shot Questions Set C

Short Answer Type Questions  

Question. Determine the A.P. whose 3rd term is 5 and the 7th term is 9.
Answer: We have,
\( a_3 = a + (3 – 1) d = a + 2d = 5 \) …(1)
and \( a_7 = a + (7 – 1) d = a + 6d = 9 \) …(2)
Solving the pair of linear equations (1) and (2), we get
\( a = 3, d = 1 \)
Hence, the required A.P. is 3, 4, 5, 6, 7,.....

Question. The sum of the first \( n \) terms of an A.P. is \( 5n – n^2 \). Find the \( n^{th} \) term of this A.P.
Answer: We have, \( S_n = 5n – n^2 \)
\( \therefore S_{n – 1} = 5(n – 1) – (n – 1)^2 \)
\( = 5n – 5 – (n^2 + 1 – 2n) = –n^2 + 7n – 6 \)
\( n^{th} \) term of A.P., \( a_n = S_n – S_{n – 1} \)
\( = 5n – n^2 – (–n^2 + 7n – 6) \)
\( = 5n – n^2 + n^2 – 7n + 6 = –2n + 6 \)

Question. In the following A.P., find the missing terms \( 3, —, —, \frac{9}{2} \).
Answer: Here, \( a = 3, a_4 = \frac{9}{2} \)
Now, \( a_4 = \frac{9}{2} \Rightarrow a + 3d = \frac{9}{2} \Rightarrow 3 + 3d = \frac{9}{2} \)
\( \Rightarrow 3d = \frac{3}{2} \Rightarrow d = \frac{1}{2} \)
\( \therefore \) Missing terms are, \( a_2 = a + d = 3 + \frac{1}{2} = \frac{7}{2} \)
and \( a_3 = a + 2d = 3 + 2 \left( \frac{1}{2} \right) = 4 \)

Question. Find how many two-digit numbers are divisible by 6.
Answer: The two-digit numbers that are divisible by 6 are 12, 18, 24, …, 96.
It forms an A.P. with \( a = 12, d = 6 \)
Since, \( n^{th} \) term of an A.P. is \( a_n = a + (n − 1)d \)
\( \therefore 12 + (n − 1) \times 6 = 96 \)
\( \Rightarrow 2 + (n – 1) = 16 \Rightarrow n = 14 + 1 = 15 \)
Thus, there are 15 two-digit numbers that are divisible by 6.

Question. Which term of the A.P. 6, 13, 20, 27, ...... is 98 more than its 24th term?
Answer: The A.P. is 6, 13, 20, 27, .....
\( a = 6, d = 7 \)
\( \therefore t_{24} = a + 23d = 6 + 23 \times 7 \)
\( \Rightarrow t_{24} = 6 + 161 \Rightarrow t_{24} = 167 \)
Now, \( t_n = t_{24} + 98 = 167 + 98 = 265 \)
\( \Rightarrow 6 + 7 (n – 1) = 265 \)
\( \Rightarrow 7 (n – 1) = 259 \Rightarrow n – 1 = 37 \Rightarrow n = 38 \)
Hence, \( t_{38} \) is the required term.

Question. A sum of ₹ 3200 is to be used to award four prizes. If each prize after the first prize is ₹ 400 less than the preceding prize, then find the value of each of the prizes.
Answer: Let the value of four prizes (in ₹) be \( a – 3d, a – d, a + d \) and \( a + d \).
Here common difference is \( 2d \).
Also, Sum = ₹ 3200 [Given]
\( \Rightarrow a – 3d + a – d + a + d + a + 3d = 3200 \)
\( \Rightarrow 4a = 3200 \Rightarrow a = 800 \)
Also, common difference = 400 [Given]
\( \Rightarrow 2d = 400 \Rightarrow d = 200 \)
\( \therefore \) Value of each prize (in ₹) is
\( 800 – 3(200), 800 – 200, 800 + 200 \) and \( 800 + 3(200) \)
or 200, 600, 1000 and 1400

Question. Which term of the A.P. 3, 14, 25, 36, … will be 99 more than its 25th term?
Answer: Let \( a \) be the first term and \( d \) be the common difference of given A.P.
\( \therefore a = 3 \) and \( d = 14 − 3 = 11 \)
Let \( a_n \) be the required term.
Then, \( a_n − a_{25} = 99 \)
\( \Rightarrow a + (n − 1)d − a − (25 − 1)d = 99 \)
\( \Rightarrow (n − 1) \times 11 − 24 \times 11 = 99 \)
\( \Rightarrow n − 25 = 9 \Rightarrow n = 34 \)
\( \therefore 34^{th} \) term will be 99 more than its \( 25^{th} \) term.

Question. Find the sum of all three digit natural numbers, which are multiples of 11.
Answer: The sequence of three digit numbers which are divisible by 11 is 110, 121, 132, …, 990.
It forms an A.P. with first term, \( a = 110 \) and common difference, \( d = 121 – 110 = 11 \).
Let there be \( n \) terms in the sequence.
\( \therefore a_n = 990 \Rightarrow a + (n – 1)d = 990 \)
\( \Rightarrow 110 + (n – 1)11 = 990 \)
\( \Rightarrow 110 + 11n – 11 = 990 \Rightarrow 11n + 99 = 990 \)
\( \Rightarrow 11n = 990 – 99 = 891 \Rightarrow n = 81 \)
Now, \( S_n = \frac{n}{2} [a + l] \)
\( \therefore \) Required sum \( = \frac{81}{2} [110 + 990] \)
\( = \frac{81}{2} \times 1100 = 44550 \)

Question. Split 69 into three parts such that they are in A.P. and the product of two smaller parts is 483.
Answer: Let the three parts of 69 which are in A.P. be \( a – d, a, a + d \).
Then, sum of three parts = 69
i.e., \( a – d + a + a + d = 69 \)
\( \Rightarrow 3a = 69 \Rightarrow a = 23 \)
Also, the product of two smaller parts = 483
\( \Rightarrow (a – d) \times a = 483 \) …(1)
Substituting \( a = 23 \) in (1), we get
\( (23 – d) \times 23 = 483 \)
\( \Rightarrow (23 – d) = \frac{483}{23} = 21 \Rightarrow d = 23 – 21 = 2 \)
Thus, the three parts of 69 are 21, 23, 25

Question. For what value of \( n \) is the \( n^{th} \) term of the following two A.P.’s is same? The two A.P.s are 1, 7, 13, 19, ..... and 69, 68, 67, .....
Answer: Clearly, 1, 7, 13, 19, ..... forms an A.P. with first term 1 and common difference 6. Therefore, its \( n^{th} \) term is given by
\( a_n = 1 + (n – 1) \times 6 = 6n – 5 \)
Also, 69, 68, 67, 66, ..... forms an A.P. with first term 69 and common difference – 1.
so, its \( n^{th} \) term is given by
\( a'_n = 69 + (n – 1) \times (– 1) = – n + 70 \)
The two A.P.’s will have identical \( n^{th} \) terms, if
\( a_n = a'_n \)
\( \Rightarrow 6n – 5 = – n + 70 \Rightarrow 7n = 75 \)
\( \Rightarrow n = 75/7 \), which is not a natural number.
Hence, there is no value of \( n \) for which the two A.P.’s will have identical terms.

Short Answer Type Questions (SA-II)

Question. The sum of the 2nd and the 7th term of an A.P. is 30. If its 15th term is 1 less than twice its 8th term, find the A.P.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the A.P.
Now, \( a_2 + a_7 = 30 \Rightarrow a + d + a + 6d = 30 \)
\( \Rightarrow 2a + 7d = 30 \) ...(i)
Also, \( a_{15} = 2a_8 – 1 \)
\( \Rightarrow a + 14d = 2(a + 7d) – 1 \Rightarrow a + 14d = 2a + 14d – 1 \)
\( \Rightarrow a = 1 \) ...(ii)
From (i) and (ii), we get
\( 2 + 7d = 30 \Rightarrow 7d = 28 \Rightarrow d = 4 \)
Hence, the A.P is formed as 1, 5, 9, ....

Question. If \( S_n \) denotes the sum of first \( n \) terms of an A.P., prove that \( S_{30} = 3[S_{20} – S_{10}] \).
Answer: Let \( a \) be the first term and \( d \) be the common difference of the A.P.
Now, sum of \( n \) terms, \( S_n = \frac{n}{2} [2a + (n – 1)d] \)
\( \therefore S_{30} = \frac{30}{2} [2a + (30 – 1)d] = 15 [2a + 29d] \)
\( = 30a + 435d \)
\( S_{20} = \frac{20}{2} [2a + (20 – 1)d] = 10 [2a + 19d] \)
\( = 20a + 190d \)
\( S_{10} = \frac{10}{2} [2a + (10 – 1)d] = 5 [2a + 9d] \)
\( = 10a + 45d \)
\( \therefore 3[S_{20} – S_{10}] = 3[20a + 190d – 10a – 45d] \)
\( = 30a + 435d = S_{30} \)

Question. If \( m \) times the \( m^{th} \) term of an A.P. is equal to \( n \) times its \( n^{th} \) term, then find the \( (m + n)^{th} \) term of the A.P.
Answer: Let the first term and the common difference of the A.P. are \( a \) and \( d \) respectively.
According to question, we have
\( m \cdot a_m = n \cdot a_n \)
\( \Rightarrow m \cdot [a + (m – 1)d] = n \cdot [a + (n – 1)d] \)
\( \Rightarrow ma + (m^2 – m)d = na + (n^2 – n)d \)
\( \Rightarrow (m – n)a + (m^2 – m – n^2 + n)d = 0 \)
\( \Rightarrow (m – n)a + [(m – n)(m + n) – (m – n)]d = 0 \)
\( \Rightarrow (m – n)a + (m – n)(m + n – 1)d = 0 \)
\( \Rightarrow (m – n)[a + (m + n – 1)d] = 0 \)
\( \Rightarrow a + (m + n – 1)d = 0 \) (\( \because m \neq n \))
\( \Rightarrow a_{m + n} = 0 \)
Thus, the \( (m + n)^{th} \) term of the A.P. is 0.

Question. Find the value of the middle term of the following A.P. : \( -6, -2, 2, ……, 58 \).
Answer: The given A.P. is \( -6, -2, 2, …, 58 \).
Here, first term, \( a = -6 \) and common difference, \( d = -2 - (-6) = - 2 + 6 = 4 \)
Last term, \( l = a_n = 58 \)
\( \Rightarrow a + (n – 1)d = 58 \Rightarrow –6 + (n – 1)4 = 58 \)
\( \Rightarrow (n – 1)4 = 64 \Rightarrow (n – 1) = 16 \Rightarrow n = 17 \)
Middle term of the A.P. \( = \left( \frac{n + 1}{2} \right)^{th} \) term
\( = \left( \frac{17 + 1}{2} \right)^{th} \text{ term} = 9^{th} \text{ term} \)
\( \therefore a_9 = a + (9 − 1) d = −6 + 8(4) = −6 + 32 = 26 \)

Question. Find the sum of first 40 positive integers divisible by 6.
Answer: Numbers divisible by 6 are 6, 12, 18, ...
The series form an A.P. with first term \( (a) = 6 \) and common difference \( (d) = 6 \) and \( n = 40 \)
Now, sum of \( n \) terms \( S_n = \frac{n}{2} \{ 2a + (n - 1)d \} \)
\( \therefore S_{40} = \frac{40}{2} \{ 2 \times 6 + (40 - 1) 6 \} = 20 \{12 + 234\} \)
\( = 20 \times 246 = 4920 \)

Question. Find the A.P. whose fourth term is 9 and the sum of its sixth term and thirteenth term is 40.
Answer: Given, \( a_4 = 9 \) and \( a_6 + a_{13} = 40 \)
Now \( a_4 = 9 \Rightarrow a + 3d = 9 \Rightarrow a = 9 − 3d \)
Also, \( a_6 + a_{13} = 40 \)
\( \Rightarrow (a + 5d) + (a + 12d) = 40 \)
\( \Rightarrow 2a + 17d = 40 \)
On substituting the value of \( a \), we get
\( 2 (9 − 3d) + 17d = 40 \)
\( \Rightarrow 18 + 11d = 40 \Rightarrow 11d = 22 \)
\( \Rightarrow d = 2 \therefore a = 9 − 3 (2) = 3 \)
Thus, the A.P. is 3, 5, 7, 9 …

Question. If the sum of \( n \) terms of an A.P. is \( (pn + qn^2) \), where \( p \) and \( q \) are constants, find the common difference.
Answer: Let \( a_1, a_2, a_3, ....., a_n \) be the given A.P., then
Sum \( = a_1 + a_2 + a_3 + ..... + a_n = pn + qn^2 \) …(1)
Clearly, \( S_1 \) is the sum of single term.
\( S_2 \) is the sum of the \( 1^{st} \) two terms
\( S_3 \) is the sum of the \( 1^{st} \) three terms
Putting \( n = 1, 2 \) in (1), we get
\( S_1 = a_1 = p \times (1) + q \times (1)^2 = p + q \) …(2)
\( S_2 = a_1 + a_2 = p \times (2) + q \times (2)^2 = 2p + 4q \)
\( \Rightarrow a_2 = 2p + 4q – a_1 \) …(3)
\( \Rightarrow a_2 = 2p + 4q – (p + q) = p + 3q \) (Using (2))
Now, the common difference,
\( d = a_2 – a_1 = p + 3q – (p + q) = 2q \)

Question. How many terms of the A.P. \( – 6, \frac{-11}{2}, –5, … \) are needed to give the sum –25? Explain the double answer.
Answer: Here \( a = –6, d = \frac{-11}{2} – (–6) = \frac{1}{2} \)
Let –25 be the sum of \( n \) terms of this A.P., where \( n \in N \).
Using \( S_n = \frac{n}{2} [2a + (n – 1)d] \), we have
\( –25 = \frac{n}{2} \left[ 2(–6) + (n – 1) \left( \frac{1}{2} \right) \right] \)
\( \Rightarrow –50 = n \left( \frac{n – 25}{2} \right) \Rightarrow –100 = n^2 – 25n \)
\( \Rightarrow n^2 – 25n + 100 = 0 \Rightarrow (n – 5) (n – 20) = 0 \)
\( \therefore n = 5, 20 \).
Both the values of \( n \) are natural numbers and therefore, admissible.
Explanation of Double Answer
\( S_{20} = – 6 – \frac{11}{2} – 5 – \frac{9}{2} – 4 – \frac{7}{2} – 3 – \frac{5}{2} – 2 – \frac{3}{2} – 1 – \frac{1}{2} \)
\( + 0 + \frac{1}{2} + 1 + \frac{3}{2} + 2 + \frac{5}{2} + 3 + \frac{7}{2} \)
\( = – 6 – \frac{11}{2} – 5 – \frac{9}{2} – 4 = S_5 \).

Question. If the sum of a certain number of terms of an A.P. 25, 22, 19, .....is 116,then find the last term.
Answer: We know that, \( S_n \) being the sum of \( n \) terms of A.P.,
\( S_n = 116 \).
\( \Rightarrow 116 = \frac{n}{2} [2 \times 25 + (n - 1) (- 3)] \)
\( \Rightarrow 232 = n [50 + 3 - 3n] \)
\( \Rightarrow n [53 - 3n] = 232 \)
\( \Rightarrow 3n^2 - 53n + 232 = 0 \)
\( \Rightarrow 3n^2 - 24n - 29n + 232 = 0 \)
\( \Rightarrow 3n (n - 8) - 29 (n - 8) = 0 \)
\( \Rightarrow (n - 8) (3n - 29) = 0 \Rightarrow n = 8 \) or \( n = 29/3 \)
But, \( n = 29/3 \) is not possible \( \Rightarrow n = 8 \)
Thus, 116 is the sum of 8 terms of the A.P.
The last term is given by
\( t_8 = a + (8 - 1)d = a + 7d = 25 + 7 \times (- 3) = 4 \)

Question. Find the sum of \( n \) terms of the series \( \left( 4 - \frac{1}{n} \right) + \left( 4 - \frac{2}{n} \right) + \left( 4 - \frac{3}{n} \right) + ... \)
Answer: We have, \( \left( 4 - \frac{1}{n} \right) + \left( 4 - \frac{2}{n} \right) + \left( 4 - \frac{3}{n} \right) + ..... \) which forms an A.P where first term \( (a) = \left( 4 - \frac{1}{n} \right) \)
Common difference \( (d) = \left( 4 - \frac{2}{n} \right) - \left( 4 - \frac{1}{n} \right) = - \frac{1}{n} \)
and last term \( (l) = \left( 4 - \frac{n}{n} \right) = (4 - 1) = 3 \) (Series has \( n \) terms)
\( \therefore \text{Sum of } n \text{ terms } (S_n) = \frac{n}{2} (a + l) \)
\( = \frac{n}{2} \left( 4 - \frac{1}{n} + 3 \right) = \frac{n}{2} \left( 7 - \frac{1}{n} \right) = \frac{7n}{2} - \frac{1}{2} = \left( \frac{7n - 1}{2} \right) \)

Question. The sum of the first seven terms of an A.P. is 182. If its 4th and the 17th terms are in the ratio 1 : 5, find the A.P.
Answer: Given, sum of first seven terms of an A.P., \( S_7 = 182 \)
i.e., \( 182 = \frac{7}{2} [2a + (7 - 1)d] \)
\( \Rightarrow 364 = 14a + 42d \Rightarrow 26 = a + 3d \) ...(i)
Also, \( \frac{a_4}{a_{17}} = \frac{1}{5} \Rightarrow \frac{a + 3d}{a + 16d} = \frac{1}{5} \)
\( \Rightarrow 5(a + 3d) = a + 16d \)
\( \Rightarrow 5a + 15d = a + 16d \)
\( \Rightarrow 4a – d = 0 \Rightarrow d = 4a \) ...(ii)
From (i) and (ii), we get
\( 26 = a + 3(4a) \Rightarrow 13a = 26 \Rightarrow a = 2 \)
\( \therefore d = 4(2) = 8 \)
Hence, the A.P. is formed as 2, 10, 18, ...

Question. Jasleen saved ₹ 4 during first week of the year and then increased her weekly savings by ₹ 1.75 each week. In which week, will her weekly savings be ₹ 19.75?
Answer: Let in the \( n^{th} \) week, Jasleen's weekly savings will be ₹ 19.75.
Clearly, Jasleen's weekly savings form an A.P. with first term, \( a = 4 \) and common difference, \( d = 1.75 \)
\( \therefore a_n = 19.75 \Rightarrow a + (n – 1)d = 19.75 \)
\( \Rightarrow 4 + (n – 1) (1.75) = 19.75 \Rightarrow (n – 1) (1.75) = 15.75 \)
\( \Rightarrow n – 1 = \frac{15.75}{1.75} \Rightarrow n – 1 = 9 \Rightarrow n = 10 \).
Hence, Jasleen's weekly savings will be ₹ 19.75 in the \( 10^{th} \) week.

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

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 three digit number be \( xyz \).
Digits are in A.P. \( \therefore x = y – d \) and \( z = y + d \) [where \( d \) is common difference]
According to question,
\( (y – d) + y + (y + d) = 15 \)
\( \Rightarrow 3y = 15 \Rightarrow y = 5 \)
Since, number obtained by reversing the digits \( (z y x) \) i.e., \( 100z + 10y + x \) is 594 less than the original number
\( \Rightarrow (100x + 10y + z) – (100z + 10y + x) = 594 \)
\( \Rightarrow (z – 100z) + (100x – x) = 594 \)
\( \Rightarrow 99x – 99z = 594 \Rightarrow x – z = 6 \)
\( \Rightarrow (y – d) – (y + d) = 6 \)
\( \Rightarrow –2d = 6 \Rightarrow d = –3 \)
So, \( x = y – d = 5 – (–3) = 8 \) and \( z = y + d = 5 – 3 = 2 \)
\( \therefore \) The number is \( xyz = 852 \).

Question. Find the number of terms of the A.P. \( 18, 15\frac{1}{2}, 13, ….., -49\frac{1}{2} \) and find the sum of all its terms.
Answer: Given, first term \( (a) = 18 \),
common difference \( (d) = 15 \frac{1}{2} - 18 = -2 \frac{1}{2} = - \frac{5}{2} \)
and last term \( (l) \) of the A.P. \( = - 49 \frac{1}{2} = - \frac{99}{2} \)
Let the A.P. has \( n \) terms.
\( \therefore a_n = a + (n − 1)d \)
\( \Rightarrow - \frac{99}{2} = 18 + (n − 1) \left( - \frac{5}{2} \right) \)
\( \Rightarrow 5(n – 1) = 135 \Rightarrow n = 27 + 1 = 28 \)
\( \therefore \) Sum of all 28 terms is, \( S_{28} = \frac{28}{2} \left[ 18 - \frac{99}{2} \right] = - 441 \)

Long Answer Type Questions 

Question. The sum of three numbers in A.P. is 12 and sum of their cubes is 288. Find the numbers.
Answer: Let the three numbers in A.P. be \( a – d, a, a + d \).
According to question, \( a – d + a + a + d = 12 \)
\( \Rightarrow 3a = 12 \Rightarrow a = 4 \)
Also, \( (4 – d)^3 + (4)^3 + (4 + d)^3 = 288 \)
\( \Rightarrow 64 – 48d + 12d^2 – d^3 + 64 + 64 + 48d + 12d^2 + d^3 = 288 \)
\( \Rightarrow 24d^2 + 192 = 288 \Rightarrow d^2 = 4 \Rightarrow d = \pm 2 \)
\( \therefore \) The numbers will be \( a – d, a, a + d \)
\( \Rightarrow 4 + 2, 4, 4 – 2 = 6, 4, 2 \), if \( d = –2 \)
or \( 4 – 2, 4, 4 + 2 = 2, 4, 6 \), if \( d = 2 \)

Question. The sum of 4th and 8th terms of an A.P. is 24 and sum of its 6th and 10th terms is 44. Find the sum of first ten terms of the A.P.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the A.P.
\( n^{th} \) term of the A.P. is \( a_n = a + (n -1)d \)
Now, we have \( a_4 + a_8 = 24 \)
\( \Rightarrow a + 3d + a + 7d = 24 \)
\( \Rightarrow 2a + 10d = 24 \Rightarrow a + 5d = 12 \) ...(i)
Also, \( a_6 + a_{10} = 44 \)
\( \Rightarrow a + 5d + a + 9d = 44 \Rightarrow 2a + 14d = 44 \)
\( \Rightarrow a + 7d = 22 \) ...(ii)
On subtracting (i) from (ii), we get
\( 2d = 10 \Rightarrow d = 5 \)
On substituting the value of \( d \) in (i), we get
\( a + 5(5) = 12 \Rightarrow a = – 13 \)
Thus, the A.P. is \( −13, −8, −3, … \)
\( \therefore \) Sum of first 10 terms, \( S_{10} = \frac{10}{2} [2a + (10 – 1)d] \)
\( = 5[2(–13) + 9 \times 5] = 5(–26 + 45) = 5 \times 19 = 95 \)

Question. The 16th term of an A.P. is 1 more than twice its 8th term. If the 12th term of the A.P. is 47, then find its nth term. Also, find the sum of its first n terms.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the A.P.
According to question, \( a_{16} = 2a_8 + 1 \)
\( \Rightarrow a + 15d = 2[a + 7d] + 1 \)
\( \Rightarrow a + 15d = 2a + 14d + 1 \)
\( \Rightarrow d = a + 1 \) ...(i)
Also, \( a_{12} = 47 \Rightarrow a + 11d = 47 \)
\( \Rightarrow a + 11(a + 1) = 47 \) [Using (i)]
\( \Rightarrow a + 11a + 11 = 47 \Rightarrow 12a = 36 \Rightarrow a = 3 \)
\( \therefore d = 3 + 1 = 4 \)
\( n^{th} \) term of the A.P., \( a_n = a + (n − 1)d \)
\( \therefore a_n = 3 + (n − 1) 4 = 3 + 4n − 4 = 4n − 1 \)
Now, \( S_n = \frac{n}{2} [a + l] = \frac{n}{2} [3 + 4n − 1] = 2n^2 + n \)

Question. How many multiples of 4 lie between 10 and 250? Also find their sum.
Answer: Multiples of 4 between 10 and 250 are 12, 16, 20,……, 248, which forms an A.P.
So, \( a = 12, d = 4, a_n = 248 \).
Let there be \( n \) terms in A.P.
\( \therefore 12 + (n − 1) \times 4 = 248 \Rightarrow (n − 1) \times 4 = 236 \)
\( \Rightarrow n − 1 = 59 \Rightarrow n = 59 + 1 = 60 \)
Now, \( S_{60} = \frac{60}{2} (12 + 248) = 30 \times 260 = 7800 \)
\( \because S_n = \frac{n}{2} (a + l) \)

Question. If the ratio of the sum of the first \( n \) terms of two A.P.’s is \( (7n + 1) : (4n + 27) \), then find the ratio of their 9th terms.
Answer: Let \( a_1, a_2 \) be the first term and \( d_1, d_2 \) be the common difference of the two A.P.’s respectively.
Given, ratio of sum of first \( n \) terms \( = \frac{7n + 1}{4n + 27} \)
\( \therefore \frac{ \frac{n}{2} \{ 2a_1 + (n - 1)d_1 \} }{ \frac{n}{2} \{ 2a_2 + (n - 1)d_2 \} } = \frac{7n + 1}{4n + 27} \) ...(i)
\( \Rightarrow \frac{ 2a_1 + (n - 1)d_1 }{ 2a_2 + (n - 1)d_2 } = \frac{7n + 1}{4n + 27} \)
\( \Rightarrow \frac{ a_1 + \left( \frac{n-1}{2} \right)d_1 }{ a_2 + \left( \frac{n-1}{2} \right)d_2 } = \frac{7n + 1}{4n + 27} \)
Putting \( \frac{n - 1}{2} = 8 \), we get
\( n – 1 = 16 \Rightarrow n = 17 \)
\( \frac{ a_1 + 8d_1 }{ a_2 + 8d_2 } = \frac{7(17) + 1}{4(17) + 27} \)
\( \Rightarrow \frac{ a_1 + 8d_1 }{ a_2 + 8d_2 } = \frac{120}{95} = \frac{24}{19} \)
\( \therefore \text{Ratio of } 9^{th} \text{ terms } = \frac{ a_1 + 8d_1 }{ a_2 + 8d_2 } = \frac{24}{19} \)