OP Malhotra Class 11 Maths Solutions Chapter 14 Sequence and Series Exercise 14 (C)

Question 1. Find the sum :
(i) 10 terms of 5 + 8 + 11 +....;
(ii) 18 terms of 57 + 49 + 41 +....;
(iii) n terms of 4 + 7 + 10 + ....;
(iv) 24 terms and n terms of \( 2\frac { 1 }{ 2 } \), \( 3\frac { 1 }{ 3 } \), \( 4\frac { 1 }{ 6 } \), \( 5 \) ....;
Answer:
(i) The series is an Arithmetic Progression (A.P.) because the difference between consecutive terms is constant. Here, the first term \( a \) is 5, the common difference \( d \) is \( 8 - 5 = 3 \), and we need the sum of \( n = 10 \) terms. We use the formula for the sum of an A.P., \( S_n = \frac{n}{2}[2a + (n-1)d] \).
\( \implies S_{10} = \frac{10}{2}[2 \times 5 + (10 - 1)3] \)
\( \implies S_{10} = 5[10 + (9 \times 3)] \)
\( \implies S_{10} = 5[10 + 27] \)
\( \implies S_{10} = 5[37] \)
\( \implies S_{10} = 185 \). So, the sum of the first 10 terms is 185.
(ii) This series is also an Arithmetic Progression. The first term \( a \) is 57, and the common difference \( d \) is \( 49 - 57 = -8 \). We need to find the sum of \( n = 18 \) terms. Applying the A.P. sum formula \( S_n = \frac{n}{2}[2a + (n-1)d] \):
\( \implies S_{18} = \frac{18}{2}[2 \times 57 + (18 - 1)(-8)] \)
\( \implies S_{18} = 9[114 + (17 \times -8)] \)
\( \implies S_{18} = 9[114 - 136] \)
\( \implies S_{18} = 9[-22] \)
\( \implies S_{18} = -198 \). Therefore, the sum of the first 18 terms is -198.
(iii) Solution not provided in the source. To solve this, we would identify the first term \( a = 4 \), the common difference \( d = 7 - 4 = 3 \), and then use the formula for the sum of \( n \) terms of an A.P., \( S_n = \frac{n}{2}[2a + (n-1)d] \). This formula would give us an expression for the sum in terms of \( n \).
(iv) This is an Arithmetic Progression with the first term \( a = 2\frac{1}{2} = \frac{5}{2} \). The common difference \( d \) is found by subtracting the first term from the second: \( d = 3\frac{1}{3} - 2\frac{1}{2} = \frac{10}{3} - \frac{5}{2} = \frac{20-15}{6} = \frac{5}{6} \). We need to find the sum of the first 24 terms and also the sum of \( n \) terms.
Using the A.P. sum formula \( S_n = \frac{n}{2}[2a + (n-1)d] \):
For \( n = 24 \):
\( S_{24} = \frac{24}{2} \left[2 \times \frac{5}{2} + (24 - 1)\frac{5}{6}\right] \)
\( \implies S_{24} = 12 \left[5 + 23 \times \frac{5}{6}\right] \)
\( \implies S_{24} = 12 \left[5 + \frac{115}{6}\right] \)
\( \implies S_{24} = 12 \left[\frac{30+115}{6}\right] \)
\( \implies S_{24} = 12 \times \frac{145}{6} \)
\( \implies S_{24} = 2 \times 145 \)
\( \implies S_{24} = 290 \).
For \( n \) terms:
\( S_n = \frac{n}{2} \left[2 \times \frac{5}{2} + (n - 1)\frac{5}{6}\right] \)
\( \implies S_n = \frac{n}{2} \left[5 + (n - 1)\frac{5}{6}\right] \)
\( \implies S_n = \frac{n}{2} \left[\frac{30 + 5(n - 1)}{6}\right] \)
\( \implies S_n = \frac{n}{12} [30 + 5n - 5] \)
\( \implies S_n = \frac{n}{12} [5n + 25] \)
\( \implies S_n = \frac{5n(n+5)}{12} \). The sum of \( n \) terms is \( \frac{5n(n+5)}{12} \).
In simple words: For (i) and (ii), we used the first number, how much it changes by, and the number of terms to find the total sum. For (iii), the solution isn't given, but we know how to find it. For (iv), we identified the pattern, then found the sum for 24 terms and a general formula for 'n' terms.

🎯 Exam Tip: Always double-check your common difference calculation, especially with negative numbers or fractions, as an error here will cascade through the entire sum calculation.

Question 2. Find the sum of all the numbers between 100 and 200 which are divisible by 7 .
Answer: First, we need to find the numbers between 100 and 200 that can be divided evenly by 7. The first number is 105 (\( 7 \times 15 \)), and the last number is 196 (\( 7 \times 28 \)). This forms an Arithmetic Progression (A.P.) with the first term \( a = 105 \), common difference \( d = 7 \), and the last term \( l = 196 \).
To find the number of terms \( n \), we use the formula \( T_n = a + (n-1)d \):
\( 196 = 105 + (n - 1)7 \)
\( \implies 196 - 105 = (n - 1)7 \)
\( \implies 91 = (n - 1)7 \)
\( \implies \frac{91}{7} = n - 1 \)
\( \implies 13 = n - 1 \)
\( \implies n = 14 \). There are 14 such numbers.
Now, we find the sum \( S_n \) using the formula \( S_n = \frac{n}{2}[a + l] \):
\( S_{14} = \frac{14}{2}[105 + 196] \)
\( \implies S_{14} = 7[301] \)
\( \implies S_{14} = 2107 \). The sum of all numbers between 100 and 200 divisible by 7 is 2107.
In simple words: We found all the numbers from 105 to 196 that can be divided by 7. There are 14 such numbers. Then we added all these numbers together, and the total sum is 2107.

🎯 Exam Tip: When finding numbers "between" two values, ensure you exclude the endpoints if they are not explicitly included (e.g., "between 100 and 200" means 100 and 200 are not counted, so the first multiple of 7 is 105 and the last is 196).

Question 3. The sum of a series of terms in A.P. is 128. If the first term is 2 and the last term is 14, 1, find the common difference.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the given A.P. We are given \( a = 2 \), the last term \( l = 14 \), and the sum \( S_n = 128 \).
First, we find the number of terms \( n \) using the formula \( S_n = \frac{n}{2}(a + l) \):
\( 128 = \frac{n}{2}(2 + 14) \)
\( \implies 128 = \frac{n}{2}(16) \)<
\( \implies 128 = 8n \)
\( \implies n = \frac{128}{8} \)
\( \implies n = 16 \). There are 16 terms in the series.
Next, we find the common difference \( d \) using the formula for the last term \( l = a + (n-1)d \):
\( 14 = 2 + (16 - 1)d \)
\( \implies 14 = 2 + 15d \)
\( \implies 14 - 2 = 15d \)
\( \implies 12 = 15d \)
\( \implies d = \frac{12}{15} \)
\( \implies d = \frac{4}{5} \). The common difference is \( \frac{4}{5} \).
In simple words: We are given the first number, the last number, and the total sum of a series. First, we found out how many numbers are in the series (16 terms). Then, using the first, last, and number of terms, we calculated how much each number changes by, which is \( \frac{4}{5} \).

🎯 Exam Tip: When finding the common difference, ensure you correctly use the first term, last term, and the number of terms. Missing a term or using the wrong count can lead to an incorrect 'd'.

Question 4. The sum of 30 terms of a series in A.P., whose last term is 98, is 1635 . Find the first term and the common difference.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the given A.P. We are given the number of terms \( n = 30 \), the sum \( S_{30} = 1635 \), and the last term \( l = 98 \).
First, we find the first term \( a \) using the sum formula \( S_n = \frac{n}{2}(a + l) \):
\( S_{30} = \frac{30}{2}(a + 98) \)
\( \implies 1635 = 15(a + 98) \)
\( \implies \frac{1635}{15} = a + 98 \)
\( \implies 109 = a + 98 \)
\( \implies a = 109 - 98 \)
\( \implies a = 11 \). So, the first term is 11.
Next, we find the common difference \( d \) using the formula for the last term \( l = a + (n-1)d \):
\( 98 = 11 + (30 - 1)d \)
\( \implies 98 = 11 + 29d \)
\( \implies 98 - 11 = 29d \)
\( \implies 87 = 29d \)
\( \implies d = \frac{87}{29} \)
\( \implies d = 3 \). Thus, the common difference is 3.
In simple words: We know the sum, the number of terms, and the last term. First, we found the starting number (which is 11). Then, using this starting number and other known facts, we figured out that each number in the list increases by 3.

🎯 Exam Tip: When both sum and last term are given, first calculate 'a' using \( S_n = \frac{n}{2}(a + l) \), then use \( l = a + (n-1)d \) to find 'd'. This sequential approach simplifies calculations.

Question 5. If the sums of the first 8 and 19 terms of an A.P. are 64 and 361 respectively, find (i) the common difference and (ii) the sum of n terms of the series.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the A.P.
We are given the sum of the first 8 terms, \( S_8 = 64 \), and the sum of the first 19 terms, \( S_{19} = 361 \). We use the sum formula \( S_n = \frac{n}{2}[2a + (n-1)d] \).
For \( S_8 = 64 \):
\( \frac{8}{2}[2a + (8 - 1)d] = 64 \)
\( \implies 4[2a + 7d] = 64 \)
\( \implies 2a + 7d = 16 \)...(1)
For \( S_{19} = 361 \):
\( \frac{19}{2}[2a + (19 - 1)d] = 361 \)
\( \implies 19[2a + 18d] = 722 \)
\( \implies 2a + 18d = \frac{722}{19} \)
\( \implies 2a + 18d = 38 \)...(2)
Now we solve equations (1) and (2) simultaneously. Subtract equation (1) from equation (2):
\( (2a + 18d) - (2a + 7d) = 38 - 16 \)
\( \implies 11d = 22 \)
\( \implies d = 2 \).
Substitute \( d = 2 \) into equation (1):
\( 2a + 7(2) = 16 \)
\( \implies 2a + 14 = 16 \)
\( \implies 2a = 2 \)
\( \implies a = 1 \).
So, the first term \( a = 1 \) and the common difference \( d = 2 \).
(i) The common difference \( d \) is 2.
(ii) The sum of \( n \) terms of the series \( S_n \) is:
\( S_n = \frac{n}{2}[2a + (n-1)d] \)
\( \implies S_n = \frac{n}{2}[2(1) + (n-1)2] \)
\( \implies S_n = \frac{n}{2}[2 + 2n - 2] \)
\( \implies S_n = \frac{n}{2}[2n] \)
\( \implies S_n = n^2 \). The sum of \( n \) terms is \( n^2 \).
In simple words: We used the given sums for 8 and 19 terms to find the starting number (1) and how much each number changes by (2). Then, we used these to write a general formula for the sum of 'n' terms, which turned out to be \( n^2 \).

🎯 Exam Tip: When given two sum values for an A.P., form two linear equations in 'a' and 'd' and solve them simultaneously. This is a standard method to find the fundamental properties of the progression.

Question 6. Find the number of terms of the series 21, 18, 15, 12, ....... which must be taken to give a sum of zero.
Answer: The given series is an Arithmetic Progression with the first term \( a = 21 \) and the common difference \( d = 18 - 21 = -3 \). We need to find the number of terms \( n \) such that the sum \( S_n = 0 \).
Using the sum formula \( S_n = \frac{n}{2}[2a + (n-1)d] \):
\( 0 = \frac{n}{2}[2(21) + (n - 1)(-3)] \)
\( \implies 0 = n[42 - 3n + 3] \)
\( \implies 0 = n[45 - 3n] \).
This gives two possibilities: \( n = 0 \) (which is not a valid number of terms for a series) or \( 45 - 3n = 0 \).
\( \implies 3n = 45 \)
\( \implies n = \frac{45}{3} \)
\( \implies n = 15 \). Therefore, 15 terms must be taken for the sum to be zero. This happens because the terms are decreasing; eventually, the positive terms are cancelled out by the negative terms.
In simple words: We have a list of numbers that go down by 3 each time. We need to find how many of these numbers we must add together so that the total sum becomes zero. The answer is 15 terms.

🎯 Exam Tip: Remember that \(n\) (number of terms) must always be a positive integer. If you get multiple solutions for \(n\), discard any that are zero, negative, or not whole numbers.

Question 7. The sum of n terms of a series is (n² + 2n) for all values of n. Find the first 3 terms of the series.
Answer: We are given the sum of \( n \) terms of a series as \( S_n = n^2 + 2n \). To find the \( n^{th} \) term \( T_n \), we use the relationship \( T_n = S_n - S_{n-1} \).
First, find \( S_{n-1} \):
\( S_{n-1} = (n-1)^2 + 2(n-1) \)
\( \implies S_{n-1} = (n^2 - 2n + 1) + (2n - 2) \)
\( \implies S_{n-1} = n^2 - 1 \).
Now, find \( T_n \):
\( T_n = (n^2 + 2n) - (n^2 - 1) \)
\( \implies T_n = n^2 + 2n - n^2 + 1 \)
\( \implies T_n = 2n + 1 \).
With the formula for the \( n^{th} \) term, we can find the first three terms:
First term \( T_1 \):
\( T_1 = 2(1) + 1 = 3 \).
Second term \( T_2 \):
\( T_2 = 2(2) + 1 = 5 \).
Third term \( T_3 \):
\( T_3 = 2(3) + 1 = 7 \).
So, the first three terms of the series are 3, 5, and 7. This series is actually an A.P. as the terms increase by a constant difference of 2.
In simple words: We have a formula that tells us the sum of 'n' terms. To find any single term, we subtract the sum of the terms before it from the sum up to that term. Using this, we found the first three numbers in the series are 3, 5, and 7.

🎯 Exam Tip: The relation \( T_n = S_n - S_{n-1} \) is crucial for finding individual terms when only the sum formula is given. Be careful with algebraic expansion and simplification when calculating \(S_{n-1}\).

Question 8. The third term of an arithmetical progression is 7, and the seventh term is 2 more than 3 times the third term. Find the first term, the common difference and the sum of the first 20 terms.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the A.P.
We are given the third term \( T_3 = 7 \). Using the formula \( T_n = a + (n-1)d \):
\( a + (3 - 1)d = 7 \)
\( \implies a + 2d = 7 \)...(1)
We are also given that the seventh term \( T_7 \) is 2 more than 3 times the third term:
\( T_7 = 2 + 3T_3 \)
\( T_7 = 2 + 3(7) \)
\( T_7 = 2 + 21 \)
\( T_7 = 23 \).
Now use the formula for \( T_7 \):
\( a + (7 - 1)d = 23 \)
\( \implies a + 6d = 23 \)...(2)
Subtract equation (1) from equation (2) to find \( d \):
\( (a + 6d) - (a + 2d) = 23 - 7 \)
\( \implies 4d = 16 \)
\( \implies d = 4 \). The common difference is 4.
Substitute \( d = 4 \) into equation (1) to find \( a \):
\( a + 2(4) = 7 \)
\( \implies a + 8 = 7 \)
\( \implies a = 7 - 8 \)
\( \implies a = -1 \). The first term is -1.
Finally, we find the sum of the first 20 terms \( S_{20} \) using the sum formula \( S_n = \frac{n}{2}[2a + (n-1)d] \):
\( S_{20} = \frac{20}{2}[2(-1) + (20 - 1)4] \)
\( \implies S_{20} = 10[-2 + 19 \times 4] \)
\( \implies S_{20} = 10[-2 + 76] \)
\( \implies S_{20} = 10[74] \)
\( \implies S_{20} = 740 \). The sum of the first 20 terms is 740.
In simple words: We used the given information about the 3rd and 7th terms to find the starting number (-1) and how much each number changes by (4). Then, we calculated the total sum of the first 20 numbers in this pattern, which is 740.

🎯 Exam Tip: Translate word problems into algebraic equations carefully. Use \( T_n = a + (n-1)d \) and \( S_n = \frac{n}{2}[2a + (n-1)d] \) correctly, and solve the system of equations for 'a' and 'd'.

Question 9. The interior angles of a polygon are in arithmetic progression. The smallest angle is 52°. Find the number of sides of the polygon.
Answer: Let \( a \) be the smallest interior angle and \( d \) be the common difference of the A.P. We are given \( a = 52^\circ \) and \( d = 8^\circ \). Let \( n \) be the number of sides of the polygon.
The sum of the interior angles of a polygon with \( n \) sides is given by \( (n-2) \times 180^\circ \).
The sum of the angles, since they form an A.P., can also be found using \( S_n = \frac{n}{2}[2a + (n-1)d] \).
So, we set the two sum formulas equal:
\( \frac{n}{2}[2(52^\circ) + (n - 1)8^\circ] = (n - 2) \times 180^\circ \)
\( \implies \frac{n}{2}[104^\circ + 8n - 8^\circ] = (n - 2) \times 180^\circ \)
\( \implies \frac{n}{2}[96^\circ + 8n] = (n - 2) \times 180^\circ \)
\( \implies n[48^\circ + 4n] = (n - 2) \times 180^\circ \)
\( \implies 4n^2 + 48n = 180n - 360 \)
\( \implies 4n^2 + 48n - 180n + 360 = 0 \)
\( \implies 4n^2 - 132n + 360 = 0 \).
Divide by 4:
\( \implies n^2 - 33n + 90 = 0 \).
Factorize the quadratic equation:
\( (n - 3)(n - 30) = 0 \).
This gives two possible values for \( n \): \( n = 3 \) or \( n = 30 \).
We need to check if these values are valid. The interior angle of any convex polygon cannot be more than \( 180^\circ \).
If \( n = 30 \), the largest angle (the \( 30^{th} \) term) would be:
\( T_{30} = a + (30 - 1)d \)
\( T_{30} = 52^\circ + 29 \times 8^\circ \)
\( T_{30} = 52^\circ + 232^\circ \)
\( T_{30} = 284^\circ \).
Since \( 284^\circ > 180^\circ \), \( n = 30 \) is not a possible number of sides for a polygon with these angle properties.
If \( n = 3 \), the angles would be \( 52^\circ, 60^\circ, 68^\circ \), which are all less than \( 180^\circ \). The sum is \( 180^\circ \), which is correct for a triangle.
Therefore, the number of sides of the polygon is 3.
In simple words: We used two ways to calculate the sum of angles in a polygon: one for A.P. and one for any polygon. Setting them equal gave us two possible numbers of sides (3 or 30). We ruled out 30 because an angle in such a polygon would be too large. So, the polygon must have 3 sides, meaning it's a triangle.

🎯 Exam Tip: Always verify your solutions in context. For polygons, remember that interior angles cannot exceed \(180^\circ\). Discard any mathematical solutions that violate geometric rules.

Question 10. Determine the sum of first 35 terms of an A.P. if t2 = 1 and t7 = 22.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the A.P.
We are given the second term \( t_2 = 1 \) and the seventh term \( t_7 = 22 \). Using the formula \( t_n = a + (n-1)d \):
For \( t_2 = 1 \):
\( a + d = 1 \)...(1)
For \( t_7 = 22 \):
\( a + 6d = 22 \)...(2)
Subtract equation (1) from equation (2):
\( (a + 6d) - (a + d) = 22 - 1 \)
\( \implies 5d = 21 \)
\( \implies d = \frac{21}{5} \). The common difference is \( \frac{21}{5} \).
Substitute \( d = \frac{21}{5} \) into equation (1) to find \( a \):
\( a + \frac{21}{5} = 1 \)
\( \implies a = 1 - \frac{21}{5} \)
\( \implies a = \frac{5 - 21}{5} \)
\( \implies a = -\frac{16}{5} \). The first term is \( -\frac{16}{5} \).
Now we find the sum of the first 35 terms \( S_{35} \) using the formula \( S_n = \frac{n}{2}[2a + (n-1)d] \):
\( S_{35} = \frac{35}{2}\left[2\left(-\frac{16}{5}\right) + (35 - 1)\left(\frac{21}{5}\right)\right] \)
\( \implies S_{35} = \frac{35}{2}\left[-\frac{32}{5} + 34 \times \frac{21}{5}\right] \)
\( \implies S_{35} = \frac{35}{2}\left[-\frac{32}{5} + \frac{714}{5}\right] \)
\( \implies S_{35} = \frac{35}{2}\left[\frac{714 - 32}{5}\right] \)
\( \implies S_{35} = \frac{35}{2}\left[\frac{682}{5}\right] \)
\( \implies S_{35} = \frac{7 \times 5}{2} \times \frac{682}{5} \)
\( \implies S_{35} = 7 \times \frac{682}{2} \)
\( \implies S_{35} = 7 \times 341 \)
\( \implies S_{35} = 2387 \). The sum of the first 35 terms is 2387.
In simple words: We used the given second and seventh terms to find the starting number and how much each number changes by. Then, we used these values to calculate the total sum of the first 35 numbers in this pattern, which is 2387.

🎯 Exam Tip: Systems of linear equations are frequently used in A.P. problems. Ensure you correctly set up equations using the term formula \( T_n = a + (n-1)d \) for given terms, then solve for 'a' and 'd'.

Question 11. Find the sum of all natural numbers between 100 and 1000 which are multiples of 5 .
Answer: The natural numbers between 100 and 1000 that are multiples of 5 form an Arithmetic Progression. The first term \( a \) is 105 (\( 5 \times 21 \)), the common difference \( d \) is 5, and the last term \( l \) is 995 (\( 5 \times 199 \)).
First, we find the number of terms \( n \) using the formula \( l = a + (n-1)d \):
\( 995 = 105 + (n - 1)5 \)
\( \implies 995 - 105 = (n - 1)5 \)
\( \implies 890 = (n - 1)5 \)
\( \implies \frac{890}{5} = n - 1 \)
\( \implies 178 = n - 1 \)
\( \implies n = 179 \). There are 179 such numbers.
Next, we find the sum \( S_n \) using the formula \( S_n = \frac{n}{2}[a + l] \):
\( S_{179} = \frac{179}{2}[105 + 995] \)
\( \implies S_{179} = \frac{179}{2}[1100] \)
\( \implies S_{179} = 179 \times 550 \)
\( \implies S_{179} = 98450 \). The sum of all such natural numbers is 98450.
In simple words: We found all the numbers between 100 and 1000 that can be divided by 5. The first is 105 and the last is 995. There are 179 such numbers. When we add them all up, the total sum is 98450.

🎯 Exam Tip: Be precise when interpreting "between" – it usually excludes the boundary numbers. Identify the first and last terms accurately, then use \(l=a+(n-1)d\) to find \(n\), and \(S_n=\frac{n}{2}(a+l)\) for the sum.

Question 12. How many terms of the A.P. 1, 4, 7, ....are needed to give the sum 715 ?
Answer: The given Arithmetic Progression has the first term \( a = 1 \) and the common difference \( d = 4 - 1 = 3 \). We need to find the number of terms \( n \) such that the sum \( S_n = 715 \).
Using the sum formula \( S_n = \frac{n}{2}[2a + (n-1)d] \):
\( 715 = \frac{n}{2}[2(1) + (n - 1)3] \)
\( \implies 1430 = n[2 + 3n - 3] \)
\( \implies 1430 = n[3n - 1] \)
\( \implies 1430 = 3n^2 - n \).
Rearrange into a quadratic equation:
\( 3n^2 - n - 1430 = 0 \).
We solve this quadratic equation for \( n \) using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\( n = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-1430)}}{2(3)} \)
\( \implies n = \frac{1 \pm \sqrt{1 + 12 \times 1430}}{6} \)
\( \implies n = \frac{1 \pm \sqrt{1 + 17160}}{6} \)
\( \implies n = \frac{1 \pm \sqrt{17161}}{6} \)
\( \implies n = \frac{1 \pm 131}{6} \).
This gives two possible values for \( n \):
\( n = \frac{1 + 131}{6} = \frac{132}{6} = 22 \)
\( n = \frac{1 - 131}{6} = \frac{-130}{6} = -\frac{65}{3} \).
Since the number of terms \( n \) must be a positive whole number (a natural number), we choose \( n = 22 \).
Thus, 22 terms are needed for the sum to be 715.
In simple words: We know the pattern of numbers (starting with 1, increasing by 3 each time). We want to find out how many numbers we need to add to get a total of 715. We used a formula that led to a special kind of equation, and solving it told us that we need 22 terms.

🎯 Exam Tip: When \(S_n\) is given and \(n\) is unknown, you will typically form a quadratic equation in \(n\). Remember to choose the positive integer solution for \(n\).

Question 13. Find the rth term of an A.P., sum of whose first n terms is \( 2n + 3n^2 \).
Answer: We are given that the sum of the first \( n \) terms of an A.P. is \( S_n = 2n + 3n^2 \). To find the \( r^{th} \) term \( T_r \), we first find the \( n^{th} \) term \( T_n \) using the relation \( T_n = S_n - S_{n-1} \).
First, calculate \( S_{n-1} \):
\( S_{n-1} = 2(n-1) + 3(n-1)^2 \)
\( \implies S_{n-1} = 2n - 2 + 3(n^2 - 2n + 1) \)
\( \implies S_{n-1} = 2n - 2 + 3n^2 - 6n + 3 \)
\( \implies S_{n-1} = 3n^2 - 4n + 1 \).
Now, calculate \( T_n \):
\( T_n = S_n - S_{n-1} \)
\( \implies T_n = (2n + 3n^2) - (3n^2 - 4n + 1) \)
\( \implies T_n = 2n + 3n^2 - 3n^2 + 4n - 1 \)
\( \implies T_n = 6n - 1 \).
Therefore, the \( r^{th} \) term of the A.P. is \( T_r = 6r - 1 \). This method is useful for finding individual terms when only the sum formula is provided.
In simple words: We have a formula for adding up 'n' numbers in a list. To find a specific number in that list (like the 'r'-th number), we subtract the sum of the numbers before it. Doing this math shows that the 'r'-th number in the list is always \( 6r - 1 \).

🎯 Exam Tip: The fundamental relationship \( T_n = S_n - S_{n-1} \) is critical here. Practice the algebraic expansion and subtraction carefully to avoid sign errors, as this is a common point of mistake.

Question 14. In an arithmetical progression, the sum of p terms is m and the sum of q terms is also m. Find the sum of (p + q) terms.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the A.P.
We are given that the sum of \( p \) terms, \( S_p = m \), and the sum of \( q \) terms, \( S_q = m \).
Using the sum formula \( S_n = \frac{n}{2}[2a + (n-1)d] \):
For \( S_p = m \):
\( \frac{p}{2}[2a + (p - 1)d] = m \)
\( \implies p[2a + (p - 1)d] = 2m \)...(1)
For \( S_q = m \):
\( \frac{q}{2}[2a + (q - 1)d] = m \)
\( \implies q[2a + (q - 1)d] = 2m \)...(2)
From (1) and (2), we have:
\( p[2a + (p - 1)d] = q[2a + (q - 1)d] \)
\( \implies 2ap + p(p - 1)d = 2aq + q(q - 1)d \)
\( \implies 2ap - 2aq = q(q - 1)d - p(p - 1)d \)
\( \implies 2a(p - q) = d[q(q - 1) - p(p - 1)] \)
\( \implies 2a(p - q) = d[q^2 - q - p^2 + p] \)
\( \implies 2a(p - q) = d[(q^2 - p^2) - (q - p)] \)
\( \implies 2a(p - q) = d[(q - p)(q + p) + (p - q)] \)
\( \implies 2a(p - q) = d[-(p - q)(q + p) + (p - q)] \)
\( \implies 2a(p - q) = d(p - q)[1 - (p + q)] \).
Since \( p \neq q \), we can divide both sides by \( (p - q) \):
\( 2a = d[1 - p - q] \)...(A)
Now, we need to find the sum of \( (p + q) \) terms, \( S_{p+q} \):
\( S_{p+q} = \frac{(p+q)}{2}[2a + (p + q - 1)d] \).
Substitute \( 2a = d[1 - p - q] \) from (A) into the expression for \( S_{p+q} \):
\( S_{p+q} = \frac{(p+q)}{2}[d(1 - p - q) + (p + q - 1)d] \)
\( \implies S_{p+q} = \frac{(p+q)}{2}[d(1 - p - q) - d(1 - p - q)] \)
\( \implies S_{p+q} = \frac{(p+q)}{2}[0] \)
\( \implies S_{p+q} = 0 \).
Thus, the sum of \( (p + q) \) terms is 0. This implies that the positive terms are exactly balanced by the negative terms in the progression up to \( p+q \) terms.
In simple words: We are told that the sum of 'p' terms is the same as the sum of 'q' terms. By using the sum formula and doing some algebra, we find a relationship between the first term and the common difference. When we use this relationship to find the sum of 'p+q' terms, we discover that the total sum is 0.

🎯 Exam Tip: When given sums for different numbers of terms, set up simultaneous equations. Algebraic manipulation is key here; look for common factors like \((p-q)\) to simplify expressions and derive core relationships.

Question 15. The sum of the first fifteen terms of an arithmetical progression is 105 and the sum of the next fifteen terms is 780 . Find the first three terms of the arithmetical progression.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the A.P.
We are given that the sum of the first 15 terms \( S_{15} = 105 \). Using the sum formula \( S_n = \frac{n}{2}[2a + (n-1)d] \):
\( \frac{15}{2}[2a + (15 - 1)d] = 105 \)
\( \implies 15[2a + 14d] = 210 \)
\( \implies 2a + 14d = \frac{210}{15} \)
\( \implies 2a + 14d = 14 \)...(1)
The sum of the *next* 15 terms (from term 16 to term 30) is 780. So, the sum of the first 30 terms, \( S_{30} \), will be the sum of the first 15 terms plus the sum of the next 15 terms:
\( S_{30} = S_{15} + 780 = 105 + 780 = 885 \).
Now use the sum formula for \( S_{30} \):
\( \frac{30}{2}[2a + (30 - 1)d] = 885 \)
\( \implies 15[2a + 29d] = 885 \)
\( \implies 2a + 29d = \frac{885}{15} \)
\( \implies 2a + 29d = 59 \)...(2)
Subtract equation (1) from equation (2) to find \( d \):
\( (2a + 29d) - (2a + 14d) = 59 - 14 \)
\( \implies 15d = 45 \)
\( \implies d = 3 \). The common difference is 3.
Substitute \( d = 3 \) into equation (1) to find \( a \):
\( 2a + 14(3) = 14 \)
\( \implies 2a + 42 = 14 \)
\( \implies 2a = 14 - 42 \)
\( \implies 2a = -28 \)
\( \implies a = -14 \). The first term is -14.
The first three terms of the A.P. are \( a \), \( a+d \), and \( a+2d \):
\( T_1 = -14 \)
\( T_2 = -14 + 3 = -11 \)
\( T_3 = -14 + 2(3) = -14 + 6 = -8 \).
Thus, the first three terms are -14, -11, and -8. This means the numbers are increasing by 3 at each step.
In simple words: We used the given sums for the first 15 terms and the next 15 terms to set up two equations. Solving these equations helped us find the starting number (-14) and how much each number changes by (3). Then we just listed the first three numbers in this pattern.

🎯 Exam Tip: Be careful with phrases like "sum of the next fifteen terms". This means finding the sum up to 30 terms and subtracting the sum of the first 15 terms to get the required sum.

Question 16. The sum of the first six terms of an arithmetic progression is 42 . The ratio of the 10th term to the 30th term of the A.P. is \( \frac { 1 }{ 3 } \). Calculate the first term and the 13th term.
Answer: Let \( a \) be the first term and \( d \) be the common difference of an A.P.
We are given that the sum of the first six terms \( S_6 = 42 \). Using the sum formula \( S_n = \frac{n}{2}[2a + (n-1)d] \):
\( \frac{6}{2}[2a + (6 - 1)d] = 42 \)
\( \implies 3[2a + 5d] = 42 \)
\( \implies 2a + 5d = 14 \)...(1)
We are also given that the ratio of the \( 10^{th} \) term to the \( 30^{th} \) term is \( \frac{1}{3} \). Using the formula \( T_n = a + (n-1)d \):
\( \frac{T_{10}}{T_{30}} = \frac{a + 9d}{a + 29d} = \frac{1}{3} \)
\( \implies 3(a + 9d) = 1(a + 29d) \)
\( \implies 3a + 27d = a + 29d \)
\( \implies 2a = 2d \)
\( \implies a = d \). The first term and common difference are equal.
Now substitute \( a = d \) into equation (1):
\( 2a + 5a = 14 \)
\( \implies 7a = 14 \)
\( \implies a = 2 \).
Since \( a = d \), then \( d = 2 \). So the first term is 2 and the common difference is 2.
Finally, we calculate the \( 13^{th} \) term \( T_{13} \):
\( T_{13} = a + (13 - 1)d \)
\( \implies T_{13} = a + 12d \)
\( \implies T_{13} = 2 + 12(2) \)
\( \implies T_{13} = 2 + 24 \)
\( \implies T_{13} = 26 \). The \( 13^{th} \) term is 26.
In simple words: We used the sum of the first six terms and the ratio of the 10th and 30th terms to find out that the first number and how much it changes by are both 2. Then, we calculated the 13th number in this pattern, which is 26.

🎯 Exam Tip: When dealing with ratios of terms, set up the fractional equation first. Cross-multiplication and simplification will reveal a relationship between 'a' and 'd' that can then be used with sum information.

Question 17. A sum of Rs. 6240 is paid off in 30 instalments, such that each instalment is Rs. 10 more than the preceeding instalment. Calculate the value of the first instalment.
Answer: Let the first instalment be \( a \). Since each subsequent instalment is Rs. 10 more than the preceding one, the instalments form an Arithmetic Progression (A.P.) with a common difference \( d = 10 \). There are \( n = 30 \) instalments, and the total sum paid is \( S_{30} = \text{Rs. } 6240 \).
We use the sum formula for an A.P., \( S_n = \frac{n}{2}[2a + (n-1)d] \):
\( 6240 = \frac{30}{2}[2a + (30 - 1)10] \)
\( \implies 6240 = 15[2a + 29 \times 10] \)
\( \implies 6240 = 15[2a + 290] \)
\( \implies \frac{6240}{15} = 2a + 290 \)
\( \implies 416 = 2a + 290 \)
\( \implies 2a = 416 - 290 \)
\( \implies 2a = 126 \)
\( \implies a = 63 \).
Therefore, the value of the first instalment is Rs. 63. This is a practical application of arithmetic progressions in finance.
In simple words: Someone paid off Rs. 6240 in 30 payments, where each payment was Rs. 10 more than the one before. We used a special formula to figure out that the very first payment made was Rs. 63.

🎯 Exam Tip: Identify the components of the A.P. from the word problem: what's the first term 'a', the common difference 'd', the number of terms 'n', and the sum 'S_n'. Then, apply the relevant formula to solve for the unknown.

Question 18. The nth term of an A.P. is p and the sum of the first n term is s. Prove that the first term is \( \frac{2 s-p n}{n} \).
Answer: Let \( a \) be the first term and \( T_n \) be the \( n^{th} \) term (which is also the last term \( l \)). We are given that \( T_n = p \) and the sum of the first \( n \) terms \( S_n = s \).
The formula for the sum of an A.P. when the first and last terms are known is \( S_n = \frac{n}{2}(a + l) \).
Substitute the given values \( S_n = s \) and \( l = p \):
\( s = \frac{n}{2}(a + p) \).
Now, we need to rearrange this equation to solve for \( a \):
\( 2s = n(a + p) \)
\( \implies \frac{2s}{n} = a + p \)
\( \implies a = \frac{2s}{n} - p \).
To combine the terms on the right side, find a common denominator:
\( a = \frac{2s - pn}{n} \).
This proves that the first term \( a \) is \( \frac{2s - pn}{n} \). This relationship is useful when you have the sum and the last term, but not the first term directly.
In simple words: We are given the last number in a list (p) and the total sum of all 'n' numbers (s). We want to show a formula for the very first number (a). By using the formula for the sum of a list and rearranging it, we found that the first number 'a' is equal to \( \frac{2s - pn}{n} \).

🎯 Exam Tip: For "prove that" questions, start with the given formulas and known relationships, then logically manipulate them step-by-step to arrive at the desired result. Be clear with each algebraic transformation.

Question 19. The sum of the first n terms of the arithmetical progression 3, \( 5\frac { 1 }{ 2 } \), 8, .... is equal to the 2nth term of the arithmetical progression \( 16\frac { 1 }{ 2 } \), \( 28\frac { 1 }{ 2 } \), \( 40\frac { 1 }{ 2 } \). Calculate the value of n.
Answer: For the first arithmetic progression: `3, \( 5\frac { 1 }{ 2 } \), 8, ....`
The first term \( a = 3 \).
The common difference \( d = 5\frac{1}{2} - 3 = \frac{11}{2} - \frac{6}{2} = \frac{5}{2} \).
The sum of the first \( n \) terms \( S_n \) is given by \( S_n = \frac{n}{2}[2a + (n-1)d] \):
\( S_n = \frac{n}{2}\left[2(3) + (n-1)\frac{5}{2}\right] \)
\( \implies S_n = \frac{n}{2}\left[6 + \frac{5n - 5}{2}\right] \)
\( \implies S_n = \frac{n}{2}\left[\frac{12 + 5n - 5}{2}\right] \)
\( \implies S_n = \frac{n}{4}[5n + 7] \)...(1)
For the second arithmetic progression: `\( 16\frac { 1 }{ 2 } \), \( 28\frac { 1 }{ 2 } \), \( 40\frac { 1 }{ 2 } \), ....`
The first term \( A = 16\frac{1}{2} = \frac{33}{2} \).
The common difference \( D = 28\frac{1}{2} - 16\frac{1}{2} = \frac{57}{2} - \frac{33}{2} = \frac{24}{2} = 12 \).
The \( 2n^{th} \) term \( T'_{2n} \) is given by \( T'_{2n} = A + (2n-1)D \):
\( T'_{2n} = \frac{33}{2} + (2n - 1)12 \)
\( \implies T'_{2n} = \frac{33}{2} + 24n - 12 \)
\( \implies T'_{2n} = 24n + \frac{33 - 24}{2} \)
\( \implies T'_{2n} = 24n + \frac{9}{2} \)...(2)
According to the question, \( S_n = T'_{2n} \). Equating (1) and (2):
\( \frac{n}{4}[5n + 7] = 24n + \frac{9}{2} \).
Multiply the entire equation by 4 to remove denominators:
\( n(5n + 7) = 4(24n) + 4\left(\frac{9}{2}\right) \)
\( \implies 5n^2 + 7n = 96n + 18 \).
Rearrange into a quadratic equation:
\( 5n^2 + 7n - 96n - 18 = 0 \)
\( \implies 5n^2 - 89n - 18 = 0 \).
Solve using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\( n = \frac{-(-89) \pm \sqrt{(-89)^2 - 4(5)(-18)}}{2(5)} \)
\( \implies n = \frac{89 \pm \sqrt{7921 + 360}}{10} \)
\( \implies n = \frac{89 \pm \sqrt{8281}}{10} \)
\( \implies n = \frac{89 \pm 91}{10} \).
This gives two possible values for \( n \):
\( n = \frac{89 + 91}{10} = \frac{180}{10} = 18 \)
\( n = \frac{89 - 91}{10} = \frac{-2}{10} = -\frac{1}{5} \).
Since the number of terms \( n \) must be a positive integer, we take \( n = 18 \).
In simple words: We have two lists of numbers. For the first list, we found a formula for the sum of 'n' terms. For the second list, we found a formula for its '2n'-th term. We set these two formulas equal to each other because the problem says they are the same. This led to a quadratic equation, and by solving it, we found that 'n' is 18.

🎯 Exam Tip: Carefully set up expressions for \(S_n\) and \(T_k\) for each A.P. Convert mixed fractions to improper fractions before calculating common differences. A common error is incorrect algebraic manipulation in solving the quadratic equation.

Question 20. If the sum of the first 4 terms of an arithmetic progression is p, the sum of the first 8 terms is q and the sum of the first 12 terms is r express \( 3p + r \) in terms of q.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the A.P.
We are given:
Sum of the first 4 terms \( S_4 = p \). Using \( S_n = \frac{n}{2}[2a + (n-1)d] \):
\( \frac{4}{2}[2a + (4 - 1)d] = p \)
\( \implies 2[2a + 3d] = p \)
\( \implies 4a + 6d = p \)...(1)
Sum of the first 8 terms \( S_8 = q \):
\( \frac{8}{2}[2a + (8 - 1)d] = q \)
\( \implies 4[2a + 7d] = q \)
\( \implies 8a + 28d = q \)...(2)
Sum of the first 12 terms \( S_{12} = r \):
\( \frac{12}{2}[2a + (12 - 1)d] = r \)
\( \implies 6[2a + 11d] = r \)
\( \implies 12a + 66d = r \)...(3)
We need to express \( 3p + r \) in terms of \( q \).
Substitute equations (1) and (3) into \( 3p + r \):
\( 3p + r = 3(4a + 6d) + (12a + 66d) \)
\( \implies 3p + r = 12a + 18d + 12a + 66d \)
\( \implies 3p + r = 24a + 84d \).
Now, factor out 3 from the right side:
\( 3p + r = 3(8a + 28d) \).
From equation (2), we know that \( 8a + 28d = q \). Substitute this into the expression:
\( 3p + r = 3(q) \)
\( \implies 3p + r = 3q \).
This shows a linear relationship between the sums of different numbers of terms in an A.P.
In simple words: We have the sums of 4, 8, and 12 terms as p, q, and r. We wrote formulas for these sums using the first term and common difference. Then, we used these formulas to show that `3p + r` is equal to `3q`.

🎯 Exam Tip: When expressing one quantity in terms of another, set up all known equations involving 'a' and 'd'. Then, manipulate them algebraically to substitute and simplify until you get the desired relationship. This problem highlights a specific property of A.P. sums: if sums \(S_x, S_y, S_z\) are given, there often exists a linear relation between them.

Question 25. If the sums of n, 2n, 3n terms of an A.S are S1, S2, S3 respectively, prove that S3 = 3(S2 S1)
Answer: Let 'a' be the first term and 'd' be the common difference of the given arithmetic progression (A.P.). We write the sum of n terms as \( S_n = \frac{n}{2}[2a + (n - 1)d] \).
So, the given sums are:
\( S_1 = \frac{n}{2}[2a + (n - 1)d] \) ...(1)
\( S_2 = \frac{2n}{2}[2a + (2n - 1)d] \) ...(2)
\( S_3 = \frac{3n}{2}[2a + (3n - 1)d] \) ...(3)
Now, let's calculate \( S_2 - S_1 \):
\( S_2 - S_1 = \frac{2n}{2}[2a + (2n - 1)d] - \frac{n}{2}[2a + (n - 1)d] \)
\( = \frac{n}{2} \left( 2[2a + (2n - 1)d] - [2a + (n - 1)d] \right) \)
\( = \frac{n}{2} [4a + (4n - 2)d - 2a - (n - 1)d] \)
\( = \frac{n}{2} [4a + 4nd - 2d - 2a - nd + d] \)
\( = \frac{n}{2} [2a + 3nd - d] \)
\( = \frac{n}{2} [2a + (3n - 1)d] \)
Next, we multiply this result by 3:
\( 3(S_2 - S_1) = 3 \cdot \frac{n}{2} [2a + (3n - 1)d] \)
\( = \frac{3n}{2} [2a + (3n - 1)d] \)
From equation (3), we know that \( S_3 = \frac{3n}{2}[2a + (3n - 1)d] \).
Therefore, \( S_3 = 3(S_2 - S_1) \). This identity is useful for checking the consistency of arithmetic series calculations.
In simple words: To prove this, we first write down the formulas for the sum of 'n', '2n', and '3n' terms of an arithmetic progression. Then, we substitute these formulas into the right side of the equation we want to prove. After simplifying, we find that the right side becomes exactly the formula for the sum of '3n' terms, which is the left side. This confirms the statement.

🎯 Exam Tip: When proving identities involving sums of APs, always start by defining the first term 'a' and common difference 'd', then write out the sum formulas. Work systematically from one side (usually the more complex side) to reach the other, being careful with algebraic simplification.

Question 26. If the sum of p terms of an A.S is q and the sum of q terms is p, show that the sum of (p + q) terms is -(p + q).
Answer: Let 'a' be the first term and 'd' be the common difference of the arithmetic progression (A.P.). Given that the sum of p terms (\(S_p\)) is q, and the sum of q terms (\(S_q\)) is p.
Using the formula for the sum of n terms, \( S_n = \frac{n}{2}[2a + (n - 1)d] \):
\( S_p = q \implies \frac{p}{2}[2a + (p - 1)d] = q \)
\( \implies 2ap + p(p - 1)d = 2q \) ...(1)
\( S_q = p \implies \frac{q}{2}[2a + (q - 1)d] = p \)
\( \implies 2aq + q(q - 1)d = 2p \) ...(2)
Subtract equation (2) from equation (1):
\( (2ap - 2aq) + [p(p - 1)d - q(q - 1)d] = 2q - 2p \)
\( 2a(p - q) + d[p^2 - p - (q^2 - q)] = 2(q - p) \)
\( 2a(p - q) + d[p^2 - q^2 - p + q] = 2(q - p) \)
\( 2a(p - q) + d[(p - q)(p + q) - (p - q)] = -2(p - q) \)
\( 2a(p - q) + d(p - q)[(p + q) - 1] = -2(p - q) \)
Since \( p \neq q \), we can divide both sides by \( (p - q) \):
\( 2a + d(p + q - 1) = -2 \) ...(3)
Now, we need to find the sum of \( (p + q) \) terms, \( S_{p+q} \):
\( S_{p+q} = \frac{p + q}{2} [2a + (p + q - 1)d] \)
Substitute the value of \( [2a + (p + q - 1)d] \) from equation (3) into this expression:
\( S_{p+q} = \frac{p + q}{2} (-2) \)
\( = -(p + q) \)
Hence proved. This problem highlights how sums in an AP can have interesting reciprocal properties.
In simple words: We are given two facts about the sums of terms in an arithmetic progression: the sum of 'p' terms is 'q', and the sum of 'q' terms is 'p'. We write these as two equations using the sum formula. By subtracting the equations, we find a simple relationship between the first term and common difference. We then use this relationship in the formula for the sum of 'p+q' terms, which simplifies to '-(p+q)'.

🎯 Exam Tip: For problems involving sums of p and q terms, remember to set up simultaneous equations. The key step is often subtracting the two sum equations to find a simplified expression for \( 2a + d(p + q - 1) \), which can then be directly substituted into the sum formula for \( S_{p+q} \).

Question 27. The ratio between the sum of n terms of two A.P.'s is (7n + 1) : (4n + 27). Find the ratio of their 11 th terms.
Answer: Let the first arithmetic progression (A.P.) have its first term as 'a' and common difference as 'd'. Let the second A.P. have its first term as 'A' and common difference as 'D'.
The ratio of the sum of n terms for the two A.P.'s is given as:
\( \frac{S_n}{S'_n} = \frac{\frac{n}{2}[2a + (n - 1)d]}{\frac{n}{2}[2A + (n - 1)D]} = \frac{7n + 1}{4n + 27} \)
This simplifies by canceling \( \frac{n}{2} \):
\( \frac{2a + (n - 1)d}{2A + (n - 1)D} = \frac{7n + 1}{4n + 27} \)
To find the ratio of their 11th terms (\(T_{11}\) and \(T'_{11}\)), we know that the formula for the k-th term is \( T_k = a + (k - 1)d \). For the 11th term, \( k = 11 \), so we need the form \( a + 10d \).
To get this form from the sum ratio, we divide the numerator and denominator of the left side by 2:
\( \frac{a + \left(\frac{n - 1}{2}\right)d}{A + \left(\frac{n - 1}{2}\right)D} = \frac{7n + 1}{4n + 27} \)
Now, we equate the coefficient of 'd' to 10 (which is \( k - 1 \) for the 11th term):
\( \frac{n - 1}{2} = 10 \)
\( \implies n - 1 = 20 \)
\( \implies n = 21 \)
Substitute \( n = 21 \) into the ratio of the sums to find the ratio of the 11th terms:
\( \frac{T_{11}}{T'_{11}} = \frac{7(21) + 1}{4(21) + 27} \)
\( = \frac{147 + 1}{84 + 27} \)
\( = \frac{148}{111} \)
We can simplify this fraction by dividing both numerator and denominator by their greatest common divisor, which is 37:
\( \frac{148}{111} = \frac{4 \times 37}{3 \times 37} = \frac{4}{3} \)
Thus, the ratio of their 11th terms is 4:3.
In simple words: We are given how the sums of 'n' terms for two different number patterns (arithmetic progressions) compare. To find how their 11th terms compare, we use a special trick: we replace 'n' in the sum ratio formula with a specific number (which turns out to be 21) that makes the sum formula look like the formula for the 11th term. Plugging this value of 'n' into the given ratio then directly gives us the ratio of the 11th terms.

🎯 Exam Tip: For problems relating the ratio of sums to the ratio of terms in an AP, remember the key insight: to find the ratio of the k-th terms, you substitute \( n = 2k - 1 \) into the given ratio of sums of n terms. This correctly transforms the sum formula's \( (n-1)d \) into \( 2(k-1)d \), allowing a direct comparison with the term formula \( a+(k-1)d \).