Arithmetic Progression

The correct answer is (C) : 23
a, A₁, A₂ …….. A_n, 100
Let d be the common difference of above A.P. then

⇒ 7a + 8d = 100 …(i)
and a + n = 33 …(ii)
and 100 = a + (n + 1)d
⇒
⇒ 800 = 8a + 7a² – 338a + 3400
⇒ 7a² – 330a + 2600 = 0

but

∴ n = 23

The correct answer is 53




d_i= 33 + 11, 9
Sum of CD’s = 33 + 11 + 9
= 53

The correct answer is 166




∴ m + n = 55 + 111 = 166

Given an arithmetic progression (A.P.) , where the sum of terms in the form , we need to find .

In an A.P., each term can be expressed as , where is the first term and is the common difference. Therefore:

We can split the sum into two parts:

1. The first sum is a geometric series:

The sum of an infinite geometric series with first term and common ratio is:

Thus, .

2. For the second sum, consider:

Decompose using the known series formula for which is :

, so

, thus .

Substituting back, we have:

The task is to find . From , it follows:

Therefore, , which fits within the provided range .

The correct answer is 2223
Given : 3,6,9,12 ..... upto 78 terms
5,9,13,17 ...... upto 59 terms
Now , last term of first series t₇₈
and last term of second series t₅₉
Now , common difference of common terms = LCM {3,4} = 12
Therefore , First common term is 9 and last common term is 225
So, series will be 9,21,33 ....... 225
⇒ 9 + (n-1)12 = 225
⇒ n = 19
∴ Sum of common on terms =

Step-by-step solution

1. Observe the pattern of set sizes.
   The n-th set contains 2n − 1 elements (odd numbers 1,3,5,…).
2. Find how many terms appear before the n^th set.
   The total number of terms in the first (n−1) sets is the sum of the first (n−1) odd numbers: So the index (in the sequence of multiples of 3) of the first term of the n^th set is
3. Apply the formula for n = 11.
   For the 11^th set: so the first term is the 101^st positive multiple of 3: The number of terms in the 11^th set is These 21 terms are consecutive multiples of 3: The last term is
4. Compute the sum of this arithmetic progression.
   The sum of an arithmetic progression with terms, first term and last term is Here . So

Final answer

The sum of the elements in the 11^th set is 6993.

Bonus — compact general formula

For the n-th set (with size ), the middle term index is , so the middle term value is Since an arithmetic block of odd length has the average equal to the middle term, the sum of the n^th set is For this gives , confirming the result.

Solution:

First, notice that the pattern appears to alternate between a coefficient that is either or an odd integer ( ) multiplied by a power of or . However, the provided solution snippet shows a regrouping approach:

The snippet then shows evaluating several known sums:

• Sum of first integers: .
• Sum of first squares: .

After carefully combining like terms (and possibly correcting sign patterns), the final numeric answer shown is:

Therefore, the sum of the first 20 terms of the given series is .

Analysis:

The series is:

The solution snippet proposes a regrouping strategy:

• Even Squares:
• Consecutive Integers:
• Group A: (This is the crucial part that's not explicitly calculated.)

Using Sum Formulas (Correctly Applied)

• Sum of Squares:
• Sum of Integers:

The Missing Calculation: Group A

The key issue is that the snippet doesn't show how to calculate Group A. This group consists of alternating terms with increasing powers of 2 and odd coefficients.

Why This Is Important

Without calculating Group A, we can't verify the final answer of 1310. It's essential to understand how the alternating terms contribute to the overall sum.

What We Know (and Don't Know)

• We know how the even squares and consecutive integers are calculated.
• We know the final answer is stated as 1310.
• We don't know the exact calculation for Group A.

The Need for a Complete Solution

To provide a fully verified solution, we would need to:

1. Derive a general form for the terms in Group A.
2. Find a method to sum the alternating series in Group A.
3. Combine the result from Group A with the sums of even squares and consecutive integers.

In conclusion: While the snippet provides a useful regrouping strategy and correctly applies sum formulas, it lacks the crucial calculation of Group A, which is necessary to confirm the final answer of 1310.

Given Arithmetic Progression: 3, 8, 13, ..., 373

Step 1: Finding the number of terms (n):

T_n = a + (n - 1)d

Substitute :

373 = 3 + (n - 1)5

Step 2: Sum of the arithmetic progression:

Substitute :

Step 3: Finding the sum of terms divisible by 3: Numbers divisible by 3 are 3, 18, 33, ..., 363.

Sum of these terms:

Substitute :

Step 4: Required sum:

The given arithmetic progressions (APs) are:



The least common multiple (LCM) of the common differences is:

The first common term of the three sequences can be found by checking the terms that satisfy all three APs. The common terms are:

The sum of the common terms is:

The correct answer is option (B): 1

The problem involves an arithmetic progression (A.P.) where the sixth term . We need to find the common difference that maximizes the product of three terms: .

Let's analyze the problem step-by-step:

1. In an arithmetic progression, the nth term can be given by the formula: .
2. Thus, the sixth term can be written as: .
3. From the above, we get: .
4. Now we know:
   • ,
   • ,
   • .
5. The product can be written as: .
6. To maximize this product, we consider the roots of these linear expressions, which hints that should be zero or near zero for one of its factors. To find the critical points, we look at the value of that could make two of the factors equalized.
7. A trial and error or mathematical optimization technique can help find that setting to an appropriate value within given options maximizes the product.

After testing the values, choosing , you find that it results in equalizing terms and maximizing the product due to symmetry. Therefore, the common difference is .

Given the sequence:

Step 1: Finding the general term
The general term is given by:

Step 2: Finding the 10th term
Simplifying:
Hence,

Step 3: Sum of 10 terms with common difference (c.d.) = 3
Simplifying:

Final Answer:

1. To determine the 20^th term from the end of the given arithmetic sequence, let's first identify the characteristics of the sequence. The sequence is given as .
2. Check the first few differences to confirm it's an arithmetic sequence:
   • Second term:
   • This indicates a common difference, , between terms (i.e., the difference between subsequent terms is negative).
3. The formula for the term of an arithmetic sequence is given by: , where:
   • is the first term, in this case.
   • is the common difference, .
4. To find the 20^th term from the end, first determine the total number of terms in the sequence. Use the first term and last given term: .
5. Solve for :
   • Rewriting as
   • Multiply by :
6. So, there are 200 terms in the sequence. To find the 20^th term from the end, calculate the 181^st term from the start:
7. Calculate :
8. Therefore, the 20^th term from the end of the sequence is -115, which is the correct answer.

To solve this problem, we need to understand the impact of computer system crashes on the completion time of the assignment.

Initially, the company plans to use computer systems to complete the assignment in 17 days. With computer crashes occurring each day after the first, we know that:

• On the first day, no computer system crashes.
• On the second day, 4 systems crash.
• On the third day, 4 more systems crash, and this pattern continues daily.

The assignment actually took 8 more days longer than planned. Hence, the assignment was completed in days.

We need to set up an equation to determine the initial number of systems .

Work Done Analysis

Let’s calculate the total work in terms of "computer-days" required to complete the assignment without crashes:

The total work required is computer-days.

Accounting for Crashes

Now consider the reduction in computer-days due to crashes:

• Day 1: Work done by systems.
• Day 2: Work done by systems.
• Day 3: Work done by systems.
• ... and so forth until the end of the 25 days, following the pattern.

Therefore, the total work done over 25 days is:

where is the day number, totaling 25.

Summation

The series can be simplified with an arithmetic series where:

• Common difference
• We are given terms.

The last term, .

The sum of the first terms of an arithmetic series is given by:

Substitute in the values to set up the equation:

The term is a sum of the first 24 natural numbers:

So the equation simplifies to:

Solving for gives:

Thus, the initial number of computer systems, , is 150.

Let the first term of the A.P. be and the common difference be . The A.P. has terms. The last term of the A.P. can be expressed as . According to the problem, this last term exceeds the first term by 27, so we have:

This simplifies to:

The sum of the odd terms consists of terms. Using the sum formula for an A.P., the sum of these terms is given as 40:

This can be rewritten as:

Similarly, the sum of the even terms also consists of terms, and the sum is given as 55:

This simplifies to:

Now, we have the following system of equations:

Simplifying the first equation, we have:

Substitute in the second equation:

From the third equation, substituting :

Subtract the second equation from the third:

This reduces to:

Solving for , we substitute:

Simplifying:

This simplifies to:

Therefore:

Thus:

Thus, the correct value of that satisfies all conditions is 5.

Let the elements of set be (since they are in A.P.) and the elements of set be .
The sum of the elements of set is given by: The product of the elements of set is: Similarly, the sum of the elements of set is: The product of the elements of set is: We are given that , so substitute into the equation for : Now, we are given the relation: Substitute the expressions for and into this relation, and solve for . After solving, we get .
Thus, the correct answer is .

To solve the problem, let's analyze the information given and make use of the formulas related to an arithmetic progression (A.P.). The problem states the following:

• The sum of the first terms of the A.P., denoted as , is 700 for some .
• The 6th term, , is 7.
• The sum of the first 7 terms, , is 7.

The sum of the first terms of an A.P. is given by:

where is the first term, is the common difference, and is the number of terms.

Let's first use the condition: . Plugging into the formula for the sum, we have:

Simplifying gives:

Dividing by 7, we get:

Now, using the condition: .

The nth term of an A.P. is given by:

For , we have:

We now have two equations:

• (1)
• (2)

Subtract equation (1) from equation (2):

Solve for :

Substitute back into equation (1):

Now, using the sum condition in the equation:

Simplifying, we get:

Solve this quadratic equation for using the quadratic formula.

After finding the feasible , we can calculate :

Substitute the appropriate values of and to find:

Thus, the correct answer is 64.

We are tasked with solving the problem involving an arithmetic progression (AP) and determining the value of the 11th term . Let us proceed step by step:

1. Representation of the AP:
The terms of the AP are represented as:

2. Sum of the First Three Terms:
The sum of the first three terms is given as 54:



Dividing through by 3:

3. Constraint on the Sum of the First 20 Terms:
The sum of the first 20 terms lies between 1600 and 1800:



Simplify the inequality:



Divide through by 10:

4. Substituting :
From Equation (i), . Substitute into :



Thus, the inequality becomes:



Subtract 36 from all sides:



Divide through by 17:





Since must be an integer, we conclude:

5. Solving for :
From Equation (i), :

6. Finding :
The general formula for the -th term of an AP is:



For :



Substitute and :

Final Answer:
The value of is .

To solve this problem, let's break it down step-by-step:

Step 1: Understanding the Arithmetic Progression (A.P.)

The first term of the A.P. is given as . We need to find the common difference (denoted as ) and eventually the sum of the first 20 terms.

Step 2: Conditions given

We are told that the sum of the first four terms is equal to one-fifth of the sum of the next four terms.

The sum of the first four terms is:

Substitute :

The next four terms would be the 5th to the 8th terms. The sum is:

Which simplifies to:

According to the problem, :

Solving this equation:

Multiply through by 5 to eliminate the fraction:

Rearrange to find :

Step 3: Find the sum of the first 20 terms

The sum of the first terms of an A.P. is given by:

For the first 20 terms:

Therefore, the sum of the first 20 terms is , which matches the given correct option.

We are given the sum: In an Arithmetic Progression (A.P.), the sum of terms equidistant from the ends is equal, so: Thus, the number of pairs is: Hence, we calculate: Now using the sum of A.P. formula, we get:

We are asked to find the remainder when is divided by 7. First, we reduce modulo 7: Thus, . Now, since , we get: Hence, the remainder when is divided by 7 is .
Therefore, the correct answer is (D) 1.