Geometric Progression

The corrrect answer is 41651

Now



= 1 + 25 × 17 (101 – 3)
= 41651

The correct answer is (A) : 21 a₁₁
Let first term of G.P. be a and common ratio is r
Then,





∴ Then,

The correct answer is 40
Let G.P. be a₁ = a, a₂ = ar, a₃ = ar², ……


⇒ 2ar² - r - 3 = 0
∴ r = -1 or
∵ a1 = a > 0 then r ≠ -1
Now,






= 40

From the given conditions:

Also, since are in arithmetic progression:

Using equation (1) and simplifying:

Substituting , we solve for and , eventually leading to:

Now compute :

Step 1: General form of terms in a geometric progression. In a geometric progression (GP), the general term is given by: where is the first term and is the common ratio.

Step 2: Using the given conditions. We are given the product of the fourth and sixth terms: Using the general form of the terms: so: We are also given the sum of the fifth and seventh terms: Using the general form of the terms: so:

Step 3: Solving equations (1) and (2). From equation (1): From equation (2): Substitute equation (4) into equation (3): Solving this:

Step 4: Finding and solving for the desired expression. From equation (3), substitute into the equation: Now, we compute the expression . The terms are: so: Summing these terms: Simplifying: Substitute : Thus: Finally:

...(i)

Given:

The terms and are in a geometric progression (GP).

• Step 1: Simplify the given equation:

• Factorize :

.

• Step 2: Substitute into the equation:

.

• Rearrange:

.

• Factorize :

.

• Solve for :

(since ).

• Step 3: Calculate the sum of the three terms:

.

• Substitute and :

.

.

Final Answer: The sum of the three terms is .

The given terms are .
Given that , we substitute this into the terms: Now, use the given sum of the 5 terms being : Now solve for : Thus, . Now, substitute into : Thus, the value of is .

The given problem is about finding the common ratio of a geometric progression (G.P.) under specific conditions. Let's solve it step by step:

Let the first term of the G.P. be and the common ratio be . The sum of all 64 terms in the G.P. can be expressed using the formula for the sum of a geometric series:

The odd terms of the G.P. form another G.P. with the first term and the common ratio . Since there are 64 terms in total, and half of them are odd terms (i.e., 32 terms), their sum is given by:

According to the problem, the total sum of all terms is 7 times the sum of the odd terms:

We can simplify the equation by canceling from both sides (assuming and ):

Cross-multiplying gives us:

Expanding and simplifying the equation:

Factoring the quadratic equation, we get:

Thus, the solutions are and . Since would result in a constant sequence (which doesn't satisfy the conditions unless specified otherwise), we discard it. Therefore, the common ratio is:

The correct answer is 6.

Define the first GP with first term and common ratio . Given , we have:

The 11th term of the first GP is:

Define the second GP with first term and common ratio . Given , we have:

The pth term of the second GP is:

Since , we have:

Dividing both sides by , we get:

Equate the exponents:

Solving for :

To solve this problem, we need to work with the properties of a geometric progression (G.P.). Let's denote the first term of the G.P. as and the common ratio as . Therefore, the terms of the sequence can be expressed as follows:

• First term:
• Second term:
• Third term:
• Fourth term:
• Fifth term:
• Sixth term:
• Eighth term:

According to the problem, the sum of the second (i.e., ) and sixth (i.e., ) terms is given by:

... (1)

The product of the third (i.e., ) and fifth (i.e., ) terms is:

Which simplifies to:

... (2)

From equation (1), factor out :

Substitute the value of from equation (2):

Plugging it into equation (1):

Multiplying through by gives:

Simplifying, this turns into:

Or:

Let . Then, the equation becomes a quadratic:

Dividing throughout by 7 gives:

Using the quadratic formula, , find the roots:

Thus, or .

(since is positive).

Substitute into the expression for :

Thus, calculate the sum of the 4th ( ), 6th ( ), and 8th ( ) terms:

The terms are:

Therefore, the required sum is:

Thus, the sum of the 4th, 6th, and 8th terms is 91.

To find the range of , observe that it can be rewritten as:

where . Since , we have .

Simplifying the expression, we get:

Finding the range involves checking critical points and endpoint behavior. Consider:

Use derivatives or direct evaluation to find that the range is , so and .

Now, compute the sum of the infinite geometric progression (G.P.) with first term and common ratio .

The sum of an infinite G.P. is:

Thus, the sum of the G.P. is 96, which indeed falls within the range [96,96].

To solve this problem, we are given two conditions based on the infinite geometric progression (G.P.):

• The sum of the infinite G.P. is .
• The sum of the cube terms of each term in the G.P. is .

We will use the formula for the sum of an infinite geometric series, which is:

where .

1. Applying the sum formula to the first series:

• .
• Let's denote this as Equation (1).

2. Applying the sum formula to the second series:

Note that the common ratio is and the first term is :

• .
• Let's denote this as Equation (2).

Now, let's solve these two equations:

From Equation (1):

• .

Substitute in Equation (2):

Simplifying gives:

Since can be cancelled, we have:\)

Simplifying, we have:

Using , we simplify:\)

Simplifying gives us:

\) , hence \) , \) .

From :

.

Finding :

.

Hence, the value of is 31, making the correct answer 31.

To find the sum of the first nine terms of the given geometric progression (G.P.), we use the information provided in the question and the properties of G.P.

Let the first term of the G.P. be and the common ratio be . Then, the terms of the G.P. can be expressed as

The second, fourth, and sixth terms of the G.P. are:

• Second term:
• Fourth term:
• Sixth term:

According to the problem, the sum of these terms is 21:

(Equation 1)

The eighth, tenth, and twelfth terms of the G.P. are:

• Eighth term:
• Tenth term:
• Twelfth term:

According to the problem, the sum of these terms is 15309:

(Equation 2)

Now, divide Equation 2 by Equation 1:

Thus,

, hence

Now, substitute back into Equation 1 to find :

Finally, we calculate the sum of the first nine terms of the G.P.:

The sum of the first terms of a G.P. is given by:

Substituting the known values , , and :

Hence, the sum of the first nine terms is 757.

Step 1: Let the GP be: Since the GP is increasing, .
Step 2: Using the given conditions:
Step 3: From (2), . Substitute in (1): Let :
Step 4: Hence,
Step 5: Required sum: