Let $a_1, a_2, a_3, ldots$ $a_1, a_2, a_3, \ldots$ be a G.P. of increasing positive numbers[cite: 655]. If $a_3a_5 = 729$ $a_3a_5 = 729$ and $a_2 + a_4 = frac1114$ $a_2 + a_4 = \frac{111}{4}$ [cite: 656], then $24(a_1 + a_2 + a_3)$ $24(a_1 + a_2 + a_3)$ is equal to[cite: 657]:

A. 131
B. 130
C. 129
D. 128

Solution & Explanation

### Related Formula For a geometric sequence configuration with first term

a

and common ratio

r

a_n = a cdot r^n-1

### Core Logic Convert information markers using parameter notations [cite: 1372, 1373, 1376]:

a_3 a_5 = (ar^2)(ar^4) = a^2 r^6 = 729 implies ar^3 = 27

[cite: 1373, 1374] From second expression block data [cite: 1376]:

a_2 + a_4 = ar + ar^3 = frac1114

[cite: 1376] Substitute

ar^3 = 27

directly into the linear equation block [cite: 1376]:

ar + 27 = frac1114 implies ar = frac1114 - 27 = frac34

[cite: 1376] ### Step 1: Finding parameters a and r Divide the calculated components to evaluate the ratio [cite: 1387]:

fracar^3ar = frac273/4 implies r^2 = 36 implies r = 6

[cite: 1387] (Choose

+6

because terms must stay strictly positive [cite: 655]). Find value for first base variable

a

[cite: 1389]:

a(6) = frac34 implies a = frac18

[cite: 1389] ### Step 2: Sum configuration resolving Now compute targeted expansion expression value [cite: 1390]:

24(a_1 + a_2 + a_3) = 24(a + ar + ar^2) = 24a(1 + r + r^2)

[cite: 1390]

= 24 cdot left(frac18right) cdot (1 + 6 + 36) = 3 cdot 43 = 129

[cite: 1390, 1391] ### Pattern Recognition Product entries like

a_3 a_5 = a_4^2

help identify the central term index value quickly in symmetric geometric progressions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

Reference Study Guides

Sequences And Series

More Sequences and Series Previous-Year Questions

Q53 2025 Arithmetic Progression

The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by

frac212

. Then the number of terms which are integers in the A.P. is :

A. 4
B. 10
C. 6
D. 8

Solution

### Related Formula

textSum of an A.P.: S_k = frack2 left( 2a + (k-1)d right)

textGeneral term of an A.P.: a_k = a_1 + (k-1)d

### Core Logic Let the A.P. have

n

terms (where

n

is even). The terms are divided into

n/2

odd-indexed terms and

n/2

even-indexed terms. ### Step 1: Set up the even and odd sums Sum of even terms:

a_2 + a_4 + dots + a_n = 30 quad text--- (1)

Sum of odd terms:

a_1 + a_3 + dots + a_n-1 = 24 quad text--- (2)

Subtracting equation (2) from (1):

(a_2 - a_1) + (a_4 - a_3) + dots + (a_n - a_n-1) = 30 - 24 = 6

Since there are

n/2

such pairs, and the difference of adjacent terms is the common difference

d

fracn2 d = 6 implies n d = 12 quad text--- (3)

### Step 2: Solve for n and d We are given that the last term exceeds the first by

frac212

a_n - a_1 = (n-1)d = frac212

n d - d = 10.5

Substitute

nd = 12

from (3):

12 - d = 10.5 implies d = 1.5 = frac32

Using this in (3):

n left(frac32right) = 12 implies n = 8

### Step 3: Solve for the first term The sum of the odd terms is:

S_textodd = frac42 left[ 2a_1 + (4-1)(2d) right] = 24

2 left[ 2a_1 + 3(3) right] = 24 implies 2a_1 + 9 = 12 implies a_1 = 1.5 = frac32

Thus, the terms are:

frac32, \, 3, \, frac92, \, 6, \, frac152, \, 9, \, frac212, \, 12

The terms that are integers are

3, 6, 9, 12

. The total number of integer terms is 4. ### Pattern Recognition Sum of even terms minus sum of odd terms in any A.P. with an even number of terms

n

is always equal to

fracn2 d

. This is an extremely useful relation to remember. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

Q72 2025 Sum of Special Series

If the sum of the first 10 terms of the series

frac4 cdot 11 + 4 cdot 1^4 + frac4 cdot 21 + 4 cdot 2^4 + frac4 cdot 31 + 4 cdot 3^4 + dots

fracmathrmmmathrmn

, where

gcd(mathrmm, mathrmn) = 1

, then

mathrmm + mathrmn

is equal to ____________.

Numerical Answer. Answer: 441 to 441

Solution

### Related Formula

textSophie Germain's algebraic factorization: 1 + 4r^4 = (2r^2 + 2r + 1)(2r^2 - 2r + 1)

textTelescoping Series representation: T_r = f(r) - f(r+1)

### Core Logic This is a telescoping series sum. We expand the denominator using Sophie Germain's algebraic identity to write the general term as a difference of two consecutive rational expressions. ### Step 1: Write down the general term and factor The general term

T_r

of the series is:

T_r = frac4r1 + 4r^4

Using the factorization of

1+4r^4

T_r = frac4r(2r^2 - 2r + 1)(2r^2 + 2r + 1)

Notice that the numerator

4r

is the exact difference of the two quadratic factors:

(2r^2 + 2r + 1) - (2r^2 - 2r + 1) = 4r

### Step 2: Split the fraction into telescoping terms Rewrite the general term

T_r

T_r = frac(2r^2 + 2r + 1) - (2r^2 - 2r + 1)(2r^2 - 2r + 1)(2r^2 + 2r + 1)

T_r = frac12r^2 - 2r + 1 - frac12r^2 + 2r + 1 = f(r) - f(r+1)

### Step 3: Expand the sum and solve Sum the first 10 terms: - For

r = 1

T_1 = frac11 - frac15

- For

r = 2

T_2 = frac15 - frac113

- ... - For

r = 10

T_10 = frac1181 - frac1221

All intermediate terms cancel out:

S_10 = 1 - frac1221 = frac220221 = fracmn

Since

220

and

221

are coprime (their greatest common divisor is 1):

m = 220 quad textand quad n = 221

m + n = 220 + 221 = 441

### Pattern Recognition Sophie Germain identity: The expansion

4r^4 + 1 = (2r^2 - 2r + 1)(2r^2 + 2r + 1)

is highly common in telescoping series. Spotting this factorization collapses the sum instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

Q 2025 Arithmetic Progression Summation

Let

a_1, a_2, a_3, ldots

be in an A.P. such that

sum_k=1^12 a_2k-1 = -frac725 a_1, a_1 neq 0

. If

sum_k=1^n a_k = 0

, then

n

is:

A. $11$ $11$
B. $10$ $10$
C. $18$ $18$
D. $17$ $17$

Solution

### Related Formula Sum of an Arithmetic Progression:

S_m = fracm2[2a_1 + (m-1)d]

### Core Logic Express the odd terms summation in terms of

a_1

and

d

, establish their linear dependency, and then solve for

n

where total sum vanishes. ### Step 1: Simplify the Given Summation The summation represents the sum of

12

terms:

a_1 + a_3 + a_5 + dots + a_23

. This is an A.P. with initial term

a_1

and common difference

2d

frac122[2a_1 + 11(2d)] = -frac725a_1

6[2a_1 + 22d] = -frac725a_1 implies 12a_1 + 132d = -frac725a_1

### Step 2: Relate initial term to common difference Multiply through by

5

to eliminate the fraction:

60a_1 + 660d = -72a_1 implies 132a_1 + 660d = 0 implies a_1 = -5d

### Step 3: Solve for n Set the general sum of

n

terms to

0

fracn2[2a_1 + (n-1)d] = 0 implies 2a_1 + (n-1)d = 0

Substitute

a_1 = -5d

2(-5d) + (n-1)d = 0 implies -10d + nd - d = 0 implies nd = 11d

Since

a_1 neq 0

d neq 0

, which yields:

n = 11

### Pattern Recognition The condition

a_1 = -5d

means the sequence starts positive/negative and counts down symmetrically. A sum of

n

terms equals zero when the middle term or balanced pairs completely wipe each other out, pointing directly to

n = 2(5) + 1 = 11

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

Q70 2025 Special Series

The sum

1 + frac1+32! + frac1+3+53! + frac1+3+5+74! + dots

is equal to

A. $6e$ $6e$
B. $4e$ $4e$
C. $3e$ $3e$
D. $2e$ $2e$

Solution

### Related Formula Sum of first

r

odd natural numbers:

sum_k=1^r (2k-1) = r^2

Exponential series expansion:

sum_r=0^infty frac1r! = e

### Core Logic Let's find the general

r

-th term of the series:

T_r = frac1 + 3 + 5 + dots + (2r-1)r! = fracr^2r! = fracr(r-1)!

### Step 1: Expressing term in terms of sum limits Let's write

r = (r-1) + 1

T_r = fracr-1+1(r-1)! = frac1(r-2)! + frac1(r-1)!

Our infinite sum is:

S = sum_r=1^infty T_r = sum_r=2^infty frac1(r-2)! + sum_r=1^infty frac1(r-1)!

Both sums are standard representations of the exponential expansion. ### Step 2: Summing the parts - First part:

sum_r=2^infty frac1(r-2)! = 1 + frac11! + frac12! + dots = e

- Second part:

sum_r=1^infty frac1(r-1)! = 1 + frac11! + frac12! + dots = e

textTotal Sum S = e + e = 2e

### Pattern Recognition The general term containing

r^2

in summation with factorials converges to

2e

. Remember the shortcut:

sum fracr^2r! = 2e

sum fracr^3r! = 5e

. It is extremely useful to memorize these common limits. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series Class 12 Mathematics: Limits, Continuity and Differentiability

Q63 2025 Arithmetico-Geometric Progression

Let

mathbfx_1, mathbfx_2, mathbfx_3, mathbfx_4

be in a geometric progression. If 2, 7, 9, 5 are subtracted respectively from

mathbfx_1, mathbfx_2, mathbfx_3, mathbfx_4

then the resulting numbers are in an arithmetic progression. Then the value of

frac124 (mathbfx_1 mathbfx_2 mathbfx_3 mathbfx_4)

is:

A. $72$ $72$
B. $18$ $18$
C. $36$ $36$
D. $216$ $216$

Solution

### Related Formula For a geometric progression, the terms can be set as

a, ar, ar^2, ar^3

. For three terms

A, B, C

to be in arithmetic progression, they must satisfy:

2B = A + C

### Core Logic Let the elements be

x_1 = a, x_2 = ar, x_3 = ar^2, x_4 = ar^3

. After the specified subtractions, the sequence becomes:

a - 2, quad ar - 7, quad ar^2 - 9, quad ar^3 - 5

Since this sequence is in AP, we form two separate common difference linear linkages:

2(ar - 7) = (a - 2) + (ar^2 - 9) implies 2ar - 14 = ar^2 + a - 11 implies ar^2 - 2ar + a + 3 = 0 quad dots (1)

2(ar^2 - 9) = (ar - 7) + (ar^3 - 5) implies 2ar^2 - 18 = ar^3 + ar - 12 implies ar^3 - 2ar^2 + ar + 6 = 0 quad dots (2)

### Step 1: Solve the Simultaneous Polynomials Multiply equation (1) by

r

ar^3 - 2ar^2 + ar + 3r = 0 quad dots (3)

Subtract equation (3) from equation (2):

(ar^3 - 2ar^2 + ar + 6) - (ar^3 - 2ar^2 + ar + 3r) = 0

6 - 3r = 0 implies 3r = 6 implies r = 2

Substitute

r = 2

back into equation (1):

a(2)^2 - 2a(2) + a + 3 = 0

4a - 4a + a + 3 = 0 implies a = -3

### Step 2: Find the Continuous Product Value The continuous product term is:

mathbfx_1mathbfx_2mathbfx_3mathbfx_4 = a cdot ar cdot ar^2 cdot ar^3 = a^4 r^6

mathbfx_1mathbfx_2mathbfx_3mathbfx_4 = (-3)^4 cdot (2)^6 = 81 times 64 = 5184

Now divide by 24 as required:

frac124(5184) = 216

### Pattern Recognition Notice that multiplying the first AP condition equation by

r

perfectly mimics the structure of the second condition equation except for the absolute scalar value, allowing direct elimination of all polynomial variable indices simultaneously. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

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Let $a_1, a_2, a_3, ldots$ $a_1, a_2, a_3, \ldots$ be a G.P. of increasing positive numbers[cite: 655]. If $a_3a_5 = 729$ $a_3a_5 = 729$ and $a_2 + a_4 = frac1114$ $a_2 + a_4 = \frac{111}{4}$ [cite: 656], then $24(a_1 + a_2 + a_3)$ $24(a_1 + a_2 + a_3)$ is equal to[cite: 657]:

Solution & Explanation

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Let a_1, a_2, a_3, ldots$a_1, a_2, a_3, \ldots$ be a G.P. of increasing positive numbers[cite: 655]. If a_3a_5 = 729$a_3a_5 = 729$ and a_2 + a_4 = frac1114$a_2 + a_4 = \frac{111}{4}$ [cite: 656], then 24(a_1 + a_2 + a_3)$24(a_1 + a_2 + a_3)$ is equal to[cite: 657]:

Solution & Explanation

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Let $a_1, a_2, a_3, ldots$ $a_1, a_2, a_3, \ldots$ be a G.P. of increasing positive numbers[cite: 655]. If $a_3a_5 = 729$ $a_3a_5 = 729$ and $a_2 + a_4 = frac1114$ $a_2 + a_4 = \frac{111}{4}$ [cite: 656], then $24(a_1 + a_2 + a_3)$ $24(a_1 + a_2 + a_3)$ is equal to[cite: 657]: