Let a1 , a2 , a3 , ..... be a G.P. of increasing positive terms such that a2 cdot a3 cdot a4 = 64 and a1 + a3 + a5 =…

Year	2026	2025	2024	Total
Questions	17	24	14	55

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

n-th term of a GP: Tₙ = a rⁿ⁻¹

Core Logic

Let the terms of the GP be $a, ar, ar², ar³,$ $a, ar, ar^2, ar^3, \dots$ Given: $a₂ \cdot a₃ \cdot a₄ = 64$ $a_2 \cdot a_3 \cdot a_4 = 64$

(ar) \cdot (ar²) \cdot (ar³) = 64

a³ r⁶ = 64 \Rightarrow (ar²)³ = 64 \Rightarrow ar² = 4

Since it is a GP of increasing positive terms, $r > 1$ $r > 1$ and $a > 0$ $a > 0$.

Step 1: Utilize the sum condition

Given: $a₁ + a₃ + a₅ = (813)/(7)$ $a_1 + a_3 + a_5 = \frac{813}{7}$

a + ar² + ar⁴ = (813)/(7)

Extract $a$ $a$ from $ar² = 4 \Rightarrow a = (4)/(r²)$ . Substitute this in:

(4)/(r²) + 4 + 4r² = (813)/(7)

4((1)/(r²) + 1 + r²) = (813)/(7)

Let $r² = t$ $r^2 = t$:

4((1)/(t) + 1 + t) = (813)/(7)

Wait, there is a much faster method by just scaling the required expression.

Step 2: Calculate the required expression

We need $a₃ + a₅ + a₇ = ar² + ar⁴ + ar⁶$ $a_3 + a_5 + a_7 = ar^2 + ar^4 + ar^6$. Notice that $ar² + ar⁴ + ar⁶ = r² (a + ar² + ar⁴)$ . So, required sum = $r² ((813)/(7))$ $r^2 \left(\frac{813}{7}\right)$.

To find $r²$ $r^2$, we solve the quadratic in $t = r²$ $t = r^2$:

4((t² + t + 1)/(t)) = (813)/(7)

28t² + 28t + 28 = 813t

28t² - 785t + 28 = 0

The roots are $t = 28$ $t = 28$ and $t = (1)/(28)$ $t = \frac{1}{28}$. Since the GP is increasing, $r > 1 \Rightarrow r² = 28$ $r > 1 \Rightarrow r^2 = 28$.

Step 3: Final evaluation

Alternatively, expand directly:

ar² (1 + r² + r⁴) = 4 (1 + 28 + (28)²) = 4(1 + 28 + 784) = 4(813) = 3252.

(Note: $r² \times (813)/(7) = 28 \times (813)/(7) = 4 \times 813 = 3252$ )

Pattern Recognition

In GP questions demanding a sum shifted by a fixed index (like $a₁+a₃+a₅$ $a_1+a_3+a_5$ to $a₃+a₅+a₇$ $a_3+a_5+a_7$), immediately look to factor out the common ratio multiplier $r^k$ $r^k$. Here it's a simple scaling by $r²$ $r^2$.

Chapter Mix

Class 11 Maths: Sequences and Series Class 10 Maths: Quadratic Equations

More Sequences and Series Previous-Year Questions

Q22 jee_main_2026_21_jan_morning Telescoping Sums and Recurrence Relations

Let

a₁ = 1

and for

n \geq 1

aₙ₊₁ = (1)/(2) aₙ + (n² - 2n - 1)/(n² (n + 1)²)

. Then

|Σn=1\infty(aₙ - (2)/(n²))|

is equal

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

Partial fraction decomposition for telescopic summing:

(2n² - (n+1)² + 1 )/( ) structures directly cancel in series expansions.

Core Logic

Given recurrence:

aₙ₊₁ - (1)/(2)aₙ = (n² - 2n - 1)/(n²(n+1)²)

Rewrite the numerator to split the fraction:

n² - 2n - 1 = 2n² - (n² + 2n + 1) = 2n² - (n+1)²

aₙ₊₁ - (1)/(2)aₙ = (2n² - (n+1)²)/(n²(n+1)²) = (2)/((n+1)²) - (1)/(n²)

Step 1: Telescope generation

Multiply both sides by appropriate powers of 2 to create a cancelling chain: For $n=1$ $n=1$: $a₂ - (1)/(2)a₁ = (2)/(2²) - (1)/(1²)$ For $n=2$ $n=2$: multiply by 2 $\Rightarrow 2[a₃ - (1)/(2)a₂ = (2)/(3²) - (1)/(2²)] \Rightarrow 2a₃ - a₂ = (2 \times 2)/(3²) - (2)/(2²)$ Wait, let's look at a cleaner telescopic scaling: $aₙ₊₁ - (2)/((n+1)²) = (1)/(2) (aₙ - (2)/(n²))$ .

Step 2: Identify Geometric Progression

Let $Vₙ = aₙ - (2)/(n²)$ $V_n = a_n - \frac{2}{n^2}$. The recurrence gives $Vₙ₊₁ = (1)/(2) Vₙ$ $V_{n+1} = \frac{1}{2} V_n$. This proves $Vₙ$ $V_n$ is a geometric progression with common ratio $r = 1/2$ $r = 1/2$. First term $V₁ = a₁ - (2)/(1²) = 1 - 2 = -1$ .

Step 3: Infinite Summation

We need $| Σn=1\infty ( aₙ - (2)/(n²) ) | = | Σn=1\infty Vₙ |$ . Since $Vₙ$ $V_n$ is an infinite GP:

S\infty = (V₁)/(1 - r) = (-1)/(1 - 1/2) = (-1)/(1/2) = -2

Taking absolute value: $|-2| = 2$ $|-2| = 2$

Pattern Recognition

When dealing with rational fraction recurrences $Aₙ₊₁ - k Aₙ = f(n) - k f(n-1)$ $A_{n+1} - k A_n = f(n) - k f(n-1)$, immediately substitute $Vₙ = Aₙ - f(n)$ $V_n = A_n - f(n)$. This substitution instantly isolates a classical Geometric Progression.

Chapter Mix

Class 11 Maths: Sequences and Series

Q12 jee_main_2026_21_jan_evening Geometric Progression

Let

a₁, (a₂)/(2), (a₃)/(2²), \dots, a₁₀2⁹

be a G.P. of common ratio \frac{1}{\sqrt{2}}. If

a₁ + a₂ + \dots + a₁₀ = 62

, then

a₁

is equal to:

A. $2(\sqrt(2) - 1)$ $2(\sqrt{2} - 1)$
B. $2 - \sqrt(2)$ $2 - \sqrt{2}$
C. $\sqrt(2) - 1$ $\sqrt{2} - 1$
D. $2(2 - \sqrt(2))$ $2(2 - \sqrt{2})$

Solution

Related Formula

Sum of G.P. Sₙ = (a(rⁿ - 1))/(r - 1) (for r > 1)

Core Logic

The given sequence is a G.P. with ratio $1\sqrt(2)$ $\frac{1}{\sqrt{2}}$.

(a₂/2)/(a₁) = 1\sqrt(2) a₂ = a₁ \sqrt(2)

(a₃/2²)/(a₂/2) = 1\sqrt(2) (a₃)/(2 a₂) = 1\sqrt(2) a₃ = a₂ \sqrt(2) = a₁ (\sqrt(2))²

Thus, $a₁, a₂, a₃, , a₁₀$ $a_1, a_2, a_3, \dots, a_{10}$ forms a standard G.P. with first term $a₁$ $a_1$ and common ratio $R = \sqrt(2)$ $R = \sqrt{2}$.

Step 1: Calculate the Sum

Sum of this new sequence is $S₁₀ = 62$ $S_{10} = 62$.

S₁₀ = a₁ ( (\sqrt(2))¹⁰ - 1 )\sqrt(2) - 1 = 62

Since $(\sqrt(2))¹⁰ = 2⁵ = 32$ $(\sqrt{2})^{10} = 2^5 = 32$:

62 = a₁ (32 - 1)\sqrt(2) - 1

62 = 31 a₁\sqrt(2) - 1

2 = a₁\sqrt(2) - 1

a₁ = 2(\sqrt(2) - 1)

Pattern Recognition

If a sequence $bₙ = aₙkⁿ⁻¹$ $b_n = \frac{a_n}{k^{n-1}}$ is a G.P. with ratio $r$ $r$, then the base sequence $aₙ$ $a_n$ is inherently a G.P. with ratio $R = kr$ $R = kr$.

Chapter Mix

Class 11 Maths: Sequence and Series

Q20 jee_main_2026_22_january_morning Arithmetic Progression

If the sum of the first four terms of an A.P. is 6 and the sum of its first six terms is 4, then the sum of its first twelve terms is

A. $-20$ $-20$
B. $-24$ $-24$
C. $-26$ $-26$
D. $-22$ $-22$

Solution

Related Formula

Sₙ = (n)/(2)[2a + (n-1)d]

Core Logic

Given the sum of first 4 terms is $S₄ = 6$ $S_4 = 6$:

(4)/(2)(2a + 3d) = 6 2a + 3d = 3 (1)

Given the sum of first 6 terms is $S₆ = 4$ $S_6 = 4$:

(6)/(2)(2a + 5d) = 4 2a + 5d = (4)/(3) (2)

Step 1: Finding Parameters a and d

Subtract equation (1) from equation (2):

(2a + 5d) - (2a + 3d) = (4)/(3) - 3

2d = -(5)/(3) d = -(5)/(6)

Substitute $d$ $d$ into equation (1):

2a + 3(-(5)/(6)) = 3

2a - (5)/(2) = 3

2a = 3 + (5)/(2) = (11)/(2) a = (11)/(4)

Step 2: Calculating Sum of 12 terms

S₁₂ = (12)/(2)[2a + 11d]

S₁₂ = 6 [2((11)/(4)) + 11(-(5)/(6))]

S₁₂ = 6 [ (11)/(2) - (55)/(6) ] = 6 [ (33 - 55)/(6) ]

S₁₂ = 33 - 55 = -22

Pattern Recognition

Standard two-variable linear equations format strictly from sum identities. Fractions easily resolve by distributing the external multiplication constant $n/2$ $n/2$ directly into the bracket.

Chapter Mix

Class 11 Maths: Sequences and Series

Q19 jee_main_2026_22_january_evening Functional Equations and Geometric Series

Let

f

and

g

be functions satisfying

f(x+y) = f(x)f(y)

f(1) = 7

and

g(x+y) = g(xy)

g(1) = 1

, for all

x, y in N

. If

Σx=1ⁿ ((f(x))/(g(x))) = 19607

, then

n

is equal to:

A. 7
B. 5
C. 6
D. 4

Solution

Related Formula

Exponential functional equation: $f(x+y) = f(x)f(y) f(x) = a^x$ $f(x+y) = f(x)f(y) \implies f(x) = a^x$. Geometric progression sum formula: $Sₙ = (a(rⁿ - 1))/(r - 1)$ $S_n = \frac{a(r^n - 1)}{r - 1}$.

Core Logic

Since $f(1) = 7$ $f(1) = 7$, $f(x) = 7^x$ $f(x) = 7^x$. For $g(x+y) = g(xy)$ $g(x+y) = g(xy)$, set $y = 1 g(x+1) = g(x)$ $y = 1 \implies g(x+1) = g(x)$. Since $g(1) = 1$ $g(1) = 1$, we have $g(1) = g(2) = = g(n) = 1$ $g(1) = g(2) = \dots = g(n) = 1$.

Step 1: Solve Summation

Σx=1ⁿ (7^x)/(1) = 19607 7 ( (7ⁿ - 1)/(7 - 1) ) = 19607

(7)/(6) (7ⁿ - 1) = 19607 7ⁿ - 1 = 16806 7ⁿ = 16807

Since $7⁵ = 16807$ $7^5 = 16807$, $n = 5$ $n = 5$.

Pattern Recognition

Recognize $f(x)=7^x$ $f(x)=7^x$ and constant function $g(x)=1$ $g(x)=1$ from given functional equations.

Chapter Mix

Class 11 Maths: Sequences and Series Class 11 Maths: Functions and Graphs

Let a₁$a_{1}$ , a₂$a_{2}$ , a₃$a_{3}$ , ..... be a G.P. of increasing positive terms such that a₂ · a₃ · a₄ = 64$a_{2} \cdot a_{3} \cdot a_{4} = 64$ and a₁ + a₃ + a₅ = (813)/(7)$a_{1} + a_{3} + a_{5} = \frac{813}{7}$ . Then a₃ + a₅ + a₇$a_{3} + a_{5} + a_{7}$ is equal to :

Solution & Explanation

Related Formula

Core Logic

Step 1: Utilize the sum condition

Step 2: Calculate the required expression

Step 3: Final evaluation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Sequences and Series Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Telescope generation

Step 2: Identify Geometric Progression

Step 3: Infinite Summation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate the Sum

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding Parameters a and d

Step 2: Calculating Sum of 12 terms

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve Summation

Pattern Recognition

Chapter Mix

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