Solution & Explanation
### Related Formula
textSum of an A.P.: S_k = frack2 left( 2a + (k-1)d right)$$\text{Sum of an A.P.: } S_k = \frac{k}{2} \left( 2a + (k-1)d \right)$$
textGeneral term of an A.P.: a_k = a_1 + (k-1)d$$\text{General term of an A.P.: } a_k = a_1 + (k-1)d$$
### Core Logic
Let the A.P. have n$n$ terms (where n$n$ is even). The terms are divided into n/2$n/2$ odd-indexed terms and n/2$n/2$ 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)$$a_2 + a_4 + \dots + a_n = 30 \quad \text{--- (1)}$$
Sum of odd terms:
a_1 + a_3 + dots + a_n-1 = 24 quad text--- (2)$$a_1 + a_3 + \dots + a_{n-1} = 24 \quad \text{--- (2)}$$
Subtracting equation (2) from (1):
(a_2 - a_1) + (a_4 - a_3) + dots + (a_n - a_n-1) = 30 - 24 = 6$$(a_2 - a_1) + (a_4 - a_3) + \dots + (a_n - a_{n-1}) = 30 - 24 = 6$$
Since there are n/2$n/2$ such pairs, and the difference of adjacent terms is the common difference d$d$:
fracn2 d = 6 implies n d = 12 quad text--- (3)$$\frac{n}{2} d = 6 \implies n d = 12 \quad \text{--- (3)}$$
### Step 2: Solve for n and d
We are given that the last term exceeds the first by frac212$\frac{21}{2}$:
a_n - a_1 = (n-1)d = frac212$$a_n - a_1 = (n-1)d = \frac{21}{2}$$
n d - d = 10.5$n d - d = 10.5$
Substitute nd = 12$nd = 12$ from (3):
12 - d = 10.5 implies d = 1.5 = frac32$$12 - d = 10.5 \implies d = 1.5 = \frac{3}{2}$$
Using this in (3):
n left(frac32right) = 12 implies n = 8$$n \left(\frac{3}{2}\right) = 12 \implies n = 8$$
### 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$$S_{\text{odd}} = \frac{4}{2} \left[ 2a_1 + (4-1)(2d) \right] = 24$$
2 left[ 2a_1 + 3(3) right] = 24 implies 2a_1 + 9 = 12 implies a_1 = 1.5 = frac32$$2 \left[ 2a_1 + 3(3) \right] = 24 \implies 2a_1 + 9 = 12 \implies a_1 = 1.5 = \frac{3}{2}$$
Thus, the terms are:
frac32, \, 3, \, frac92, \, 6, \, frac152, \, 9, \, frac212, \, 12$$\frac{3}{2}, \, 3, \, \frac{9}{2}, \, 6, \, \frac{15}{2}, \, 9, \, \frac{21}{2}, \, 12$$
The terms that are integers are 3, 6, 9, 12$3, 6, 9, 12$. 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$n$ is always equal to fracn2 d$\frac{n}{2} d$. This is an extremely useful relation to remember.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequences and Series
More Sequences and Series Previous-Year Questions
Q6
jee_main_2026_21_jan_morning
Properties of Geometric Progression
Let
a_1$a_{1}$ ,
a_2$a_{2}$ ,
a_3$a_{3}$ , ..... be a
G.P. of increasing positive terms such that
a_2 cdot a_3 cdot a_4 = 64$a_{2} \cdot a_{3} \cdot a_{4} = 64$ and
a_1 + a_3 + a_5 = frac8137$a_{1} + a_{3} + a_{5} = \frac{813}{7}$ .
Then
a_3 + a_5 + a_7$a_{3} + a_{5} + a_{7}$ is equal to :
- A. 3256
- B. 3252
- C. 3244
- D. 3248
Solution
### Related Formula
ntext-th term of a GP: T_n = a r^n-1$$n\text{-th term of a GP}: T_n = a r^{n-1}$$
### Core Logic
Let the terms of the GP be a, ar, ar^2, ar^3, dots$a, ar, ar^2, ar^3, \dots$
Given: a_2 cdot a_3 cdot a_4 = 64$a_2 \cdot a_3 \cdot a_4 = 64$
(ar) cdot (ar^2) cdot (ar^3) = 64$$(ar) \cdot (ar^2) \cdot (ar^3) = 64$$
a^3 r^6 = 64 Rightarrow (ar^2)^3 = 64 Rightarrow ar^2 = 4$$a^3 r^6 = 64 \Rightarrow (ar^2)^3 = 64 \Rightarrow ar^2 = 4$$
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: a_1 + a_3 + a_5 = frac8137$a_1 + a_3 + a_5 = \frac{813}{7}$
a + ar^2 + ar^4 = frac8137$$a + ar^2 + ar^4 = \frac{813}{7}$$
Extract a$a$ from ar^2 = 4 Rightarrow a = frac4r^2$ar^2 = 4 \Rightarrow a = \frac{4}{r^2}$. Substitute this in:
frac4r^2 + 4 + 4r^2 = frac8137$$\frac{4}{r^2} + 4 + 4r^2 = \frac{813}{7}$$
4left(frac1r^2 + 1 + r^2right) = frac8137$$4\left(\frac{1}{r^2} + 1 + r^2\right) = \frac{813}{7}$$
Let r^2 = t$r^2 = t$:
4left(frac1t + 1 + tright) = frac8137$$4\left(\frac{1}{t} + 1 + t\right) = \frac{813}{7}$$
Wait, there is a much faster method by just scaling the required expression.
### Step 2: Calculate the required expression
We need a_3 + a_5 + a_7 = ar^2 + ar^4 + ar^6$a_3 + a_5 + a_7 = ar^2 + ar^4 + ar^6$.
Notice that ar^2 + ar^4 + ar^6 = r^2 (a + ar^2 + ar^4)$ar^2 + ar^4 + ar^6 = r^2 (a + ar^2 + ar^4)$.
So, required sum = r^2 left(frac8137right)$r^2 \left(\frac{813}{7}\right)$.
To find r^2$r^2$, we solve the quadratic in t = r^2$t = r^2$:
4left(fract^2 + t + 1tright) = frac8137$$4\left(\frac{t^2 + t + 1}{t}\right) = \frac{813}{7}$$
28t^2 + 28t + 28 = 813t$$28t^2 + 28t + 28 = 813t$$
28t^2 - 785t + 28 = 0$$28t^2 - 785t + 28 = 0$$
The roots are t = 28$t = 28$ and t = frac128$t = \frac{1}{28}$.
Since the GP is increasing, r > 1 Rightarrow r^2 = 28$r > 1 \Rightarrow r^2 = 28$.
### Step 3: Final evaluation
Alternatively, expand directly:
ar^2 (1 + r^2 + r^4) = 4 (1 + 28 + (28)^2) = 4(1 + 28 + 784) = 4(813) = 3252.$$ar^2 (1 + r^2 + r^4) = 4 (1 + 28 + (28)^2) = 4(1 + 28 + 784) = 4(813) = 3252.$$
(Note: r^2 times frac8137 = 28 times frac8137 = 4 times 813 = 3252$r^2 \times \frac{813}{7} = 28 \times \frac{813}{7} = 4 \times 813 = 3252$)
### Pattern Recognition
In GP questions demanding a sum shifted by a fixed index (like a_1+a_3+a_5$a_1+a_3+a_5$ to a_3+a_5+a_7$a_3+a_5+a_7$), immediately look to factor out the common ratio multiplier r^k$r^k$. Here it's a simple scaling by r^2$r^2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Sequences and Series
Class 10 Maths: Quadratic Equations
Q22
jee_main_2026_21_jan_morning
Telescoping Sums and Recurrence Relations
Let a_1 = 1$a_1 = 1$ and for n geq 1$n \geq 1$ , a_n+1 = frac12 a_n + fracn^2 - 2n - 1n^2 (n + 1)^2$a_{n+1} = \frac{1}{2} a_n + \frac{n^2 - 2n - 1}{n^2 (n + 1)^2}$ . Then left|sum_n=1^inftyleft(a_n - frac2n^2right)right|$\left|\sum_{n=1}^{\infty}\left(a_n - \frac{2}{n^2}\right)\right|$ is equal
Numerical Answer. Answer: 2 to 2
Solution
### Related Formula
Partial fraction decomposition for telescopic summing:
frac2n^2 - (n+1)^2 + 1 dotsdots text structures directly cancel in series expansions.$$\frac{2n^2 - (n+1)^2 + 1 \dots}{\dots} \text{ structures directly cancel in series expansions.}$$
### Core Logic
Given recurrence:
a_n+1 - frac12a_n = fracn^2 - 2n - 1n^2(n+1)^2$$a_{n+1} - \frac{1}{2}a_n = \frac{n^2 - 2n - 1}{n^2(n+1)^2}$$
Rewrite the numerator to split the fraction:
n^2 - 2n - 1 = 2n^2 - (n^2 + 2n + 1) = 2n^2 - (n+1)^2$$n^2 - 2n - 1 = 2n^2 - (n^2 + 2n + 1) = 2n^2 - (n+1)^2$$
a_n+1 - frac12a_n = frac2n^2 - (n+1)^2n^2(n+1)^2 = frac2(n+1)^2 - frac1n^2$$a_{n+1} - \frac{1}{2}a_n = \frac{2n^2 - (n+1)^2}{n^2(n+1)^2} = \frac{2}{(n+1)^2} - \frac{1}{n^2}$$
### Step 1: Telescope generation
Multiply both sides by appropriate powers of 2 to create a cancelling chain:
For n=1$n=1$: a_2 - frac12a_1 = frac22^2 - frac11^2$a_2 - \frac{1}{2}a_1 = \frac{2}{2^2} - \frac{1}{1^2}$
For n=2$n=2$: multiply by 2 Rightarrow 2left[a_3 - frac12a_2 = frac23^2 - frac12^2right] Rightarrow 2a_3 - a_2 = frac2 times 23^2 - frac22^2$\Rightarrow 2\left[a_3 - \frac{1}{2}a_2 = \frac{2}{3^2} - \frac{1}{2^2}\right] \Rightarrow 2a_3 - a_2 = \frac{2 \times 2}{3^2} - \frac{2}{2^2}$
Wait, let's look at a cleaner telescopic scaling:
a_n+1 - frac2(n+1)^2 = frac12 left(a_n - frac2n^2right)$a_{n+1} - \frac{2}{(n+1)^2} = \frac{1}{2} \left(a_n - \frac{2}{n^2}\right)$.
### Step 2: Identify Geometric Progression
Let V_n = a_n - frac2n^2$V_n = a_n - \frac{2}{n^2}$.
The recurrence gives V_n+1 = frac12 V_n$V_{n+1} = \frac{1}{2} V_n$.
This proves V_n$V_n$ is a geometric progression with common ratio r = 1/2$r = 1/2$.
First term V_1 = a_1 - frac21^2 = 1 - 2 = -1$V_1 = a_1 - \frac{2}{1^2} = 1 - 2 = -1$.
### Step 3: Infinite Summation
We need left| sum_n=1^infty left( a_n - frac2n^2 right) right| = left| sum_n=1^infty V_n right|$\left| \sum_{n=1}^{\infty} \left( a_n - \frac{2}{n^2} \right) \right| = \left| \sum_{n=1}^{\infty} V_n \right|$.
Since V_n$V_n$ is an infinite GP:
S_infty = fracV_11 - r = frac-11 - 1/2 = frac-11/2 = -2$$S_{\infty} = \frac{V_1}{1 - r} = \frac{-1}{1 - 1/2} = \frac{-1}{1/2} = -2$$
Taking absolute value:
|-2| = 2$|-2| = 2$
### Pattern Recognition
When dealing with rational fraction recurrences A_n+1 - k A_n = f(n) - k f(n-1)$A_{n+1} - k A_n = f(n) - k f(n-1)$, immediately substitute V_n = A_n - f(n)$V_n = A_n - f(n)$. This substitution instantly isolates a classical Geometric Progression.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Sequences and Series
Q12
jee_main_2026_21_jan_evening
Geometric Progression
Let
a_1, fraca_22, fraca_32^2, ldots, fraca_102^9$a_1, \frac{a_2}{2}, \frac{a_3}{2^2}, \ldots, \frac{a_{10}}{2^9}$ be a
G.P. of common ratio \frac{1}{\sqrt{2}}. If
a_1 + a_2 + ldots + a_10 = 62$a_1 + a_2 + \ldots + a_{10} = 62$, then
a_1$a_1$ is equal to:
- A. 2(sqrt2 - 1)$2(\sqrt{2} - 1)$
- B. 2 - sqrt2$2 - \sqrt{2}$
- C. sqrt2 - 1$\sqrt{2} - 1$
- D. 2(2 - sqrt2)$2(2 - \sqrt{2})$
Solution
### Related Formula
textSum of G.P. S_n = fraca(r^n - 1)r - 1 text (for r > 1)$$\text{Sum of G.P. } S_n = \frac{a(r^n - 1)}{r - 1} \text{ (for } r > 1)$$
### Core Logic
The given sequence is a G.P. with ratio frac1sqrt2$\frac{1}{\sqrt{2}}$.
fraca_2/2a_1 = frac1sqrt2 implies a_2 = a_1 sqrt2$$\frac{a_2/2}{a_1} = \frac{1}{\sqrt{2}} \implies a_2 = a_1 \sqrt{2}$$
fraca_3/2^2a_2/2 = frac1sqrt2 implies fraca_32 a_2 = frac1sqrt2 implies a_3 = a_2 sqrt2 = a_1 (sqrt2)^2$$\frac{a_3/2^2}{a_2/2} = \frac{1}{\sqrt{2}} \implies \frac{a_3}{2 a_2} = \frac{1}{\sqrt{2}} \implies a_3 = a_2 \sqrt{2} = a_1 (\sqrt{2})^2$$
Thus, a_1, a_2, a_3, dots, a_10$a_1, a_2, a_3, \dots, a_{10}$ forms a standard G.P. with first term a_1$a_1$ and common ratio R = sqrt2$R = \sqrt{2}$.
### Step 1: Calculate the Sum
Sum of this new sequence is S_10 = 62$S_{10} = 62$.
S_10 = fraca_1 left( (sqrt2)^10 - 1 right)sqrt2 - 1 = 62$$S_{10} = \frac{a_1 \left( (\sqrt{2})^{10} - 1 \right)}{\sqrt{2} - 1} = 62$$
Since (sqrt2)^10 = 2^5 = 32$(\sqrt{2})^{10} = 2^5 = 32$:
62 = fraca_1 (32 - 1)sqrt2 - 1$$62 = \frac{a_1 (32 - 1)}{\sqrt{2} - 1}$$
62 = frac31 a_1sqrt2 - 1$$62 = \frac{31 a_1}{\sqrt{2} - 1}$$
2 = fraca_1sqrt2 - 1$$2 = \frac{a_1}{\sqrt{2} - 1}$$
a_1 = 2(sqrt2 - 1)$$a_1 = 2(\sqrt{2} - 1)$$
### Pattern Recognition
If a sequence b_n = fraca_nk^n-1$b_n = \frac{a_n}{k^{n-1}}$ is a G.P. with ratio r$r$, then the base sequence a_n$a_n$ is inherently a G.P. with ratio R = kr$R = kr$.
### Evaluation Rubric / Model Answer
null
### 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
S_n = fracn2[2a + (n-1)d]$$S_n = \frac{n}{2}[2a + (n-1)d]$$
### Core Logic
Given the sum of first 4 terms is S_4 = 6$S_4 = 6$:
frac42(2a + 3d) = 6 implies 2a + 3d = 3 quad dots (1)$$\frac{4}{2}(2a + 3d) = 6 \implies 2a + 3d = 3 \quad \dots (1)$$
Given the sum of first 6 terms is S_6 = 4$S_6 = 4$:
frac62(2a + 5d) = 4 implies 2a + 5d = frac43 quad dots (2)$$\frac{6}{2}(2a + 5d) = 4 \implies 2a + 5d = \frac{4}{3} \quad \dots (2)$$
### Step 1: Finding Parameters a and d
Subtract equation (1) from equation (2):
(2a + 5d) - (2a + 3d) = frac43 - 3$$(2a + 5d) - (2a + 3d) = \frac{4}{3} - 3$$
2d = -frac53 implies d = -frac56$$2d = -\frac{5}{3} \implies d = -\frac{5}{6}$$
Substitute d$d$ into equation (1):
2a + 3left(-frac56right) = 3$$2a + 3\left(-\frac{5}{6}\right) = 3$$
2a - frac52 = 3$$2a - \frac{5}{2} = 3$$
2a = 3 + frac52 = frac112 implies a = frac114$$2a = 3 + \frac{5}{2} = \frac{11}{2} \implies a = \frac{11}{4}$$
### Step 2: Calculating Sum of 12 terms
S_12 = frac122[2a + 11d]$$S_{12} = \frac{12}{2}[2a + 11d]$$
S_12 = 6 left[2left(frac114right) + 11left(-frac56right)right]$$S_{12} = 6 \left[2\left(\frac{11}{4}\right) + 11\left(-\frac{5}{6}\right)\right]$$
S_12 = 6 left[ frac112 - frac556 right] = 6 left[ frac33 - 556 right]$$S_{12} = 6 \left[ \frac{11}{2} - \frac{55}{6} \right] = 6 \left[ \frac{33 - 55}{6} \right]$$
S_12 = 33 - 55 = -22$$S_{12} = 33 - 55 = -22$$
### 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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Sequences and Series
Q19
jee_main_2026_22_january_evening
Functional Equations and Geometric Series
Let f$f$ and g$g$ be functions satisfying f(x+y) = f(x)f(y)$f(x+y) = f(x)f(y)$, f(1) = 7$f(1) = 7$ and g(x+y) = g(xy)$g(x+y) = g(xy)$, g(1) = 1$g(1) = 1$, for all x, y in mathbbN$x, y \in \mathbb{N}$. If sum_x=1^n left(fracf(x)g(x)right) = 19607$\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$, then n$n$ is equal to:
Solution
### Related Formula
Exponential functional equation: f(x+y) = f(x)f(y) implies f(x) = a^x$f(x+y) = f(x)f(y) \implies f(x) = a^x$.
Geometric progression sum formula: S_n = fraca(r^n - 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 implies 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) = dots = g(n) = 1$g(1) = g(2) = \dots = g(n) = 1$.
### Step 1: Solve Summation
sum_x=1^n frac7^x1 = 19607 implies 7 left( frac7^n - 17 - 1 right) = 19607$$\sum_{x=1}^{n} \frac{7^x}{1} = 19607 \implies 7 \left( \frac{7^n - 1}{7 - 1} \right) = 19607$$
frac76 (7^n - 1) = 19607 implies 7^n - 1 = 16806 implies 7^n = 16807$$\frac{7}{6} (7^n - 1) = 19607 \implies 7^n - 1 = 16806 \implies 7^n = 16807$$
Since 7^5 = 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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Sequences and Series
Class 11 Maths: Functions and Graphs
