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
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### Chapter Mix
Class 11 Mathematics: Sequences and Series
More Sequences and Series Previous-Year Questions — Page 5
Q59
jee_main_2025_24_jan_evening
Arithmetic Progression Sum
In an arithmetic progression, if S_40=1030$S_{40}=1030$ and S_12=57,$S_{12}=57,$ then S_30-S_10$S_{30}-S_{10}$ is equal to: [cite: 3294, 3295, 3296]
- A. 510$510$
- B. 515$515$
- C. 525$525$
- D. 505$505$
Solution
### Related Formula
Sum of first n$n$ terms of an Arithmetic Progression:
S_n = fracn2[2a + (n-1)d]$$S_n = \frac{n}{2}[2a + (n-1)d]$$
### Core Logic
Set up linear expressions for the given sums :
S_40 = frac402[2a + 39d] = 1030 Rightarrow 2a + 39d = 51.5$$S_{40} = \frac{40}{2}[2a + 39d] = 1030 \Rightarrow 2a + 39d = 51.5$$
S_12 = frac122[2a + 11d] = 57 Rightarrow 2a + 11d = 9.5$$S_{12} = \frac{12}{2}[2a + 11d] = 57 \Rightarrow 2a + 11d = 9.5$$
### Step 1: Solve for a$a$ and d$d$
Subtract the second equation from the first :
(2a + 39d) - (2a + 11d) = 51.5 - 9.5$$(2a + 39d) - (2a + 11d) = 51.5 - 9.5$$
28d = 42 Rightarrow d = frac4228 = frac32 = 1.5$$28d = 42 \Rightarrow d = \frac{42}{28} = \frac{3}{2} = 1.5$$
Substitute d = 1.5$d = 1.5$ back to find a$a$:
2a + 11(1.5) = 9.5 Rightarrow 2a + 16.5 = 9.5 Rightarrow 2a = -7 Rightarrow a = -3.5$$2a + 11(1.5) = 9.5 \Rightarrow 2a + 16.5 = 9.5 \Rightarrow 2a = -7 \Rightarrow a = -3.5$$
### Step 2: Evaluate S_30 - S_10$S_{30} - S_{10}$
Write out the formula for the target subtraction :
S_30 - S_10 = frac302[2a + 29d] - frac102[2a + 9d]$$S_{30} - S_{10} = \frac{30}{2}[2a + 29d] - \frac{10}{2}[2a + 9d]$$
= 15(2a + 29d) - 5(2a + 9d) = 30a + 435d - 10a - 45d = 20a + 390d$$= 15(2a + 29d) - 5(2a + 9d) = 30a + 435d - 10a - 45d = 20a + 390d$$ [cite: 3964, 3965]
Substitute the values of a$a$ and d$d$ :
= 20(-3.5) + 390(1.5) = -70 + 585 = 515$$= 20(-3.5) + 390(1.5) = -70 + 585 = 515$$
### Pattern Recognition
Notice that S_30 - S_10$S_{30} - S_{10}$ represents the \sum of terms from T_11$T_{11}$ to T_30$T_{30}$, which can also be formulated as 20 times A_20.5$20 \times A_{20.5}$, saving algebraic steps if calculated symmetrically.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequences and Series
Q60
jee_main_2025_24_jan_evening
Arithmetico-Geometric Progression
If 7=5+frac17(5+alpha)+frac17^2(5+2alpha)+frac17^3(5+3alpha)+dotsdotsinfty$7=5+\frac{1}{7}(5+\alpha)+\frac{1}{7^{2}}(5+2\alpha)+\frac{1}{7^{3}}(5+3\alpha)+\dots\dots\infty$, then the value of alpha$\alpha$ is: [cite: 3301, 3302]
- A. 1$1$
- B. frac67$\frac{6}{7}$
- C. 6$6$
- D. frac17$\frac{1}{7}$
Solution
### Related Formula
Sum of an infinite geometric progression:
S_infty = fraca1-r quad textfor |r| < 1$$S_{\infty} = \frac{a}{1-r} \quad \text{for } |r| < 1$$
### Core Logic
The given expression is an infinite Arithmetico-Geometric Progression (AGP) :
S = 5 + frac5+alpha7 + frac5+2alpha7^2 + frac5+3alpha7^3 + dots infty$$S = 5 + \frac{5+\alpha}{7} + \frac{5+2\alpha}{7^2} + \frac{5+3\alpha}{7^3} + \dots \infty$$
### Step 1: Shift and Subtract
Multiply the equation by the common ratio frac17$\frac{1}{7}$ and shift it by one position :
frac17S = frac57 + frac5+alpha7^2 + frac5+2alpha7^3 + dots infty$$\frac{1}{7}S = \frac{5}{7} + \frac{5+\alpha}{7^2} + \frac{5+2\alpha}{7^3} + \dots \infty$$
Subtract this from the original equation:
S - frac17S = 5 + left(frac5+alpha-57right) + left(frac5+2alpha-(5+alpha)7^2right) + dots$$S - \frac{1}{7}S = 5 + \left(\frac{5+\alpha-5}{7}\right) + \left(\frac{5+2\alpha-(5+\alpha)}{7^2}\right) + \dots$$
frac67S = 5 + fracalpha7 + fracalpha7^2 + fracalpha7^3 + dots$$\frac{6}{7}S = 5 + \frac{\alpha}{7} + \frac{\alpha}{7^2} + \frac{\alpha}{7^3} + \dots$$
### Step 2: Sum the Infinite Geometric Series
Apply the infinite GP formula to the terms involving alpha$\alpha$ :
frac67S = 5 + fracalpha7left(frac11 - frac17right) = 5 + fracalpha7left(frac76right) = 5 + fracalpha6$$\frac{6}{7}S = 5 + \frac{\alpha}{7}\left(\frac{1}{1 - \frac{1}{7}}\right) = 5 + \frac{\alpha}{7}\left(\frac{7}{6}\right) = 5 + \frac{\alpha}{6}$$
Given that S = 7$S = 7$ :
frac67(7) = 5 + fracalpha6 Rightarrow 6 = 5 + fracalpha6$$\frac{6}{7}(7) = 5 + \frac{\alpha}{6} \Rightarrow 6 = 5 + \frac{\alpha}{6}$$
1 = fracalpha6 Rightarrow alpha = 6$$1 = \frac{\alpha}{6} \Rightarrow \alpha = 6$$
### Pattern Recognition
Standard trick for infinite AGPs: Multiply by the common ratio r$r$, shift, and subtract to condense the arithmetic progression component into a straightforward infinite geometric progression.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequences and Series
Q54
jee_main_2025_24_jan_morning
Sum to n terms of Special Series
Let S_n = frac12 + frac16 + frac112 + frac120 + dots$S_{n} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \dots$ up to n$n$ terms. If the sum of the first six terms of an A.P. with first term -p$-p$ and common difference p$p$ is sqrt2026S_2025$\sqrt{2026S_{2025}}$, then the absolute difference between 20^textth$20^{\text{th}}$ and 15^textth$15^{\text{th}}$ terms of the A.P. is :
- A. 25$25$
- B. 90$90$
- C. 20$20$
- D. 45$45$
Solution
### Related Formula
The general term for the provided series is:
T_k = frac1k(k+1) = frac1k - frac1k+1$$T_k = \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$$
This sets up a standard telescoping summation sequence.
### Core Logic
Express the sum S_2025$S_{2025}$ via telescoping fractions:
S_2025 = sum_k=1^2025 left( frac1k - frac1k+1 right) = left(1 - frac12right) + left(frac12 - frac13right) + dots + left(frac12025 - frac12026right)$$S_{2025} = \sum_{k=1}^{2025} \left( \frac{1}{k} - \frac{1}{k+1} \right) = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{2025} - \frac{1}{2026}\right)$$
S_2025 = 1 - frac12026 = frac20252026$$S_{2025} = 1 - \frac{1}{2026} = \frac{2025}{2026}$$
### Step 1: Compute the boundary expression value
Substitute S_2025$S_{2025}$ into the expression value:
sqrt2026 cdot S_2025 = sqrt2026 cdot frac20252026 = sqrt2025 = 45$$\sqrt{2026 \cdot S_{2025}} = \sqrt{2026 \cdot \frac{2025}{2026}} = \sqrt{2025} = 45$$
### Step 2: Apply Arithmetic Progression Summation
The sum of the first 6 terms of the A.P. with a = -p$a = -p$ and d = p$d = p$ is equal to 45:
Sigma_6 = frac62 [2a + (6-1)d] = 45$$\Sigma_6 = \frac{6}{2} [2a + (6-1)d] = 45$$
3 [2(-p) + 5p] = 45$$3 [2(-p) + 5p] = 45$$
3 [3p] = 45 implies 9p = 45 implies p = 5$$3 [3p] = 45 \implies 9p = 45 \implies p = 5$$
### Step 3: Calculate target absolute term difference
The absolute difference between the 20^textth$20^{\text{th}}$ and 15^textth$15^{\text{th}}$ terms of any A.P. depends strictly on the common difference:
|A_20 - A_15| = |(a + 19p) - (a + 14p)| = 5p$$|A_{20} - A_{15}| = |(a + 19p) - (a + 14p)| = 5p$$
5p = 5(5) = 25$5p = 5(5) = 25$
### Pattern Recognition
The series sequence frac12 + frac16 + frac112 + dots$\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \dots$ is the well-known telescoping series sum frac1n(n+1)$\sum \frac{1}{n(n+1)}$. Its sum to n$n$ terms is identically given by fracnn+1$\frac{n}{n+1}$ without requiring manual re-derivation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequences and Series
Q63
jee_main_2025_28_jan_evening
Telescopic Series Summation
For positive integers n, if 4a_n=(n^2+5n+6)$4a_{n}=(n^{2}+5n+6)$ and S_n=sum_k=1^nleft(frac1a_kright)$S_{n}=\sum_{k=1}^{n}\left(\frac{1}{a_{k}}\right)$ then the value of 507 S_2025$S_{2025}$ is:
- A. 540$540$
- B. 1350$1350$
- C. 675$675$
- D. 135$135$
Solution
### Related Formula
Telescopic series decomposition via method of differences:
frac1(k+2)(k+3) = frac1k+2 - frac1k+3$$\frac{1}{(k+2)(k+3)} = \frac{1}{k+2} - \frac{1}{k+3}$$
### Core Logic
Given:
a_n = fracn^2+5n+64 = frac(n+2)(n+3)4$$a_n = \frac{n^2+5n+6}{4} = \frac{(n+2)(n+3)}{4}$$
Therefore, the reciprocal term is:
frac1a_k = frac4(k+2)(k+3) = 4 left[ frac1k+2 - frac1k+3 right]$$\frac{1}{a_k} = \frac{4}{(k+2)(k+3)} = 4 \left[ \frac{1}{k+2} - \frac{1}{k+3} \right]$$
### Step 1: Compute the Partial Sum
S_n = sum_k=1^n frac1a_k = 4 sum_k=1^n left( frac1k+2 - frac1k+3 right)$$S_n = \sum_{k=1}^{n} \frac{1}{a_k} = 4 \sum_{k=1}^{n} \left( \frac{1}{k+2} - \frac{1}{k+3} \right)$$
Expanding the sum terms:
S_n = 4 left[ left(frac13 - frac14right) + left(frac14 - frac15right) + dots + left(frac1n+2 - frac1n+3right) right]$$S_n = 4 \left[ \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) + \dots + \left(\frac{1}{n+2} - \frac{1}{n+3}\right) \right]$$
All intermediate terms cancel out:
S_n = 4 left[ frac13 - frac1n+3 right] = 4 left[ fracn+3 - 33(n+3) right] = frac4n3(n+3)$$S_n = 4 \left[ \frac{1}{3} - \frac{1}{n+3} \right] = 4 \left[ \frac{n+3 - 3}{3(n+3)} \right] = \frac{4n}{3(n+3)}$$
### Step 2: Calculate for n = 2025
For n = 2025$n = 2025$:
S_2025 = frac4 times 20253 times (2025 + 3) = frac4 times 20253 times 2028$$S_{2025} = \frac{4 \times 2025}{3 \times (2025 + 3)} = \frac{4 \times 2025}{3 \times 2028}$$
We need to find 507 times S_2025$507 \times S_{2025}$:
507 times S_2025 = 507 times frac4 times 20253 times 2028$$507 \times S_{2025} = 507 \times \frac{4 \times 2025}{3 \times 2028}$$
Notice that 2028 = 4 times 507$2028 = 4 \times 507$:
507 times S_2025 = 507 times frac4 times 20253 times (4 times 507) = frac20253 = 675$$507 \times S_{2025} = 507 \times \frac{4 \times 2025}{3 \times (4 \times 507)} = \frac{2025}{3} = 675$$
### Pattern Recognition
Always look for arithmetic factor groupings at the end of large number sequence questions in JEE. Here recognizing 2028 = 4 times 507$2028 = 4 \times 507$ avoids large multi-digit multiplication.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequences and Series
Q73
jee_main_2025_28_jan_evening
Arithmetic Progression Applications
The interior angles of a polygon with n sides, are in an A.P. with common difference 6^circ$6^{\circ}$ If the largest interior angle of the polygon is 219^circ$219^{\circ}$, then n is equal to
Numerical Answer. Answer: 20 to 20
Solution
### Related Formula
Sum of interior angles of an n$n$-sided polygon:
S_n = (n - 2) times 180^circ$$S_n = (n - 2) \times 180^\circ$$
Sum of an Arithmetic Progression:
S_n = fracn2 left[ 2a + (n-1)d right]$$S_n = \frac{n}{2} \left[ 2a + (n-1)d \right]$$
### Core Logic
The angles form an AP with common difference d = 6^circ$d = 6^\circ$.
The largest angle is the last term: T_n = 219^circ$T_n = 219^\circ$.
a + (n-1)6 = 219 implies a = 219 - 6n + 6 = 225 - 6n$$a + (n-1)6 = 219 \implies a = 219 - 6n + 6 = 225 - 6n$$
### Step 1: Set up the sum equation
Equating the two forms for the sum of angles:
fracn2 left[ 2a + (n-1)6 right] = (n - 2) times 180$$\frac{n}{2} \left[ 2a + (n-1)6 \right] = (n - 2) \times 180$$
Substitute a = 225 - 6n$a = 225 - 6n$:
fracn2 left[ 2(225 - 6n) + 6n - 6 right] = 180n - 360$$\frac{n}{2} \left[ 2(225 - 6n) + 6n - 6 \right] = 180n - 360$$
fracn2 left[ 450 - 12n + 6n - 6 right] = 180n - 360$$\frac{n}{2} \left[ 450 - 12n + 6n - 6 \right] = 180n - 360$$
fracn2 left[ 444 - 6n right] = 180n - 360$$\frac{n}{2} \left[ 444 - 6n \right] = 180n - 360$$
n(222 - 3n) = 180n - 360$$n(222 - 3n) = 180n - 360$$
222n - 3n^2 = 180n - 360$$222n - 3n^2 = 180n - 360$$
3n^2 - 42n - 360 = 0$$3n^2 - 42n - 360 = 0$$
### Step 2: Solve the Quadratic Equation
Divide by 3:
n^2 - 14n - 120 = 0$$n^2 - 14n - 120 = 0$$
(n - 20)(n + 6) = 0$$(n - 20)(n + 6) = 0$$
Since number of sides n$n$ must be positive, n = 20$n = 20$.
### Pattern Recognition
Always remember that any interior angle of a convex polygon must be less than 180^circ$180^\circ$. Let's check the smallest angle for n=20$n=20$: a = 225 - 120 = 105^circ$a = 225 - 120 = 105^\circ$, which is completely valid.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequences and Series
