Let Vr denote the sum of first r terms...

MathematicsArithmetic Progression (A.P.)JEE Advanced 2007Moderate

Comprehension Passage

Let

V_{r}

denote the sum of first

r

terms of an arithmetic progression (A.P.) whose first term is

r

and the common difference is

(2 r - 1)

. Let

T_{r} = V_{r + 1} - V_{r} - 2

and

Q_{r} = T_{r + 1} - T_{r}

for

r = 1, 2, \dots

Question 1:

The sum $V_{1} + V_{2} + \dots + V_{n}$ is

(A)

\frac{1}{12} n (n + 1) (3 n^{2} - n + 1)

(B)

\frac{1}{12} n (n + 1) (3 n^{2} + n + 2)

(C)

\frac{1}{3} n (2 n^{2} - n + 1)

(D)

\frac{1}{3} (2 n^{3} - 2 n + 3)

Answer:B

Question 2:

$T_{r}$ is always

(A)

an odd number

(B)

an even number

(C)

a prime number

(D)

a composite number

Answer:D

Question 3:

Which one of the following is a correct statement ?

(A)

Q_{1}, Q_{2}, Q_{3}, \dots

are in A.P. with common difference 5

(B)

Q_{1}, Q_{2}, Q_{3}, \dots

are in A.P. with common difference 6

(C)

Q_{1}, Q_{2}, Q_{3}, \dots

are in A.P. with common difference 11

(D)

Q_{1} = Q_{2} = Q_{3} = \dots

Answer:B

Visualized Solution (English)

Defining the A.P. Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V_{r}

Given an A.P. where:
First term $a = r$
Common difference $d = 2 r - 1$
Number of terms $n = r$
Sum of first $r$ terms is $V_{r} = \frac{r}{2} [2 a + (r - 1) d]$

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V_{r} Algebraically

Substitute values: $V_{r} = \frac{r}{2} [2 (r) + (r - 1) (2 r - 1)]$
Expand the product: $(r - 1) (2 r - 1) = 2 r^{2} - 3 r + 1$
Simplify inside: $V_{r} = \frac{r}{2} [2 r + 2 r^{2} - 3 r + 1] = \frac{r}{2} [2 r^{2} - r + 1]$
Final polynomial form: $V_{r} = r^{3} - \frac{r ^{2}}{2} + \frac{r}{2}$

Summing the Series .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \sum V_{r}

Required sum: $S = \sum_{r = 1}^{n} V_{r} = \sum_{r = 1}^{n} (r^{3} - \frac{r ^{2}}{2} + \frac{r}{2})$
Using linearity: $S = \sum r^{3} - \frac{1}{2} \sum r^{2} + \frac{1}{2} \sum r$
Standard formulas:
$\sum r^{3} = \frac{n ^{2} ( n + 1 ) ^{2}}{4}$
$\sum r^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$
$\sum r = \frac{n ( n + 1 )}{2}$

Final Calculation for Question 1

Substitute formulas: $S = \frac{n ^{2} ( n + 1 ) ^{2}}{4} - \frac{n ( n + 1 ) ( 2 n + 1 )}{12} + \frac{n ( n + 1 )}{4}$
Factor out $\frac{n ( n + 1 )}{12}$ :
$S = \frac{n ( n + 1 )}{12} [3 n (n + 1) - (2 n + 1) + 3]$
Simplify bracket: $3 n^{2} + 3 n - 2 n - 1 + 3 = 3 n^{2} + n + 2$
Final Answer: $\frac{n ( n + 1 ) ( 3 n ^{2} + n + 2 )}{12}$ (Option B)

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T_{r} and its Nature

Definition: $T_{r} = V_{r + 1} - V_{r} - 2$
Calculate $V_{r + 1} - V_{r}$ :
$(r + 1)^{3} - r^{3} - \frac{1}{2} ((r + 1)^{2} - r^{2}) + \frac{1}{2} (r + 1 - r)$
$= (3 r^{2} + 3 r + 1) - \frac{1}{2} (2 r + 1) + \frac{1}{2} = 3 r^{2} + 2 r + 1$
Then $T_{r} = (3 r^{2} + 2 r + 1) - 2 = 3 r^{2} + 2 r - 1$
Factorization: $T_{r} = (3 r - 1) (r + 1)$
Since $r \geq 1$ , both factors are $> 1 ⟹ T_{r}$ is composite (Option D)

Analyzing the Sequence .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q_{r}

Definition: $Q_{r} = T_{r + 1} - T_{r}$
Substitute $T_{r}$ : $Q_{r} = [3 (r + 1)^{2} + 2 (r + 1) - 1] - [3 r^{2} + 2 r - 1]$
Simplify: $Q_{r} = (3 r^{2} + 6 r + 3 + 2 r + 2 - 1) - (3 r^{2} + 2 r - 1)$
$Q_{r} = 6 r + 5$
Check common difference: $Q_{r + 1} - Q_{r} = [6 (r + 1) + 5] - [6 r + 5] = 6$
Conclusion: $Q_{r}$ is an A.P. with $d = 6$ (Option B)

Summary and Key Takeaways

Key Takeaways:
1. $V_{r}$ is a cubic polynomial because $a$ and $d$ depend linearly on $r$ .
2. Sum of $V_{r}$ involves $\sum r^{3}, \sum r^{2}, \sum r$ .
3. $T_{r}$ is composite because it factorizes into two terms $> 1$ .
4. $Q_{r}$ is linear in $r$ , making it an Arithmetic Progression.

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Conceptually Similar Problems

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Statement-1: The sum of the series $1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + \dots + (361 + 380 + 400)$ is $8000$ . Statement-2: $\sum_{k = 1}^{n} (k^{3} - (k - 1)^{3}) = n^{3}$ , for any natural number $n$ .

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Comprehension Passage

The sum $V_{1} + V_{2} + \dots + V_{n}$ is

$T_{r}$ is always

Which one of the following is a correct statement ?

Visualized Solution (English)

Defining the A.P. Sum $V_{r}$

Simplifying $V_{r}$ Algebraically

Summing the Series $\sum V_{r}$

Final Calculation for Question 1

Finding $T_{r}$ and its Nature

Analyzing the Sequence $Q_{r}$

Summary and Key Takeaways

Conceptually Similar Problems

Let $T_{r}$ be the $r$ th term of an A.P. whose first term is $a$ and common difference is $d$ . If for some positive integers $m, n, m \neq = n, T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $a - d$ equals

Let $T_{r}$ be the $r$ th term of an A.P., for $r = 1, 2, 3, \dots$ . If for some positive integers $m, n$ we have $T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $T_{mn}$ equals

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In the sum of first $n$ terms of an A.P. is $c n^{2}$ , then the sum of squares of these $n$ terms is

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Statement-1: The sum of the series $1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + \dots + (361 + 380 + 400)$ is $8000$ . Statement-2: $\sum_{k = 1}^{n} (k^{3} - (k - 1)^{3}) = n^{3}$ , for any natural number $n$ .

Comprehension Passage

The sum $V_{1} + V_{2} + \dots + V_{n}$ is

$T_{r}$ is always

Which one of the following is a correct statement ?

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In the sum of first $n$ terms of an A.P. is $c n^{2}$ , then the sum of squares of these $n$ terms is

If the sum of the first $2 n$ terms of the A.P. $2, 5, 8, \dots$ is equal to the sum of the first $n$ terms of the A.P. $57, 59, 61, \dots$ , then $n$ equals

If the first and the $(2 n - 1)$ st terms of an A.P., a G.P. and an H.P. are equal and their $n$ th terms are $a, b$ and $c$ respectively, then

Let $a_{1}, a_{2}, a_{3}, \dots, a_{100}$ be an arithmetic progression with $a_{1} = 3$ and $S_{p} = \sum_{i = 1}^{p} a_{i}, 1 \leq p \leq 100$ . For any integer $n$ with $1 \leq n \leq 20$ , let $m = 5 n$ . If $S_{m} / S_{n}$ does not depend on $n$ , then $a_{2}$ is \dots.

Statement-1: The sum of the series $1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + \dots + (361 + 380 + 400)$ is $8000$ . Statement-2: $\sum_{k = 1}^{n} (k^{3} - (k - 1)^{3}) = n^{3}$ , for any natural number $n$ .

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum V1​+V2​+⋯+Vn​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tr​ is always

Which one of the following is a correct statement ?

Visualized Solution (English)

Defining the A.P. Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vr​

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vr​ Algebraically

Summing the Series .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑Vr​

Final Calculation for Question 1

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tr​ and its Nature

Analyzing the Sequence .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Qr​

Summary and Key Takeaways

Conceptually Similar Problems

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Tr​ be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m,n,m=n,Tm​=1/n and Tn​=1/m, then a−d equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Tr​ be the rth term of an A.P., for r=1,2,3,…. If for some positive integers m,n we have Tm​=1/n and Tn​=1/m, then Tmn​ equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sum of the first n terms of the series 21​+43​+87​+1615​+… is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the sum of first n terms of an A.P. is cn2, then the sum of squares of these n terms is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the sum of the first 2n terms of the A.P. 2,5,8,… is equal to the sum of the first n terms of the A.P. 57,59,61,…, then n equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the first and the (2n−1)st terms of an A.P., a G.P. and an H.P. are equal and their nth terms are a,b and c respectively, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Statement-1: The sum of the series 1+(1+2+4)+(4+6+9)+(9+12+16)+⋯+(361+380+400) is 8000. Statement-2: ∑k=1n​(k3−(k−1)3)=n3, for any natural number n.

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum V1​+V2​+⋯+Vn​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tr​ is always

Which one of the following is a correct statement ?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Tr​ be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m,n,m=n,Tm​=1/n and Tn​=1/m, then a−d equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Tr​ be the rth term of an A.P., for r=1,2,3,…. If for some positive integers m,n we have Tm​=1/n and Tn​=1/m, then Tmn​ equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sum of the first n terms of the series 21​+43​+87​+1615​+… is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the sum of first n terms of an A.P. is cn2, then the sum of squares of these n terms is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the sum of the first 2n terms of the A.P. 2,5,8,… is equal to the sum of the first n terms of the A.P. 57,59,61,…, then n equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the first and the (2n−1)st terms of an A.P., a G.P. and an H.P. are equal and their nth terms are a,b and c respectively, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Statement-1: The sum of the series 1+(1+2+4)+(4+6+9)+(9+12+16)+⋯+(361+380+400) is 8000. Statement-2: ∑k=1n​(k3−(k−1)3)=n3, for any natural number n.

The sum $V_{1} + V_{2} + \dots + V_{n}$ is

$T_{r}$ is always

Defining the A.P. Sum $V_{r}$

Simplifying $V_{r}$ Algebraically

Summing the Series $\sum V_{r}$

Finding $T_{r}$ and its Nature

Analyzing the Sequence $Q_{r}$

Let $T_{r}$ be the $r$ th term of an A.P. whose first term is $a$ and common difference is $d$ . If for some positive integers $m, n, m \neq = n, T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $a - d$ equals

Let $T_{r}$ be the $r$ th term of an A.P., for $r = 1, 2, 3, \dots$ . If for some positive integers $m, n$ we have $T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $T_{mn}$ equals

Sum of the first $n$ terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots$ is equal to

In the sum of first $n$ terms of an A.P. is $c n^{2}$ , then the sum of squares of these $n$ terms is

If the sum of the first $2 n$ terms of the A.P. $2, 5, 8, \dots$ is equal to the sum of the first $n$ terms of the A.P. $57, 59, 61, \dots$ , then $n$ equals

If the first and the $(2 n - 1)$ st terms of an A.P., a G.P. and an H.P. are equal and their $n$ th terms are $a, b$ and $c$ respectively, then

Statement-1: The sum of the series $1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + \dots + (361 + 380 + 400)$ is $8000$ . Statement-2: $\sum_{k = 1}^{n} (k^{3} - (k - 1)^{3}) = n^{3}$ , for any natural number $n$ .

The sum $V_{1} + V_{2} + \dots + V_{n}$ is

$T_{r}$ is always

Let $T_{r}$ be the $r$ th term of an A.P. whose first term is $a$ and common difference is $d$ . If for some positive integers $m, n, m \neq = n, T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $a - d$ equals

Let $T_{r}$ be the $r$ th term of an A.P., for $r = 1, 2, 3, \dots$ . If for some positive integers $m, n$ we have $T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $T_{mn}$ equals

Sum of the first $n$ terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots$ is equal to

In the sum of first $n$ terms of an A.P. is $c n^{2}$ , then the sum of squares of these $n$ terms is

If the sum of the first $2 n$ terms of the A.P. $2, 5, 8, \dots$ is equal to the sum of the first $n$ terms of the A.P. $57, 59, 61, \dots$ , then $n$ equals

If the first and the $(2 n - 1)$ st terms of an A.P., a G.P. and an H.P. are equal and their $n$ th terms are $a, b$ and $c$ respectively, then

Statement-1: The sum of the series $1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + \dots + (361 + 380 + 400)$ is $8000$ . Statement-2: $\sum_{k = 1}^{n} (k^{3} - (k - 1)^{3}) = n^{3}$ , for any natural number $n$ .