If $t_n$ denotes the $n$th term of the series $2 + 3 + 6 + 11 + 18 + \ldots$ then $t_{50}$ is (A) $49^2 - 1$ (B) $49^2$ (C) $50^2 + 1$ (D) $49^2 + 2$

If tₙ denotes the nth term of the series 2 + 3 + 6 – 1st PUC Mathematics

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Concept understanding — Sequence Term Evaluation

Sequence Term Evaluation

Picture a staircase where the height of each step follows a rule: step 1 is 1 unit high, step 2 is 4 units, step 3 is 9 units, step 4 is 16 units. The pattern is height = (step number) $^2$ . If someone asks for the height of step 20, you don't climb up counting -- you compute $20^2 = 400$ directly. That single act of substitution is what sequence term evaluation means.

The Core Idea

A sequence is an ordered list of numbers, and each number is called a term. Its position -- first, second, third, ... -- is the index, usually written $n$ (starting at $n=1$ unless told otherwise). The term at position $n$ is written $a_n$ .

Think of the sequence as a machine: feed it an index $n$ , and it returns the term $a_n$ .

Input $n=1$ -> output $a_1 = 1^2 = 1$
Input $n=2$ -> output $a_2 = 2^2 = 4$
Input $n=10$ -> output $a_{10} = 10^2 = 100$

You are not solving anything here -- you are purely substituting a number into a formula and simplifying.

The Precise Statement

a_n = f(n)

A sequence defined this way is a function whose domain is the positive integers. Evaluating a term means computing $f(n)$ for one specific value of $n$ .

Example 1 -- sequence $a_n = 3n + 2$ :

a_1 = 3(1)+2 = 5, \qquad a_2 = 3(2)+2 = 8, \qquad a_5 = 3(5)+2 = 17

Example 2 -- sequence $a_n = \dfrac{(-1)^n}{n}$ :

a_1 = \frac{(-1)^1}{1} = -1, \qquad a_2 = \frac{(-1)^2}{2} = \frac{1}{2}, \qquad a_3 = \frac{(-1)^3}{3} = -\frac{1}{3}

Watch out

A common mistake is confusing the index with the term's value. For $a_n = 2n$ , the 5th term is $a_5 = 2 \times 5 = 10$ -- the index $n=5$ only tells you which term to compute, while $10$ is the value sitting at that position. Writing $a_5 = 5$ mixes up the position number with the answer; always finish the substitution before reading off the result.

Quick Check

Given $a_n = \dfrac{n^2 - 1}{n+1}$ , find $a_4$ .

$a_4 = \frac{4^2 - 1}{4 + 1} = \frac{16-1}{5} = \frac{15}{5} = 3$ …

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The successive differences are $1, 3, 5, 7, \ldots$ (odd numbers), giving $t_n = 2 + (n-1)^2$ , so $t_{50} = 49^2 + 2$ — option (D).

For the series $2 + 3 + 6 + 11 + 18 + \ldots$ , the first differences are

3-2=1,\quad 6-3=3,\quad 11-6=5,\quad 18-11=7,

i.e. the consecutive odd numbers $1, 3, 5, 7, \ldots$

Building up the $n$ th term from the first:

$t_n = t_1 + \sum_{k=1}^{n-1}(2k-1) = 2 + (n-1)^2,$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2025Set A-11 markMCQ
$A$
›Reveal solutionSolution

Pull the scalar $3$ out of a $3\times3$ matrix as $3^3$ , then use $\det(A^{-1}) = 1/\det A$ .

Step 1 — The two properties needed.

For a square matrix of order $n$ and a scalar $k$ :
$\det(kA) = k^{n}\,\det(A).$
The exponent is $n$ , not $1$ , because multiplying the matrix by $k$ scales every one of the $n$ rows by $k$ , and each row-scaling multiplies the determinant by $k$ once.

Second, from $A A^{-1} = I$ and the multiplicative property $\det(AB) = \det A\,\det B$ :
$\det(A)\cdot\det(A^{-1}) = \det(I) = 1 \;\Longrightarrow\; \det(A^{-1}) = \frac{1}{\det A}.$
Step 2 — Apply with $n = 3$ , $k = 3$ .
$\det\left(3A^{-1}\right) = 3^{3}\cdot \det\left(A^{-1}\right) = 27 \cdot \frac{1}{\det A}.$
Step 3 — Substitute $\det A = 3$ .

$\det\left(3A^{-1}\right) = 27 \cdot \frac{1}{3} = 9.$ …
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KCET 2022Set C-41 markMCQ
$A = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}$
›Reveal solutionSolution
The key idea is that $A$ is its own inverse ( $A^2 = I$ ), so $A^3 = A$ , and its inverse is $A$ itself. The correct option is (D).

We start with the matrix
$A = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}.$
The question asks for the inverse of $A^3$ . Instead of computing $A^3$ directly and then finding its inverse, we can look for a pattern — often such matrices have a simple property.

Check if $A$ is involutory — that is, whether $A^2 = I$ . Compute $A^2$ :

$A^2 = A \cdot A = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}.$
Multiply:

First row, first column: $2\cdot2 + (-1)\cdot3 = 4 - 3 = 1$ .

First row, second column: $2\cdot(-1) + (-1)\cdot(-2) = -2 + 2 = 0$ .

Second row, first column: $3\cdot2 + (-2)\cdot3 = 6 - 6 = 0$ .

Second row, second column: $3\cdot(-1) + (-2)\cdot(-2) = -3 + 4 = 1$ .

So
$A^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I.$
Important

$A^2 = I$ means $A$ is its own inverse: $A^{-1} = A$ .

Now find $A^3$ . Since $A^2 = I$ , we have

$A^3 = A^2 \cdot A = I \cdot A = A.$
So $A^3$ is simply $A$ itself. …
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COMEDK 2022Set 20221 markMCQ
$U_{n+1}=3 U_n-2 U_{n-1}$
›Reveal solutionSolution

Check: U(2) = 3(3) − 2(2) = 5 = 2² + 1 ✓; U(3) = 3(5) − 2(3) = 9 = 2³ + 1 ✓.

Concept: Linear homogeneous recurrence with constant coefficients — solve via the characteristic equation.

U(n+1) = 3U(n) − 2U(n−1) → characteristic equation x² = 3x − 2, i.e. x² − 3x + 2 = 0 → (x−1)(x−2) = 0 → x = 1, 2.

So U(n) = A·1ⁿ + B·2ⁿ = A + B·2ⁿ.

Apply initial conditions:

U(0) = A + B = 2

U(1) = A + 2B = 3 …
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KCET 2020Set A-11 markMCQ
$a_1, a_2, a_3, ......., a_9$
›Reveal solutionSolution

Consecutive terms of an A.P. make the three rows linearly dependent ( $R_1 + R_3 = 2R_2$ ), forcing the determinant to be $0$ — and the only option equal to $0$ is $\log_e(\log_e e)$ .

Step 1 — Write the terms of the A.P.

Let the first term be $a$ and the common difference be $d$ . Then $a_k = a + (k-1)d$ , so the determinant is
$\Delta = \begin{vmatrix} a & a+d & a+2d \\ a+3d & a+4d & a+5d \\ a+6d & a+7d & a+8d \end{vmatrix}$
Step 2 — Spot the row dependence.

Add row 1 and row 3, element by element:
$R_1 + R_3 = \big(2a+6d,\; 2a+8d,\; 2a+10d\big)$
and double row 2:
$2R_2 = \big(2a+6d,\; 2a+8d,\; 2a+10d\big)$
These are identical:
$\boxed{R_1 + R_3 = 2R_2}$
This is just the A.P. property in row form — the middle row is the average of the outer two, exactly as $a_5$ is the average of $a_2$ and $a_8$ , etc.

Step 3 — Conclude the determinant is zero.

Apply the row operation $R_1 \to R_1 + R_3 - 2R_2$ (a determinant is unchanged by adding a linear combination of other rows to a row):
$\Delta = \begin{vmatrix} 0 & 0 & 0 \\ a+3d & a+4d & a+5d \\ a+6d & a+7d & a+8d \end{vmatrix} = 0$
A determinant with an all-zero row is zero. (Equivalently: the rows are linearly dependent, so the matrix is singular.)
$\Delta = 0$
Step 4 — Which option equals $0$ ?

Evaluate each: …
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Q.If tnt_ntn​ denotes the nnnth term of the series 2+3+6+11+18+…2 + 3 + 6 + 11 + 18 + \ldots2+3+6+11+18+… then t50t_{50}t50​ is (A) 492−149^2 - 1492−1 (B) 49249^2492 (C) 502+150^2 + 1502+1 (D) 492+249^2 + 2492+2

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Q.If $t_n$ denotes the $n$ th term of the series $2 + 3 + 6 + 11 + 18 + \ldots$ then $t_{50}$ is
(A) $49^2 - 1$
(B) $49^2$
(C) $50^2 + 1$
(D) $49^2 + 2$

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