(A) Given the $n^{th}$ term is $a_{n} = n(n+2)$.Substituting $n = 1, 2, 3, 4, 5$:For $n = 1: a_{1} = 1(1+2) = 3$For $n = 2: a_{2} = 2(2+2) = 8$For $n

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) Given the $n^{th}$ term is $a_{n} = n(n+2)$.Substituting $n = 1, 2, 3, 4, 5$:For $n = 1: a_{1} = 1(1+2) = 3$For $n = 2: a_{2} = 2(2+2) = 8$For $n = 3: a_{3} = 3(3+2) = 15$For $n = 4: a_{4} = 4(4+2) = 24$For $n = 5: a_{5} = 5(5+2) = 35$Thus,the first five terms are $3, 8, 15, 24, 35$.

Class 11 Mathematics Sequences and Series nth term of special series, Sum to n terms and Infinite number of terms

Easy

Write the first five terms of the sequence whose $n^{th}$ term is $a_{n} = n(n+2)$.

A
$3, 8, 15, 24, 35$
B
$3, 8, 15, 24, 36$
C
$2, 8, 15, 24, 35$
D
$3, 8, 16, 24, 35$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

Sequences and Series

Subtopic

nth term of special series, Sum to n terms and Infinite number of terms

Find the sum to $n$ terms of the series $1 \cdot 3 \cdot 5 + 3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \dots$

Medium

View Solution

If $\sum_{k=1}^{n} a_k = 6n^3$, then $\sum_{k=1}^{6} \left(\frac{a_{k+1}-a_k}{36}\right)^2$ is equal to . . . . . . .

DifficultJEE MAIN 2026

View Solution

If the $n^{th}$ term of a sequence is $T_n = 2n - 1$,then the sum of $n$ terms $S_n = \dots$

Easy

View Solution

Show that $\frac{1 \times 2^{2}+2 \times 3^{2}+\ldots+n \times(n+1)^{2}}{1^{2} \times 2+2^{2} \times 3+\ldots+n^{2} \times(n+1)}=\frac{3 n+5}{3 n+1}$

Difficult

View Solution

What is the sum of the series $1^2 + 2.2^2 + 3^2 + 2.4^2 + 5^2 + 2.6^2 + \dots + 2(2m)^2$?

Difficult

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Write the first five terms of the sequence whose $n^{th}$ term is $a_{n} = n(n+2)$.

Similar Questions

Find the sum to $n$ terms of the series $1 \cdot 3 \cdot 5 + 3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \dots$

If $\sum_{k=1}^{n} a_k = 6n^3$, then $\sum_{k=1}^{6} \left(\frac{a_{k+1}-a_k}{36}\right)^2$ is equal to . . . . . . .

If the $n^{th}$ term of a sequence is $T_n = 2n - 1$,then the sum of $n$ terms $S_n = \dots$

Show that $\frac{1 \times 2^{2}+2 \times 3^{2}+\ldots+n \times(n+1)^{2}}{1^{2} \times 2+2^{2} \times 3+\ldots+n^{2} \times(n+1)}=\frac{3 n+5}{3 n+1}$

What is the sum of the series $1^2 + 2.2^2 + 3^2 + 2.4^2 + 5^2 + 2.6^2 + \dots + 2(2m)^2$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module