New chat PYQ Library Doubts Rank report

Sign in to see Recents

Sign in to save progress →

Sequences and Series question

2025 · 29 Jan · Shift 2 · Q50

Guest · try one question or challengeGuest · try one question or challenge

PYQ Library
JEE Main
Mathematics
Sequences and Series
2025 · 29 Jan · Shift 2 · Q50

Sequences and Series question

2025 · 29 Jan · Shift 2 · Q50

JEE MainMathematicsSequences and SeriesNumerical+4 / −1

Let

a_1, a_2, \ldots, a_{2024}

be an Arithmetic Progression such that

a_1+\left(a_5+a_{10}+a_{15}+\ldots+a_{2020}\right)+a_{2024}=2233

. Then

a_1+a_2+a_3+\ldots+a_{2024}

is equal to

\underline{\hspace{2cm}}

.

Numerical answer

More from Sequences and Series

Let $3, a, b, c$ be in A.P. and $3, a - 1, b + 1, c + 9$ be in G.P. Then, the arithmetic mean of $a, b$ and $c$ is :2024 · MCQ
Let $3, 7, 11, 15, \dots, 403$ and $2, 5, 8, 11, \dots, 404$ be two arithmetic progressions. Then the sum, of the common terms in them, is equal to .2024 · Numerical
Let $S_{n}$ denote the sum of the first $n$ terms of an arithmetic progression. If $S_{10} = 390$ and the ratio of the tenth and the fifth terms is $15 : 7$ , then $S_{15} - S_{5}$ is equal to :2024 · MCQ
If three successive terms of a G.P. with common ratio $r (r > 1)$ are the lengths of the sides of a triangle and $[r]$ denotes the greatest integer less than or equal to $r$ , then $3 [r] + [- r]$ is equal to …2024 · Numerical
Let the first three terms 2, p and q, with $q e q 2$ , of a G.P. be respectively the $7^{th}, 8^{th}$ and $1 3^{th}$ terms of an A.P. If the $5^{th}$ term of the G.P. is the $n^{th}$ term of the…2024 · MCQ
The value of $\frac{1 \times 2 ^{2} + 2 \times 3 ^{2} + \dots + 100 \times ( 101 ) ^{2}}{1 ^{2} \times 2 + 2 ^{2} \times 3 + \dots . + 10 0 ^{2} \times 101}$ is2024 · MCQ
Let three real numbers $a, b, c$ be in arithmetic progression and $a + 1, b, c + 3$ be in geometric progression. If $a > 10$ and the arithmetic mean of $a, b$ and $c$ is 8, then the cube of the geometric mean of $a, b$ and $c$ is2024 · MCQ
If $\frac{1}{1 + 2} + \frac{1}{2 + 3} + \dots + \frac{1}{99 + 100} = m$ and $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{99 \cdot 100} = n$ , then the point $(m, n)$ lies…2024 · MCQ