$\begin{aligned}
& \mathrm{S}3=3 \mathrm{a}+3 \mathrm{d}=54 \
& \Rightarrow \mathrm{a}+\mathrm{d}=18 \
& \mathrm{S}{20}=10(2 \mathrm{a}+19 \mathrm{d}) \
& \Rightarrow 10(36+17 \mathrm{d}) \
& \Rightarrow 1600 < 10(36+17 \mathrm{d}) < 1800 \
& \Rightarrow 160 < 36+17 \mathrm{d} < 180 \
& \Rightarrow 124 < 17 \mathrm{~d} < 144 \
& \Rightarrow 7 \frac{5}{17} < \mathrm{d} < 8 \frac{8}{17}
\end{aligned}Commondifferencewillbenaturalnumber\begin{aligned}
& \Rightarrow d=8 \Rightarrow a=10 \
& \Rightarrow a_{11}=10+10 \times 8=90
\end{aligned}$
Consider an A. P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800.…
JEE Main 2025 — Mathematics Algebra
Consider an A. P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its 11th term is :
Held on 29 Jan 2025 · Verified 6 Jul 2026.
Options
- A
90
- B
84
- C
122
- D
108
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
JEE Main Algebra in other years
More JEE Main Algebra questions
The number of elements in the set $S = \left\{(r, k) : k \in \mathbb{Z} \text{ and } {}^{36}C_{r+1} = \dfrac{6\left({}^{35}C_r\right)}{(k^2 - 3)}\right\}$, is :
Suppose $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in A.P. and $\mathrm{a}^{2}, 2 \mathrm{~b}^{2}, \mathrm{c}^{2}$ are in G.P. If $\mathrm{a}<\mathrm{b}<\mathrm{c}$ and $\mathrm{a}+\mathrm{b}+\mathrm{c}=1$, then $9\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}\right)$ is equal to $\_\_\_\_$ .
Let $S=\left\{z \in \mathbb{C}: 4 z^{2}+\bar{z}=0\right\}$. Then $\sum_{z \in S}|z|^{2}$ is equal to:
Let $S=\{z: 3 \leqslant|2 z-3(1+i)| \leqslant 7\}$ be a set of complex numbers. Then $\min _{z \in S}\left|\left(z+\frac{1}{2}(5+3 i)\right)\right|$ is equal to :
Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{bmatrix}$ and $B = [b_{ij}]$, $1 \leq i, j \leq 3$. If $B = A^{99} - I$, then the value of $\dfrac{b_{31} - b_{21}}{b_{32}}$ is :
Other Mathematics topics for JEE Main
Keep practicing Algebra
Work through every JEE Main Algebra PYQ, year by year.