If we write the elements of A+A, we can certainly find 39 distinct elements as 1+1,1+a1,1+a2,….1 +a18,1+77,a1+77,a2+77,……a18+77,77+77. It means all other sums are already present in these 39 values, which is only possible in case when all numbers are in A.P. Let the common difference be ' d '. 77=1+19d⇒d=4 So, i=1∑18a1=218[2a1+17d]=9[10+68]=702
Let A=1,a_1,a_2… a_18,77 be a set of integers with 1<a_1<a_2<… ..<a_18<77. Let the set A+A=x+y:x,y∈ A contain exactly 39 elements. Then, the value of…
JEE Main 2022 — Mathematics Algebra
Let A=1,a1,a2……a18,77 be a set of integers with 1<a1<a2<…..<a18<77. Let the set A+A=x+y:x,y∈A contain exactly 39 elements. Then, the value of a1+a2+…..+a18 is equal to ______.
Held on 28 Jun 2022 · Verified 6 Jul 2026.
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.