120022003200424052806…….…….……… Sum =11040 120+80+160+40+200+240+…=11040 2n[2a+(n−1)d]+80+40=11040 2n[240+(n−1)40]=10920 n[6+n−1]=546 n(n+5)=546 n=21
JEE Main 2011 — Mathematics Algebra
A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months his saving increases by Rs. 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs. 11040 after
Held on 30 Apr 2011 · Verified 6 Jul 2026.
19 months
20 months
21 months
18 months
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Let the set of all values of $k \in \mathbb{R}$ such that the equation $z(\bar{z} + 2 + i) + k(2 + 3i) = 0$, $z \in \mathbb{C}$, has at least one solution, be the interval $[\alpha, \beta]$. Then $9(\alpha + \beta)$ is equal to:
If $\alpha=1$ and $\beta=1+i\sqrt{2}$, where $i=\sqrt{-1}$ are two roots of the equation $x^3+ax^2+bx+c=0$, $a,b,c \in \mathbb{R}$, then $\int_{-1}^{1}(x^3+ax^2+bx+c)dx$ is equal to:
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$\frac{6}{3^{26}}+\frac{10 \cdot 1}{3^{25}}+\frac{10 \cdot 2}{3^{24}}+\frac{10 \cdot 2^{2}}{3^{23}}+\ldots+\frac{10 \cdot 2^{24}}{3}$ is equal to :
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Work through every JEE Main Algebra PYQ, year by year.