JEE MAIN Progressions Previous Year Questions (PYQs) – Page 1 of 13
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Let $a_1, \frac{a_2}{2}, \frac{a_3}{2^2},; \ldots, \frac{a_{10}}{2^9}$ be a G.P. of common ratio $\frac{1}{\sqrt{2}}$. If $a_1 + a_2 + \cdots + a_{10} = 62$, then $a_1$ is equal to:
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Solution
$\frac{a_2}{2a_1} = \frac{a_3}{2a_2} = \cdots = \frac{a_{10}}{2a_9} = \frac{1}{\sqrt{2}}$
$\Rightarrow a_1, a_2, a_3, \ldots, a_{10}$ are in G.P. with common ratio $\sqrt{2}$
$\sum_{i=1}^{10} a_i = a_1 \cdot \frac{(\sqrt{2})^{10} - 1}{\sqrt{2} - 1} = 62$
$\Rightarrow a_1 = 2(\sqrt{2}-1)$
If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is:
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Solution
First term $a=3$, common difference $d$
Sum of first 4 terms:
$S_4=\frac{4}{2}[2a+3d]=2(6+3d)=12+6d$
Sum of first 8 terms:
$S_8=\frac{8}{2}[2a+7d]=4(6+7d)=24+28d$
Sum of next 4 terms:
$=S_8-S_4=(24+28d)-(12+6d)=12+22d$
Given:
$12+6d=\frac{1}{5}(12+22d)$
$60+30d=12+22d$
$8d=-48 \Rightarrow d=-6$
Now sum of first 20 terms:
$S_{20}=\frac{20}{2}[2a+19d]=10(6+19(-6))$
$=10(6-114)=10(-108)=-1080$
$\boxed{-1080}$
Sum of first 4 terms:
$S_4=\frac{4}{2}[2a+3d]=2(6+3d)=12+6d$
Sum of first 8 terms:
$S_8=\frac{8}{2}[2a+7d]=4(6+7d)=24+28d$
Sum of next 4 terms:
$=S_8-S_4=(24+28d)-(12+6d)=12+22d$
Given:
$12+6d=\frac{1}{5}(12+22d)$
$60+30d=12+22d$
$8d=-48 \Rightarrow d=-6$
Now sum of first 20 terms:
$S_{20}=\frac{20}{2}[2a+19d]=10(6+19(-6))$
$=10(6-114)=10(-108)=-1080$
$\boxed{-1080}$
The value of $\dfrac{1\times2^2+2\times3^2+\ldots+100\times(101)^2}{1\times3+2\times4+3\times5+\ldots+100\times101}$ is:
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Solution
Let $9=x_{1} < x_{2} < \ldots < x_{7}$ be in an A.P. with common difference d. If the standard deviation of $x_{1}, x_{2}..., x_{7}$ is 4 and the mean is $\bar{x}$, then $\bar{x}+x_{6}$ is equal to :
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Solution
If $a_{1}, a_{2}, a_{3}, \dots$ are in A.P. such that $a_{1} + a_{7} + a_{16} = 40$,
then the sum of the first 15 terms of this A.P. is:
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Solution
If $\alpha, \beta$ and $\gamma$ are three consecutive terms of a non-constant G.P. such that the equations
$\alpha x^{2} + 2\beta x + \gamma = 0$ and $x^{2} + x - 1 = 0$
have a common root, then $\alpha(\beta + \gamma)$ is equal to:
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Solution
If the sum of the first $20$ terms of the series $\dfrac{4\cdot1}{4+3\cdot1^{2}+1^{4}}+\dfrac{4\cdot2}{4+3\cdot2^{2}+2^{4}}+\dfrac{4\cdot3}{4+3\cdot3^{2}+3^{4}}+\dfrac{4\cdot4}{4+3\cdot4^{2}+4^{4}}+\cdots$ is $\dfrac{m}{n}$, where $m$ and $n$ are coprime, then $m+n$ is equal to:
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Solution
Three positive numbers form an increasing G.P. If the middle term in this G.P.
is doubled, the new numbers are in A.P. then the common ratio of the G.P. is :
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Solution
If $a,b,c$ are in AP and $a+1,; b,; c+3$ are in GP. Given $a>10$ and the arithmetic mean of $a,b,c$ is $8$, then the cube of the geometric mean of $a,b,c$ is:
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Solution
Consider two sets $A$ and $B$, each containing three numbers in A.P. Let the sum and the product of the elements of $A$ be $36$ and $p$ respectively and the sum and the product of the elements of $B$ be $36$ and $q$ respectively. Let $d$ and $D$ be the common differences of the A.P.s in $A$ and $B$ respectively such that $D=d+3$, $d>0$. If $\dfrac{p+q}{p-q}=\dfrac{19}{5}$, then $p-q$ is equal to:
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Solution
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