NIMCET 2027 Progressions Previous Year Questions (PYQs) – Page 5 of 5
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Showing 43 questions
Let $A_k$ be the arithmetic mean of squares of $k$ natural numbers. If $\sum_{k=1}^{n}(6A_k-3k)=31$, find the value of $n$.
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4
Solution
$A_k=\frac{1^2+2^2+3^2+\cdots+k^2}{k}$
$A_k=\frac{\frac{k(k+1)(2k+1)}{6}}{k}$
$A_k=\frac{(k+1)(2k+1)}{6}$
Now,
$6A_k-3k=(k+1)(2k+1)-3k$
$=2k^2+3k+1-3k$
$=2k^2+1$
Given,
$\sum_{k=1}^{n}(6A_k-3k)=31$
$\sum_{k=1}^{n}(2k^2+1)=31$
For $n=3$,
$(2\cdot1^2+1)+(2\cdot2^2+1)+(2\cdot3^2+1)$
$=3+9+19=31$
Therefore,
$n=3$
If $a_1,a_2,\ldots,a_n$ are in A.P. and $a_1=0$ then the value of
$\left(\dfrac{a_3}{a_2}+\dfrac{a_4}{a_3}+\cdots+\dfrac{a_n}{a_{n-1}}\right)-a_2\left(\dfrac{1}{a_2}+\dfrac{1}{a_3}+\cdots+\dfrac{1}{a_{n-2}}\right)$
is equal to
1
2
3
4
Solution
If a, b, c, d are in HP and arithmetic mean of ab, bc, cd is 9 then which of the following number is the value of ad?
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Solution
$a, b, c, d$ are in H.P.
$\Rightarrow \frac{2}{b} = \frac{1}{a} + \frac{1}{c}$ and $\frac{2}{c} = \frac{1}{b} + \frac{1}{d}$
But $ab + bc + cd = 27$:
$\Rightarrow \frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d}$ are in A.P.
$\Rightarrow \frac{2}{b} = \frac{1}{a} + \frac{1}{c}$ and $\frac{2}{c} = \frac{1}{b} + \frac{1}{d}$
Given A.M. of $ab, bc, cd$ is $9$:
$\frac{ab + bc + cd}{3} = 9$
$\Rightarrow ab + bc + cd = 27$
Now multiply:
$\frac{2}{b} = \frac{1}{a} + \frac{1}{c} $
$\Rightarrow 2ac = b(a + c)$
$\frac{2}{c} = \frac{1}{b} + \frac{1}{d} \Rightarrow 2bd = c(b + d)$
Multiply both:
$4abcd = bc(a + c)(b + d)$
Cancel $bc$:
$4ad = (a + c)(b + d)$
Expand RHS:
$4ad = ab + ad + bc + cd$
$\Rightarrow 3ad = ab + bc + cd$
But $ab + bc + cd = 27$:
$\Rightarrow 3ad = 27$
$\Rightarrow ad = 9$
$\boxed{9}$
NIMCET 2027
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NIMCET 2027
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and More.
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