NIMCET 2027 Progressions Previous Year Questions (PYQs) – Page 1 of 5
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If $H$ is the harmonic mean between $P$ and $Q$, then $\dfrac{H}{P} + \dfrac{H}{Q}$ is
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Solution
Harmonic mean of $P$ and $Q$ is
$H = \dfrac{2PQ}{P+Q}$
Now,
$\dfrac{H}{P} + \dfrac{H}{Q} = H\left(\dfrac{1}{P} + \dfrac{1}{Q}\right)
= H \cdot \dfrac{P+Q}{PQ}$
Substitute
$H = \dfrac{2PQ}{P+Q}$:
$\dfrac{H}{P} + \dfrac{H}{Q}
= \dfrac{2PQ}{P+Q} \cdot \dfrac{P+Q}{PQ}
= 2$
So the correct Answer is $2$.
$a,b,c$ are positive and $c>a$ and in H.P.
Compute $\log(a+c)+\log(a-2b+c)$.
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Solution
$a,b,c$ in H.P. ⇒
$\dfrac{1}{b} = \dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{c}\right)$
⇒ $2bc = ac + ab$
Simplify gives identity:
$(a+c)(a-2b+c) = (c-b)^2$
Take log:
$\log(a+c) + \log(a-2b+c) $
$= \log\big((c-b)^2\big)$
$=2\log(c-b)$
Find $k$ in the equation
$x^3 - 6x^2 + kx + 64 = 0$
if roots are in geometric progression.
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Solution
Let roots be $a, ar, ar^2$.
Sum of roots:
$a + ar + ar^2 = 6$
$a(1+r+r^2) = 6$ ...(1)
Product of roots:
$ar \cdot ar^2 \cdot a = a^3 r^3 = -64$
$\Rightarrow (ar)^3 = -64$
$\Rightarrow ar = -4$ ...(2)
Middle coefficient relation:
Sum of pairwise products:
$k = a(ar) + ar(ar^2) + ar^2(a)$
$k = a^2 r + a^2 r^3 + a^2 r^2$
Factor:
$k = a^2 (r + r^2 + r^3)$
$k = ar \cdot a(r + r^2 + r^3)$
Using (2):
$ar = -4$
Also:
$r + r^2 + r^3 = r(1 + r + r^2)$
Thus:
$k = -4a \cdot r(1 + r + r^2)$
But from (1):
$a(1+r+r^2) = 6$
So:
$k = -4r \cdot 6 = -24r$
Now solve $ar = -4$ and equation (1).
Standard GP root problem yields $r = 1$ or $r = -1$.
Check sign consistency → $r = 1$.
So:
$k = -24(1)$
$k = -24$
The sum of infinite terms of a decreasing GP is equal to the greatest value of the function $f(x)=x^3+3x-9$ in the interval [-2,3] and the difference between the first two terms is $f'(0)$. Then the common ratio of GP is
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Solution
Given:
\(f(x)=x^3+3x-9\) on \([-2,3]\).
Since \(f'(x)=3x^2+3>0\), \(f\) is increasing, so the greatest value is at \(x=3\):
\(f(3)=27\).
Sum to infinity of decreasing GP:
\(S=\dfrac{a}{1-r}=27\) with
Also \(f'(0)=3\Rightarrow\) difference of first two terms:
\(a-ar=a(1-r)=3\).
From \(a=\;27(1-r)\),
plug into \(a(1-r)=3\): \(27(1-r)^2=3\Rightarrow(1-r)^2=\dfrac{1}{9}\Rightarrow 1-r=\dfrac{1}{3}\) (take positive)
Common ratio: \(r=1-\dfrac{1}{3}=\boxed{\dfrac{2}{3}}\).
$ \text{The sum of } 11^{2} + 12^{2} + \cdots + 30^{2} \text{ is} $
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Solution
$ \sum_{k=11}^{30} k^{2} = \frac{30\cdot 31 \cdot 61}{6} - \frac{10\cdot 11 \cdot 21}{6} = 9455 - 1385 = 8070 $
In a G.P. consisting of positive terms, each term equals the sum of the next two terms. Then the
common ratio of the G.P. is
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Solution
If a, b, c are in GP and log a - log 2b, log 2b - log 3c and log 3c - log a are in AP, then a, b, c are the lengths of the sides of a triangle which
is
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Solution
If \(x, 2x + 2, 3x + 3\) are the first three terms of a geometric progression, then the 4th term in the geometric progression is:
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Solution
The sum of $n$ terms of an arithmetic series is 216. The value of the first term is $n$ and the value of the
$n^{th}$ term is $2n$. The common difference, $d$ is.
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Solution
Given:
Sum of \(n\) terms = 216
First term \(a = n\)
\(n^{th}\) term \(= 2n\)
Use nth term formula
\[ a_n = a + (n-1)d \] \[ 2n = n + (n-1)d \] \[ n = (n-1)d \quad \cdots (1) \]
Use Sum formula
\[ S_n = \frac{n}{2}(a + a_n) \] \[ 216 = \frac{n}{2}(n + 2n) \] \[ 216 = \frac{n}{2}(3n) \] \[ 216 = \frac{3n^2}{2} \] \[ n^2 = \frac{432}{3} = 144 \] \[ n = 12 \]
Find common difference \(d\)
From equation (1):
\[ n = (n-1)d \] \[ 12 = (12-1)d \] \[ 12 = 11d \] \[ d = \frac{12}{11} \]
Answer: \( d = \dfrac{12}{11} \)
Sum of \(n\) terms = 216
First term \(a = n\)
\(n^{th}\) term \(= 2n\)
Use nth term formula
\[ a_n = a + (n-1)d \] \[ 2n = n + (n-1)d \] \[ n = (n-1)d \quad \cdots (1) \]
Use Sum formula
\[ S_n = \frac{n}{2}(a + a_n) \] \[ 216 = \frac{n}{2}(n + 2n) \] \[ 216 = \frac{n}{2}(3n) \] \[ 216 = \frac{3n^2}{2} \] \[ n^2 = \frac{432}{3} = 144 \] \[ n = 12 \]
Find common difference \(d\)
From equation (1):
\[ n = (n-1)d \] \[ 12 = (12-1)d \] \[ 12 = 11d \] \[ d = \frac{12}{11} \]
Answer: \( d = \dfrac{12}{11} \)
If $a, a, a_2, ., a_{2n-1},b$ are in AP, $a, b_1, b_2,...b_{2n-1}, b $are in GP and $a, c_1, c_2,... c_{2n-1}, b $ are in HP, where a, b are positive, then the
equation $a_n x^2-b_n+c_n$ has its roots
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Solution
NIMCET 2027
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and More.
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