Jamia Millia Islamia Progressions Previous Year Question Paper and Solution with PDF

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Previous Year Questions
Jamia Millia Islamia

Jamia Millia Islamia Previous Year Questions (PYQs)

Jamia Millia Islamia Progressions PYQ

Qus : 1 Jamia Millia Islamia PYQ

A.P. whose nth term is $2n - 1$ is

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2018 PYQ

Solution

Substituting $n=1,2,3,…$ gives terms 1, 3, 5,…

Qus : 2 Jamia Millia Islamia PYQ

In an A.P. the $p^{th}$ term is $q$ and the $(p+q)^{th}$ term is $0$. Then the $q^{th}$ term is:

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2021 PYQ

Solution

Let first term = $a$, common difference = $d$. Then, $a + (p-1)d = q$ $a + (p+q-1)d = 0$ Subtracting: $(a + (p+q-1)d) - (a + (p-1)d) = 0 - q$ $\Rightarrow qd = -q \Rightarrow d = -1$ Now $a + (p-1)d = q \Rightarrow a = q + p - 1$. $q^{th}$ term = $a + (q-1)d = (q + p - 1) - (q - 1) = p$. $\boxed{\text{Answer: (B) }p}$

Qus : 3 Jamia Millia Islamia PYQ

If 20th term of an A.P. is 30 and its 30th term is 20, then its 10th term is

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2017 PYQ

Solution

Let the first term be $a$ and common difference $d$. $ a + 19d = 30 $ $ a + 29d = 20 $ Subtract → $10d = -10 \Rightarrow d = -1$ Then $a = 49$ 10th term = $a + 9d = 49 - 9 = 40$

Qus : 4 Jamia Millia Islamia PYQ

Let sum of $n$ terms of an A.P. is $2n(n-1)$, then sum of their squares is

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2017 PYQ

Solution

Sum $S_n = 2n(n-1)$ → $a = 2$, $d = 4$ Sum of squares = $\sum (a + (r-1)d)^2$ $= n[a^2 + (n-1)(a+d)^2 + …]$ Simplifying gives $\dfrac{8n(n+1)(2n+1)}{3}$

Qus : 5 Jamia Millia Islamia PYQ

If the ratio of sum of $m$ terms and $n$ terms of an A.P. be $m^2:n^2$, then the ratio of its $m$th and $n$th terms will be

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2017 PYQ

Solution

$S_m/S_n = m^2/n^2$ We know $S_n = \dfrac{n}{2}[2a+(n-1)d]$ $\Rightarrow \dfrac{m[2a+(m-1)d]}{n[2a+(n-1)d]} = \dfrac{m^2}{n^2}$ Simplify → $\dfrac{2a+(m-1)d}{2a+(n-1)d} = \dfrac{m}{n}$ $\Rightarrow \text{ratio of } t_m : t_n = (2m-1):(2n-1)$

Qus : 6 Jamia Millia Islamia PYQ

If ‘a’ is the arithmetic mean of ‘b’ and ‘c’, and $G_1$ and $G_2$ are the two geometric means between them, then $G_1^3 + G_2^3$ is equal to —

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2023 PYQ

Solution

Given $a = \dfrac{b + c}{2}$ and $G_1, G_2$ are geometric means between $b$ and $c$. So $b, G_1, G_2, c$ are in G.P. Let common ratio = $r$. Then $G_1 = br$ and $G_2 = br^2$, $c = br^3$. Hence $a = \dfrac{b + c}{2} = \dfrac{b + br^3}{2} = \dfrac{b(1 + r^3)}{2}.$ Now, $G_1^3 + G_2^3 = b^3(r^3 + r^6) = b^3r^3(1 + r^3).$ But $c = br^3$ and $(1 + r^3) = \dfrac{2a}{b}$. So, $G_1^3 + G_2^3 = b^3r^3 \times \dfrac{2a}{b} = 2ab^2r^3 = 2abc.$

Qus : 7 Jamia Millia Islamia PYQ

Let the sequence be $1\times2, 3\times2^2, 5\times2^3, 7\times2^4, 9\times2^5, \ldots$ Then this sequence is...

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2020 PYQ

Solution

The coefficients $1, 3, 5, 7, 9, \ldots$ form an arithmetic sequence, and the powers of 2 form a geometric sequence. Hence, the overall sequence is arithmetico-geometric.

Qus : 8 Jamia Millia Islamia PYQ

The sum of 3 numbers in A.P. is $-3$ and their product is $8$. Then the sum of squares of the numbers is:

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2022 PYQ

Solution

Let the numbers be $a - d,\ a,\ a + d$. $3a = -3 \Rightarrow a = -1$ $(a - d)a(a + d) = 8$ $\Rightarrow -1(1 - d^2) = 8 \Rightarrow d^2 = 9$ Sum of squares $= 3a^2 + 2d^2 = 3(1) + 2(9) = 21$

Qus : 9 Jamia Millia Islamia PYQ

If $x,\ 2x + 2,\ 3x + 3$ are in G.P., then the 4th term is:

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2022 PYQ

Solution

For G.P., $(2x + 2)^2 = x(3x + 3)$ $4x^2 + 8x + 4 = 3x^2 + 3x$ $\Rightarrow x^2 + 5x + 4 = 0$ $\Rightarrow x = -1$ or $x = -4$ For non-trivial case, $x = -4$ Then terms are $-4, -6, -9$ and ratio $r = \dfrac{3}{2}$ 4th term $= -4 \times \left(\dfrac{3}{2}\right)^3 = -13.5$

Qus : 10 Jamia Millia Islamia PYQ

In a sequence of $21$ terms, the first $11$ terms are in A.P. with common difference $2$ and the last $11$ terms are in G.P. with common ratio $2$. If the middle term of A.P. is equal to the middle term of G.P., then the middle term of the entire sequence is

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2022 PYQ

Solution

Let $t_1$ be the first term. For the A.P.: $t_6=t_1+10$, $t_{11}=t_1+20$. For the G.P. (terms $11$ to $21$ with ratio $2$): $t_{16}=t_{11}\cdot 2^5=32t_{11}$. Given $t_6=t_{16}$: $t_1+10=32(t_1+20)\Rightarrow 31t_1=-630\Rightarrow t_1=-\dfrac{630}{31}$. Middle term of entire sequence is $t_{11}=t_1+20=-\dfrac{630}{31}+20=-\dfrac{10}{31}$.

Qus : 11 Jamia Millia Islamia PYQ

If $m^{\text{th}}$ term of an A.P. is $n$ and $n^{\text{th}}$ term is $m$, then its $10^{\text{th}}$ term is …

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2016 PYQ

Solution

Let first term $a$, common difference $d$. $a+(m-1)d=n \quad\text{and}\quad a+(n-1)d=m$. Subtract: $(m-n)d=n-m \Rightarrow d=-1$. Then $a=n+(m-1)=m+n-1$. $t_{10}=a+9d=(m+n-1)+9(-1)=m+n-10$.

Qus : 12 Jamia Millia Islamia PYQ

Let sum of $n$ terms of an A.P. be $S_n=3n^2+5$. If $T_n$ (the $n$th term) of this series is $159$, then $n$ is …

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2016 PYQ

Solution

$T_n=S_n-S_{n-1}=(3n^2+5)-[3(n-1)^2+5]=6n-3$. Set $6n-3=159 \Rightarrow 6n=162 \Rightarrow n=27$.

Qus : 13 Jamia Millia Islamia PYQ

If the roots of $x^3-9x^2+23x-15=0$ are in A.P., then their common difference is …

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2016 PYQ

Solution

Let roots be $a-d,\;a,\;a+d$. Then sum $=3a=9 \Rightarrow a=3$. Sum of pairwise products $=3a^2-d^2=23 \Rightarrow 27-d^2=23 \Rightarrow d^2=4$. Hence $d=\pm2$.

Qus : 14 Jamia Millia Islamia PYQ

If sum of $n$ terms of a series is $3n^2 + 4n$, then the series is …

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2016 PYQ

Solution

$S_n = 3n^2 + 4n$ Then, $T_n = S_n - S_{n-1} = (3n^2 + 4n) - [3(n-1)^2 + 4(n-1)] = 6n + 1$. Since $T_n$ is linear in $n$, the series is an A.P.

Qus : 15 Jamia Millia Islamia PYQ

The sum of integers from 1 to 100 that are divisible by 2 or 5 is:

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2019 PYQ

Solution

Sum of numbers divisible by 2 up to 100: $2 + 4 + \dots + 100 = 2(1 + 2 + \dots + 50) = 2 \times \frac{50 \times 51}{2} = 2550.$ Sum of numbers divisible by 5 up to 100: $5 + 10 + \dots + 100 = 5(1 + 2 + \dots + 20) = 5 \times \frac{20 \times 21}{2} = 1050.$ Sum of numbers divisible by both 2 and 5 (i.e., by 10): $10 + 20 + \dots + 100 = 10(1 + 2 + \dots + 10) = 10 \times 55 = 550.$ Required sum = $2550 + 1050 - 550 = 3050.$

Qus : 16 Jamia Millia Islamia PYQ

If $a_n = 4n + 6$, find the $15^{th}$ term of the sequence.

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2024 PYQ

Solution

$a_{15} = 4(15) + 6 = 60 + 6 = 66$

Qus : 17 Jamia Millia Islamia PYQ

Which term of the A.P. 92, 88, 84, 80, ... is 0?

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2018 PYQ

Solution

(a = 92, d = −4,

0 = 92 + (n−1)(−4) → 4n = 96 → n = 24)

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