Jamia Millia Islamia MCA Progressions Previous Year Questions (PYQs) – Page 1 of 2

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

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 = $a + (q-1)d = (q + p - 1) - (q - 1) = p$. $\boxed{\text{Answer: (B) }p}$

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 = $a + 9d = 49 - 9 = 40$

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 $\dfrac{8n(n+1)(2n+1)}{3}$

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

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, $G_1^3 + G_2^3 = b^3r^3 \times \dfrac{2a}{b} = 2ab^2r^3 = 2abc.$

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.

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 $= 3a^2 + 2d^2 = 3(1) + 2(9) = 21$

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 $= -4 \times \left(\dfrac{3}{2}\right)^3 = -13.5$

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 $t_{11}=t_1+20=-\dfrac{630}{31}+20=-\dfrac{10}{31}$.