Jamia Millia Islamia MCA Progressions Previous Year Questions (PYQs) – Page 1 of 2
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A.P. whose nth term is $2n - 1$ is
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Solution
Substituting $n=1,2,3,…$ gives terms 1, 3, 5,…
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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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 = $a + (q-1)d = (q + p - 1) - (q - 1) = p$.
$\boxed{\text{Answer: (B) }p}$
If 20th term of an A.P. is 30 and its 30th term is 20, then its 10th term is
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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 = $a + 9d = 49 - 9 = 40$
Let sum of $n$ terms of an A.P. is $2n(n-1)$, then sum of their squares is
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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 $\dfrac{8n(n+1)(2n+1)}{3}$
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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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 → $\dfrac{2a+(m-1)d}{2a+(n-1)d} = \dfrac{m}{n}$
$\Rightarrow \text{ratio of } t_m : t_n = (2m-1):(2n-1)$
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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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,
$G_1^3 + G_2^3 = b^3r^3 \times \dfrac{2a}{b} = 2ab^2r^3 = 2abc.$
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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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.
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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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 $= 3a^2 + 2d^2 = 3(1) + 2(9) = 21$
If $x,\ 2x + 2,\ 3x + 3$ are in G.P., then the 4th term is:
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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 $= -4 \times \left(\dfrac{3}{2}\right)^3 = -13.5$
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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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 $t_{11}=t_1+20=-\dfrac{630}{31}+20=-\dfrac{10}{31}$.
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