Jamia Millia Islamia 2016 Previous Year Questions (PYQs) – Page 3 of 10
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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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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 $a=n+(m-1)=m+n-1$.
$t_{10}=a+9d=(m+n-1)+9(-1)=m+n-10$.
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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Solution
$T_n=S_n-S_{n-1}=(3n^2+5)-[3(n-1)^2+5]=6n-3$.
Set $6n-3=159 \Rightarrow 6n=162 \Rightarrow n=27$.
If the roots of $x^3-9x^2+23x-15=0$ are in A.P., then their common difference is …
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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 $d=\pm2$.
If $\log_{2}(5\cdot2^{x}+1),\ \log_{4}(2^{\,1-x}+1)$ and $1$ are in A.P., then $x$ equals …
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Solution
For A.P.: $2\log_{4}(2^{1-x}+1)=\log_{2}(5\cdot2^{x}+1)+1$.
Since $\log_{4}y=\tfrac12\log_{2}y$,
$\log_{2}(2^{1-x}+1)=\log_{2}\!\big(2(5\cdot2^{x}+1)\big)$.
Thus $2^{1-x}+1=10\cdot2^{x}+2$.
Let $t=2^{x}>0$. Then $\frac{2}{t}+1=10t+2 \Rightarrow 10t^{2}+t-2=0$.
$t=\frac{-1+9}{20}=\frac{2}{5}$ (positive root). Hence $2^{x}=\frac{2}{5}$,
so $x=\log_{2}\!\left(\frac{2}{5}\right)=1-\log_{2}5$.
If sum of $n$ terms of a series is $3n^2 + 4n$, then the series is …
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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.
If ${}^8C_r - {}^7C_3 = {}^7C_2$, then $r$ is equal to …
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Solution
We know ${}^8C_r = {}^7C_{r} + {}^7C_{r-1}$.
So, ${}^8C_r - {}^7C_3 = {}^7C_{r} + {}^7C_{r-1} - {}^7C_3 = {}^7C_2$.
Comparing, ${}^7C_{r-1} = {}^7C_3 \Rightarrow r - 1 = 3 \Rightarrow r = 4$.
The number of arrangements of the letters of the word BANANA in which two N’s do not appear adjacently is …
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Solution
Total letters = 6 (B, A, N, A, N, A).
Total arrangements = $\dfrac{6}{32} = 60$.
If two N’s are together, treat NN as one letter → letters = (NN, B, A, A, A).
Arrangements = $\dfrac{5}{3} = 20$.
Hence, required = $60 - 20 = 40$.
The number of numbers greater than 23000 that can be formed from the digits 1, 2, 3, 4, 5 is …
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Solution
We have 5 digits → total 5! = 120 numbers.
Numbers greater than 23000 must start with 3, 4, or 5.
For each case:
Start with 3,4,5 → remaining 4 digits can be arranged in 4! = 24 ways each.
Total = $3 \times 24 = 72$.
Additionally, numbers starting with 2 are not all smaller (only those starting 23,24,25). For 24 and 25 → also possible 24 numbers.
Total = $72 + 18 = 90$.
The coefficient of $x^8 y^{10}$ in $(x + y)^{18}$ is …
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Solution
General term = ${}^{18}C_r x^{18-r} y^r$.
We need $x^8 y^{10} \Rightarrow 18 - r = 8 \Rightarrow r = 10$.
Coefficient = ${}^{18}C_{10}$.
The coefficient of $x^4$ in expansion of $(1 + x + x^2 + x^3)^{11}$ is …
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Solution
We can write generating function $f(x) = (1 + x + x^2 + x^3)^{11}$.
Coefficient of $x^4 =$ number of ways to get power 4 as sum of 11 terms each $0,1,2,3$.
By multinomial expansion, coefficient of $x^4 = {}^{11}C_4 + 10{}^{11}C_3 + 6{}^{11}C_2 + {}^{11}C_1 = 990$.
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