\displaystyle \textbf{Question 1. }\text{Write the common difference of an A.P. whose } \\ n\text{th term is }xn+y.
\displaystyle \text{Answer:}
\displaystyle a_n=xn+y
\displaystyle a_{n+1}=x(n+1)+y=xn+x+y
\displaystyle \therefore \text{Common difference}=a_{n+1}-a_n
\displaystyle =xn+x+y-(xn+y)=x
\\

\displaystyle \textbf{Question 2. }\text{Write the common difference of an A.P. the sum of whose first } \\ n\text{ terms is }\frac{P}{2}n^2+Qn.
\displaystyle \text{Answer:}
\displaystyle S_n=\frac{P}{2}n^2+Qn
\displaystyle a_n=S_n-S_{n-1}
\displaystyle =\left(\frac{P}{2}n^2+Qn\right)-\left(\frac{P}{2}(n-1)^2+Q(n-1)\right)
\displaystyle =Pn+Q-\frac{P}{2}
\displaystyle \therefore a_{n+1}=P(n+1)+Q-\frac{P}{2}
\displaystyle \therefore \text{Common difference}=a_{n+1}-a_n=P
\\

\displaystyle \textbf{Question 3. }\text{If the sum of }n\text{ terms of an A.P. is }2n^2+3n,\text{ then write its }n\text{th term.}
\displaystyle \text{Answer:}
\displaystyle S_n=2n^2+3n
\displaystyle S_{n-1}=2(n-1)^2+3(n-1)
\displaystyle =2n^2-4n+2+3n-3
\displaystyle =2n^2-n-1
\displaystyle a_n=S_n-S_{n-1}
\displaystyle =(2n^2+3n)-(2n^2-n-1)
\displaystyle =4n+1
\\

\displaystyle \textbf{Question 4. }\text{If }\log 2,\ \log(2^x-1)\text{ and }\log(2^x+3)\text{ are in A.P., write the value of }x.
\displaystyle \text{Answer:}
\displaystyle \text{Since the given terms are in A.P.,}
\displaystyle 2\log(2^x-1)=\log2+\log(2^x+3)
\displaystyle \log(2^x-1)^2=\log\{2(2^x+3)\}
\displaystyle (2^x-1)^2=2(2^x+3)
\displaystyle 2^{2x}-2\cdot2^x+1=2\cdot2^x+6
\displaystyle 2^{2x}-4\cdot2^x-5=0
\displaystyle \text{Let }2^x=y.
\displaystyle y^2-4y-5=0
\displaystyle (y-5)(y+1)=0
\displaystyle y=5
\displaystyle \therefore 2^x=5
\displaystyle x=\log_2 5
\\

\displaystyle \textbf{Question 5. }\text{If the sums of }n\text{ terms of two arithmetic progressions are in the ratio }
\displaystyle 2n+5:3n+4,  \text{then write the ratio of their }m\text{th terms.}
\displaystyle \text{Answer:}
\displaystyle \frac{S_n}{S'_n}=\frac{2n+5}{3n+4}
\displaystyle \frac{a_m}{a'_m}=\frac{S_{2m-1}}{S'_{2m-1}}
\displaystyle =\frac{2(2m-1)+5}{3(2m-1)+4}
\displaystyle =\frac{4m+3}{6m+1}
\\

\displaystyle \textbf{Question 6. }\text{Write the sum of first }n\text{ odd natural numbers.}
\displaystyle \text{Answer:}
\displaystyle 1+3+5+\cdots+(2n-1)=n^2
\\

\displaystyle \textbf{Question 7. }\text{Write the sum of first }n\text{ even natural numbers.}
\displaystyle \text{Answer:}
\displaystyle 2+4+6+\cdots+2n=n(n+1)
\\

\displaystyle \textbf{Question 8. }\text{Write the value of }n\text{ for which }n\text{th terms of the A.P.s }3,10,17,\ldots
\displaystyle \text{and }63,65,67,\ldots\text{ are equal.}
\displaystyle \text{Answer:}
\displaystyle \text{For }3,10,17,\ldots,\quad a_n=3+(n-1)7=7n-4
\displaystyle \text{For }63,65,67,\ldots,\quad a_n=63+(n-1)2=2n+61
\displaystyle 7n-4=2n+61
\displaystyle 5n=65
\displaystyle n=13
\\

\displaystyle \textbf{Question 9. }\text{If }\frac{3+5+7+\cdots\text{ upto }n\text{ terms}}{5+8+11+\cdots\text{ upto }10\text{ terms}}=7,\text{ then find the value of }n.
\displaystyle \text{Answer:}
\displaystyle 3+5+7+\cdots\text{ upto }n\text{ terms}=\frac{n}{2}\{2\cdot3+(n-1)2\}
\displaystyle =\frac{n}{2}(2n+4)=n(n+2)
\displaystyle 5+8+11+\cdots\text{ upto }10\text{ terms}=\frac{10}{2}\{2\cdot5+9\cdot3\}
\displaystyle =5(10+27)=185
\displaystyle \frac{n(n+2)}{185}=7
\displaystyle n(n+2)=1295
\displaystyle n^2+2n-1295=0
\displaystyle (n-35)(n+37)=0
\displaystyle \therefore n=35
\\

\displaystyle \textbf{Question 10. }\text{If }m\text{th term of an A.P. is }n\text{ and }n\text{th term is }m,\text{ then write its }p\text{th term.}
\displaystyle \text{Answer:}
\displaystyle a+(m-1)d=n
\displaystyle a+(n-1)d=m
\displaystyle \text{Subtracting, }(m-n)d=n-m
\displaystyle d=-1
\displaystyle a+(m-1)(-1)=n
\displaystyle a=n+m-1
\displaystyle \therefore T_p=a+(p-1)d
\displaystyle =m+n-1-(p-1)=m+n-p
\\

\displaystyle \textbf{Question 11. }\text{If the sums of }n\text{ terms of two A.P.'s are in the ratio }
\displaystyle (3n+2):(2n+3), \text{find the ratio of their }12\text{th terms.}
\displaystyle \text{Answer:}
\displaystyle \frac{S_n}{S'_n}=\frac{3n+2}{2n+3}
\displaystyle \frac{a_{12}}{a'_{12}}=\frac{S_{2\cdot12-1}}{S'_{2\cdot12-1}}=\frac{S_{23}}{S'_{23}}
\displaystyle =\frac{3(23)+2}{2(23)+3}
\displaystyle =\frac{71}{49}
\displaystyle \therefore \text{Required ratio}=71:49
\\

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