Class 10: Arithmetic Progressions

$\displaystyle \textbf{Question 1: }\text{The sequence }\frac{2a-6b}{3b},\frac{2a-3b}{3b},\frac{2a}{3b},$
$\displaystyle \frac{2a+3b}{3b},\frac{2a+6b}{3b},\ldots\text{ is an A.P. with common difference }\underline{\hspace{1cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle d=\frac{2a-3b}{3b}-\frac{2a-6b}{3b}=\frac{3b}{3b}=1$
$\displaystyle \therefore \boxed{1}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{If the sequence }\frac{a-3b}{b},\frac{3a-3b}{b},\frac{5a-3b}{b},$
$\displaystyle \frac{7a-3b}{b},\ldots\text{ is an A.P. with common difference }3,\text{ then }a/b=\underline{\hspace{1cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle d=\frac{3a-3b}{b}-\frac{a-3b}{b}=\frac{2a}{b}$
$\displaystyle \frac{2a}{b}=3\Rightarrow\frac{a}{b}=\frac{3}{2}$
$\displaystyle \therefore \boxed{\frac{3}{2}}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{The }11\text{th term of the A.P. }-5,-\frac{5}{2},0,\frac{5}{2},\ldots$
$\displaystyle \text{is }\underline{\hspace{1cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle a=-5,\quad d=\frac{5}{2}$
$\displaystyle a_{11}=a+10d=-5+10\left(\frac{5}{2}\right)=20$
$\displaystyle \therefore \boxed{20}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{The }21\text{st term of the A.P. whose first two terms are }-3$
$\displaystyle \text{and }4\text{ is }\underline{\hspace{1cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle a=-3,\quad d=4-(-3)=7$
$\displaystyle a_{21}=a+20d=-3+20(7)=137$
$\displaystyle \therefore \boxed{137}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{The famous Mathematician associated with finding the sum}$
$\displaystyle \text{of the first }100\text{ natural numbers was }\underline{\hspace{1cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle \therefore \boxed{\text{Carl Friedrich Gauss}}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{If }n-2,4n-1\text{ and }5n+2\text{ are in A.P., then }n=\underline{\hspace{1cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle 2(4n-1)=(n-2)+(5n+2)$
$\displaystyle 8n-2=6n\Rightarrow2n=2\Rightarrow n=1$
$\displaystyle \therefore \boxed{1}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{The sum of first }50\text{ odd natural numbers is }\underline{\hspace{1cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Sum of first }n\text{ odd natural numbers}=n^2$
$\displaystyle \therefore 50^2=\boxed{2500}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{The sum of first }n\text{ odd natural numbers is }\underline{\hspace{1cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Sum of first }n\text{ odd natural numbers}=n^2$
$\displaystyle \therefore \boxed{n^2}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{If the common difference of an A.P. is }5,\text{ then }a_{18}-a_{13}=\underline{\hspace{1cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle a_{18}-a_{13}=(18-13)d=5(5)=25$
$\displaystyle \therefore \boxed{25}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{If }a_1,a_2,a_3,\ldots,a_n,\ldots\text{ is an A.P. such that}$
$\displaystyle a_{18}-a_{14}=32,\text{ then its common difference is }\underline{\hspace{1cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle a_{18}-a_{14}=(18-14)d$
$\displaystyle 32=4d\Rightarrow d=8$
$\displaystyle \therefore \boxed{8}$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{If }5,a_2,a_3,\ldots,a_{20},145\text{ is an A.P., then}$
$\displaystyle a_2+a_{20}=\underline{\hspace{1cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle a_1+a_{21}=a_2+a_{20}$
$\displaystyle a_2+a_{20}=5+145=150$
$\displaystyle \therefore \boxed{150}$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{The value of the middle term of the A.P. }-11,-7,-3,\ldots,49,53$
$\displaystyle \text{is }\underline{\hspace{1cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle a=-11,\quad d=4,\quad l=53$
$\displaystyle 53=-11+(n-1)4\Rightarrow n=17$
$\displaystyle \text{Middle term}=a_9=-11+8(4)=21$
$\displaystyle \therefore \boxed{21}$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{If }9\text{th term of an A.P. is zero, then its }29\text{th and }19\text{th}$
$\displaystyle \text{terms are in the ratio }\underline{\hspace{1cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle a_9=a+8d=0$
$\displaystyle a_{29}=20d,\quad a_{19}=10d$
$\displaystyle \therefore \boxed{a_{29}:a_{19}=2:1}$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{If }a_n\text{ denotes the }n\text{th term of the A.P. }3,8,13,18,\ldots,$
$\displaystyle \text{then the value of }a_{30}-a_{20}\text{ is }\underline{\hspace{1cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle d=5$
$\displaystyle a_{30}-a_{20}=(30-20)d=10(5)=50$
$\displaystyle \therefore \boxed{50}$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{In an A.P. }a_1,a_2,a_3,\ldots,a_n,\ldots,\text{ if }a_1=1,$
$\displaystyle a_n=20\text{ and }S_n=399,\text{ then the value of }n\text{ is }\underline{\hspace{1cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle S_n=\frac{n}{2}(a_1+a_n)$
$\displaystyle 399=\frac{n}{2}(1+20)=\frac{21n}{2}$
$\displaystyle \therefore \boxed{n=38}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{If }7\text{ times the }7\text{th term of an A.P. is equal to }11$
$\displaystyle \text{times its }11\text{th term, then the value of its }18\text{th term is }\underline{\hspace{1cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle 7(a+6d)=11(a+10d)$
$\displaystyle 7a+42d=11a+110d\Rightarrow a+17d=0$
$\displaystyle \therefore \boxed{a_{18}=a+17d=0}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{Two arithmetic progressions have the same common difference.}$
$\displaystyle \text{Their first terms are }A\text{ and }B\text{ respectively. The difference between}$
$\displaystyle \text{their }n\text{th terms is }\underline{\hspace{1cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle a_n=A+(n-1)d,\quad b_n=B+(n-1)d$
$\displaystyle a_n-b_n=A-B$
$\displaystyle \therefore \text{The required difference is }\boxed{A-B}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{If the }n\text{th terms of two A.P.s }9,7,5,\ldots\text{ and}$
$\displaystyle 24,21,18,\ldots\text{ are the same, then the value of }n\text{ is }\underline{\hspace{1cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle 9+(n-1)(-2)=24+(n-1)(-3)$
$\displaystyle 11-2n=27-3n\Rightarrow n=16$
$\displaystyle \therefore \boxed{n=16}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{If }S_n=n(4n+1)\text{ is the sum of }n\text{ terms of an A.P.,}$
$\displaystyle \text{then its common difference is }\underline{\hspace{1cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle S_n=4n^2+n$
$\displaystyle a_n=S_n-S_{n-1}=8n-3$
$\displaystyle d=a_{n+1}-a_n=8$
$\displaystyle \therefore \boxed{d=8}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{If the ratio of the sums of first }n\text{ terms of two A.P.s is}$
$\displaystyle \frac{5n+13}{7n+27},\text{ then the ratio of their }4\text{th terms is }\underline{\hspace{1cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the ratio of }4\text{th terms, put }n=2(4)-1=7.$
$\displaystyle \frac{a_4}{b_4}=\frac{5(7)+13}{7(7)+27}=\frac{48}{76}=\frac{12}{19}$
$\displaystyle \therefore \boxed{a_4:b_4=12:19}.$
$\displaystyle \\$