Class 10: Arithmetic Progressions – Very Short Answers Type Questions

$\displaystyle \textbf{Question 1: }\text{Find the common difference of the Arithmetic progression}$
$\displaystyle \frac{1}{a},\frac{3-a}{3a},\frac{3-2a}{3a},\ldots\quad(a\ne0)\hfill\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle d=\frac{3-a}{3a}-\frac{1}{a}=\frac{3-a-3}{3a}=-\frac{1}{3}$
$\displaystyle \therefore d=-\frac{1}{3}.$
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$\displaystyle \textbf{Question 2: }\text{In an A.P., if the common difference }d=-4,\text{ and the}$
$\displaystyle \text{seventh term }a_7\text{ is }4,\text{ then find the first term.}\hfill\text{[CBSE 2018]}$
$\displaystyle \text{Answer:}$
$\displaystyle a_7=a+6d$
$\displaystyle 4=a+6(-4)\Rightarrow a=28$
$\displaystyle \therefore \text{The first term is }28.$
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$\displaystyle \textbf{Question 3: }\text{Write the }n\text{th term of the A.P. }$
$\displaystyle \frac{1}{m},\frac{1+m}{m},\frac{1+2m}{m},\ldots\hfill\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle a=\frac{1}{m},\quad d=\frac{1+m}{m}-\frac{1}{m}=1$
$\displaystyle a_n=a+(n-1)d=\frac{1}{m}+n-1$
$\displaystyle \therefore a_n=\frac{1+(n-1)m}{m}.$
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$\displaystyle \textbf{Question 4: }\text{For what value of }k\text{ will the consecutive terms }2k+1,$
$\displaystyle 3k+3\text{ and }5k-1\text{ form an A.P.?}\hfill\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle 2(3k+3)=(2k+1)+(5k-1)$
$\displaystyle 6k+6=7k\Rightarrow k=6$
$\displaystyle \therefore k=6.$
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$\displaystyle \textbf{Question 5: }\text{Find the }9\text{th term from the end of the A.P. }5,9,13,\ldots,185.$
$\displaystyle \hfill\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle l=185,\quad d=4$
$\displaystyle \text{9th term from the end}=l-(9-1)d$
$\displaystyle =185-8(4)=153$
$\displaystyle \therefore \text{The }9\text{th term from the end is }153.$
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$\displaystyle \textbf{Question 6: }\text{If }\frac{4}{5},a,2\text{ are three consecutive terms of an A.P.,}$
$\displaystyle \text{then find the value of }a.\hfill\text{[CBSE 2009]}$
$\displaystyle \text{Answer:}$
$\displaystyle 2a=\frac{4}{5}+2=\frac{14}{5}$
$\displaystyle \therefore a=\frac{7}{5}.$
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$\displaystyle \textbf{Question 7: }\text{For what value of }p\text{ are }2p+1,13,5p-3\text{ three}$
$\displaystyle \text{consecutive terms of an A.P.?}\hfill\text{[CBSE 2009]}$
$\displaystyle \text{Answer:}$
$\displaystyle 2(13)=(2p+1)+(5p-3)$
$\displaystyle 26=7p-2\Rightarrow 7p=28\Rightarrow p=4$
$\displaystyle \therefore p=4.$
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$\displaystyle \textbf{Question 8: }\text{The first term of an A.P. is }p\text{ and its common difference is }q.$
$\displaystyle \text{Find its }10\text{th term.}\hfill\text{[CBSE 2008]}$
$\displaystyle \text{Answer:}$
$\displaystyle a_{10}=p+(10-1)q$
$\displaystyle \therefore a_{10}=p+9q.$
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$\displaystyle \textbf{Question 9: }\text{Write the value of }x\text{ for which }2x,x+10\text{ and }3x+2$
$\displaystyle \text{are in A.P.}$
$\displaystyle \text{Answer:}$
$\displaystyle 2(x+10)=2x+(3x+2)$
$\displaystyle 2x+20=5x+2\Rightarrow3x=18\Rightarrow x=6$
$\displaystyle \therefore x=6.$
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$\displaystyle \textbf{Question 10: }\text{Write }5\text{th term from the end of the A.P. }3,5,7,9,\ldots,201.$
$\displaystyle \text{Answer:}$
$\displaystyle l=201,\quad d=2$
$\displaystyle \text{5th term from the end}=l-(5-1)d$
$\displaystyle =201-4(2)=193$
$\displaystyle \therefore \text{The }5\text{th term from the end is }193.$
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$\displaystyle \textbf{Question 11: }\text{Write the common difference of an A.P. whose }n\text{th term is }a_n=3n+7.$
$\displaystyle \text{Answer:}$
$\displaystyle a_{n+1}=3(n+1)+7=3n+10$
$\displaystyle d=a_{n+1}-a_n=(3n+10)-(3n+7)=3$
$\displaystyle \therefore d=3.$
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$\displaystyle \textbf{Question 12: }\text{Define an arithmetic progression.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{An arithmetic progression is a sequence in which the difference between}$
$\displaystyle \text{any term and its preceding term is constant.}$
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$\displaystyle \textbf{Question 13: }\text{Which term of the sequence }114,109,104,\ldots\text{ is the first}$
$\displaystyle \text{negative term?}$
$\displaystyle \text{Answer:}$
$\displaystyle a=114,\quad d=-5$
$\displaystyle a_n=114+(n-1)(-5)=119-5n$
$\displaystyle a_n<0\Rightarrow119-5n<0\Rightarrow n>\frac{119}{5}=23.8$
$\displaystyle \therefore \text{The }24\text{th term is the first negative term.}$
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$\displaystyle \textbf{Question 14: }\text{Write the value of }a_{30}-a_{10}\text{ for the A.P. }4,9,14,19,\ldots$
$\displaystyle \text{Answer:}$
$\displaystyle a_n-a_k=(n-k)d$
$\displaystyle a_{30}-a_{10}=(30-10)(5)=100$
$\displaystyle \therefore a_{30}-a_{10}=100.$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{Write the }n\text{th term of an A.P. the sum of whose }n\text{ terms is }S_n.$
$\displaystyle \text{Answer:}$
$\displaystyle a_n=S_n-S_{n-1}$
$\displaystyle \therefore a_n=S_n-S_{n-1}.$
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$\displaystyle \textbf{Question 16: }\text{Write the sum of first }n\text{ odd natural numbers.}$
$\displaystyle \text{Answer:}$
$\displaystyle 1+3+5+\ldots+(2n-1)=n^2$
$\displaystyle \therefore \text{The required sum is }n^2.$
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$\displaystyle \textbf{Question 17: }\text{Write the sum of first }n\text{ even natural numbers.}$
$\displaystyle \text{Answer:}$
$\displaystyle 2+4+6+\ldots+2n=n(n+1)$
$\displaystyle \therefore \text{The required sum is }n(n+1).$
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$\displaystyle \textbf{Question 18: }\text{If the sum of }n\text{ terms of an A.P. is }S_n=3n^2+5n.$
$\displaystyle \text{Write its common difference.}$
$\displaystyle \text{Answer:}$
$\displaystyle a_n=S_n-S_{n-1}$
$\displaystyle =(3n^2+5n)-[3(n-1)^2+5(n-1)]=6n+2$
$\displaystyle d=a_{n+1}-a_n=6$
$\displaystyle \therefore d=6.$
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$\displaystyle \textbf{Question 19: }\text{Write the expression for the common difference of an A.P.}$
$\displaystyle \text{whose first term is }a\text{ and }n\text{th term is }b.$
$\displaystyle \text{Answer:}$
$\displaystyle b=a+(n-1)d$
$\displaystyle \therefore d=\frac{b-a}{n-1}.$
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$\displaystyle \textbf{Question 20: }\text{If the sum of first }p\text{ terms of an A.P. is }ap^2+bp,$
$\displaystyle \text{find its common difference.}\hfill\text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle S_p=ap^2+bp$
$\displaystyle a_p=S_p-S_{p-1}=2ap-a+b$
$\displaystyle d=a_{p+1}-a_p=2a$
$\displaystyle \therefore d=2a.$
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$\displaystyle \textbf{Question 21: }\text{Find the sum of the first }10\text{ multiples of }3.$
$\displaystyle \hfill\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle S_{10}=3+6+9+\ldots+30$
$\displaystyle =\frac{10}{2}(3+30)=5(33)=165$
$\displaystyle \therefore \text{The required sum is }165.$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{How many two digit numbers are divisible by }3?$
$\displaystyle \hfill\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle 12,15,18,\ldots,99\text{ form an A.P. with }a=12,\quad d=3.$
$\displaystyle 99=12+(n-1)3$
$\displaystyle 3(n-1)=87\Rightarrow n=30$
$\displaystyle \therefore \text{There are }30\text{ two digit numbers divisible by }3.$
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$\displaystyle \textbf{Question 23: }\text{In an A.P., if the sum of third and seventh term is zero,}$
$\displaystyle \text{find its }5\text{th term.}\hfill\text{[CBSE 2021]}$
$\displaystyle \text{Answer:}$
$\displaystyle a_3+a_7=0$
$\displaystyle (a+2d)+(a+6d)=0$
$\displaystyle 2a+8d=0\Rightarrow a+4d=0$
$\displaystyle \therefore a_5=0.$
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$\displaystyle \textbf{Question 24: }\text{Determine the A.P. whose third term is }5\text{ and seventh}$
$\displaystyle \text{term is }9.\hfill\text{[CBSE 2022]}$
$\displaystyle \text{Answer:}$
$\displaystyle a+2d=5\qquad\ldots\text{(i)}$
$\displaystyle a+6d=9\qquad\ldots\text{(ii)}$
$\displaystyle \text{Subtracting (i) from (ii), }4d=4\Rightarrow d=1$
$\displaystyle \text{From (i), }a+2=5\Rightarrow a=3$
$\displaystyle \therefore \text{The A.P. is }3,4,5,6,7,\ldots$
$\displaystyle \\$