Question 11 Mark
The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is given by $\frac{\text{l}^2-\text{a}^2}{\text{k}-(\text{l}+\text{a})},$ then $\text{k}=$
- S
- 2S
- 3S
- None of these
Answer
- 2S
Solution:
Given:
$\text{S}=\frac{\text{n}}{2}(\text{l}+\text{a})$
$\Rightarrow(\text{l}+\text{a})=\frac{2\text{S}}{\text{n}}$
Also, $\text{d}=\frac{\text{l}^2-\text{a}^2}{\text{k}(\text{l}+\text{a})}$
$\Rightarrow\text{d}=\frac{(\text{l}+\text{a})(\text{l}-\text{a})}{\text{k}-(\text{l}+\text{a})}$
$\Rightarrow\text{d}=\frac{[(\text{n}-1)\text{d}]\times\frac{2\text{S}}{\text{n}}}{\text{k}-\frac{2\text{S}}{\text{n}}}$
$\Rightarrow\text{k}-\frac{2\text{S}}{\text{n}}=(\text{n}-1)\frac{2\text{S}}{\text{n}}$
$\Rightarrow\text{k}=\frac{2\text{S}}{\text{n}}(\text{n}-1+1)$
$\Rightarrow\text{k}=2\text{S}$ View full question & answer→Question 21 Mark
If the sum of p terms of an A.P. is q and the sum of q terms is p, then the sum of p + q terms will be:
- 0
- p - q
- p + q
- -(p + q)
Answer
- -(p + q)
Solution:
$\text{S}_\text{p}=\text{q}$
$\Rightarrow\frac{\text{p}}{2}\big\{2\text{a}+(\text{p}-1)\text{d}\big\}=\text{q}$
$2\text{ap}+(\text{p}-1)\text{pd}=2\text{q}\ .....(1)$
$\text{S}_\text{q}=\text{p}$
$\Rightarrow\frac{\text{q}}{2}\big\{2\text{a}+(\text{q}-1)\text{d}\big\}=\text{p}$
$\Rightarrow2\text{aq}+(\text{q}-1)\text{pd}=2\text{p}\ .....(2)$
Multiplying equation (1) by q and equation (2) by p and then solving, we get:
$\text{d}=\frac{-2(\text{p}-\text{q})}{\text{pq}}$
Now, $\text{S}_{\text{p}+\text{q}}=\frac{(\text{p}+\text{q})}{2}[2\text{a}+(\text{p}+\text{q}-1)\text{d}]$
$=\frac{\text{p}}{\text{2}}[2\text{a}+(\text{p}-1)\text{d}+\text{pd}]+\frac{\text{q}}{2}[2\text{a}+(\text{q}-1)\text{d}+\text{pd}]$
$=\text{S}_\text{p}+\frac{\text{pqd}}{2}+\text{S}_\text{q}+\frac{\text{pqd}}{2}$
$=\text{p}+\text{q}+\text{pqd}$
$=\text{p}+\text{q}-\frac{2(\text{p}+\text{q})\text{pq}}{\text{pq}}$
$=-(\text{p}+\text{q})$ View full question & answer→Question 31 Mark
In the arithmetic progression whose common difference is non-zero, the sum of first 3n terms is equal to the sum of next n terms. Then the ratio of the sum of the first 2n terms to the next 2n terms is:
- $\frac{1}{5}$
- $\frac{2}{3}$
- $\frac{3}{4}$
- None of these
Answer
- $\frac{1}{5}$
Solution:
$\text{S}_{3\text{n}}=\text{S}_{4\text{n}}-\text{S}_{3\text{n}}$
$\Rightarrow2\text{S}_{3\text{n}}=\text{S}_{4\text{n}}$
$\Rightarrow2\times\frac{3\text{n}}{2}\{2\text{a}+(3\text{n}-1)\text{d}\}=\frac{4\text{n}}{2}\{2\text{a}+(4\text{n}-1)\text{d}\}$
$\Rightarrow3\{2\text{a}+(3\text{n}-1)\text{d}\}=2\{2\text{a}+(4\text{n}-1)\text{d}\}$
$\Rightarrow6\text{a}+9\text{nd}-3\text{d}=4\text{a}+8\text{nd}-2\text{d}$
$\Rightarrow2\text{a}+\text{nd}-\text{d}=0$
$\Rightarrow2\text{a}+(\text{n}-1)\text{d}=0\ .....(1)$
Required ratio: $\frac{\text{S}_{2\text{n}}}{\text{S}_{4\text{n}}-\text{S}_{2\text{n}}}$
$\frac{\text{S}_{2\text{n}}}{\text{S}_{4\text{n}}-\text{S}_{2\text{n}}}=\frac{\frac{2n}{2}\{2a+(2\text{n}-1)\text{d}\}}{\frac{\text{n}4}{2}\{2\text{a}+(4\text{n}-1)\text{d}\}-\frac{2\text{n}}{2}\{2\text{a}+(2\text{n}-1)\}\text{d}}$
$=\frac{\text{n}(\text{nd})}{2\text{n}(3\text{nd})-\text{n}(\text{nd})}$
$=\frac{1}{6-1}$
$=\frac{1}{5}$ View full question & answer→Question 41 Mark
Write the sum of first n even natural numbers.
AnswerWe have,
2 + 4 + 6 + ... up to n terms
$\therefore\text{S}_\text{n}=\frac{\text{n}}{2}[2\times2+(\text{n}-1)\times2]$
$=\frac{\text{n}}{2}[4+2\text{n}-2]$
$=\frac{\text{n}}{2}[2\text{n}+2]$
$=\frac{\text{n}}{2}\times2(\text{n}+1)$
$\text{n}(\text{n}+1)$
$\therefore2+4+6+\ ...$ up to n terms = n (n + 1)
Hence, the sum of first n even natural numbers is n (n + 1)
View full question & answer→MCQ 51 Mark
If n arithmetic means are inserted between $1$ and $31$ such that the ratio of the first mean and $n^{th}$ mean is $3 : 29$, then the value of $n$ is:
AnswerThe given series is $1, . . . . . . . . . . . , 31$
There are $n\ A.M.s$ between $1$ and $31: 1, A_1, A_2, A_3, . . . . ., A_n,$
Common difference, $\text{d}=\frac{31-1}{\text{n}+1}=\frac{30}{\text{n}+1}$
Here, we have:
$\frac{\text{A}_1}{\text{A}_\text{n}}=\frac{3}{29}$
$\Rightarrow\frac{1+\text{d}}{1+\text{nd}}=\frac{3}{29}$
$\Rightarrow\frac{1+\frac{30}{\text{n}+1}}{1+\text{n}\times\frac{30}{\text{n}+1}}=\frac{3}{30}$
$\Rightarrow\frac{\text{n}+1+30}{\text{n}+1+30\text{n}}=\frac{3}{29}$
$\Rightarrow\frac{\text{n}+31}{31\text{n}+1}=\frac{3}{29}$
$\Rightarrow29\text{n}+899=93\text{n}+3$
$\Rightarrow64\text{n}=896$
$\Rightarrow\text{n}=14$
View full question & answer→MCQ 61 Mark
If the sum of $n$ terms of an $A.P.,$ is $3n^2 + 5n$ then which of its terms is $164$?
- A
$26^{th}$
- ✓
$27^{th}$
- C
$28^{th}$
- D
AnswerCorrect option: B. $27^{th}$
$\text{S}_\text{n}=3\text{n}^2+5\text{n}$
$\text{S}_1=3(1)^2+5(1)=8$
$\therefore\text{a}_1=8$
$\text{S}_2=3(2)^2+5(2)=22$
$\therefore\text{a}_1+\text{a}_2=22$
$\Rightarrow\text{a}_2=14$
Common difference, $\text{d}=14 -8 =6$
Also, $\text{a}_\text{n}=164$
$\Rightarrow\text{a}+(\text{n}-1)\text{d}=164$
$\Rightarrow8+(\text{n}-1)6=164$
$\Rightarrow\text{n}=27$
View full question & answer→MCQ 71 Mark
If, $S_1$ is the sum of an arithmetic progression of $'n\ '$ odd number of terms and $S_2$ the sum of the terms of the series in odd places, then $\frac{\text{S}_1}{\text{S}_2}=$
- ✓
$\frac{2\text{n}}{\text{n}+1}$
- B
$\frac{\text{n}}{\text{n}+1}$
- C
$\frac{\text{n}+1}{2\text{n}}$
- D
$\frac{\text{n}+1}{\text{n}}$
AnswerCorrect option: A. $\frac{2\text{n}}{\text{n}+1}$
Let $n$ be an odd number.
Given:
$S_1 =$ Sum of odd number of terms
$=\frac{\text{n}}{2}\{2\text{a}+(\text{n}-1)\text{d}\}\ .....(1)$
Since $n$ is odd, the number of odd places $=\frac{\text{n}+1}{2}$
$S_2 =$ Sum of the terms of a series in odd places
$=\frac{\Big(\frac{\text{n}+1}{2}\Big)}{2}\Big\{2\text{a}\Big(\frac{\text{n}+1}{2}-1\Big)2\text{d}\Big\}$
From equations $(1)$ and $(2)$ we have:
$\frac{\text{S}_1}{\text{S}_2}=\frac{\frac{\text{n}}{2}\{2\text{a}+(\text{n}-1)\text{d}\}}{\frac{\text{n}+1}{4}\{2\text{a}+(\text{n}-1)\text{d}\}}$
$=\frac{2\text{n}}{\text{n}+1}$
View full question & answer→Question 81 Mark
If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers, then k =
- $\frac{1}{\text{n}}$
- $\frac{\text{n}-1}{\text{n}}$
- $\frac{\text{n}+1}{2\text{n}}$
- $\frac{\text{n}+1}{\text{n}}$
Answer
- $\frac{\text{n}+1}{\text{n}}$
Solution:
Given:
Sum of the even natural numbers = k × Sum of the odd natural numbers
$\frac{\text{n}}{2}\{2\text{a}+(\text{n}-1)\text{d}\}=\text{k}\times\frac{\text{n}}{2}\{2\text{a}+(\text{n}-1)\text{d}\}$
$\Rightarrow\{2\times2+(\text{n}-1)2\}=\text{k}\times\{2\times1+(\text{n}-1)2\}$
$\Rightarrow\frac{4+(\text{n}-1)2}{2+(\text{n}-1)2}=\text{k}$
$\Rightarrow\frac{\text{n}+1}{\text{n}}=\text{k}$ View full question & answer→Question 91 Mark
If the sums of n terms of two arithmetic progressions are in the ratio $2n + 5 : 3n + 4$, then write the ratio of their mth terms.
AnswerIt is given that
$\frac{\text{S}_\text{n}}{\text{s}_\text{n}}=\frac{2\text{n}+5}{3\text{n}+4}$
Let, $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}_1+(\text{n}-1)\text{d}_1]$
and, $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}_2+(\text{n}-1)]\text{d}_2$
$\therefore\frac{\text{S}_\text{n}}{\text{S}_\text{n}}=\frac{\frac{\text{n}}{2}[2\text{a}_1+(\text{n}-1)\text{d}_2]}{\frac{\text{n}}{2}[2\text{a}_\text{n}+(\text{n}-1)\text{d}_2]}$
$\Rightarrow\frac{2\text{a}_1+(\text{n}-1)\text{d}_1}{2\text{a}_2+(\text{n}-1)\text{d}_2}=\frac{2\text{n}+5}{3\text{n}+4}$
$\Rightarrow\frac{\text{a}_1\Big(\frac{\text{n}-1}{2}\text{}\Big)\text{d}_1}{\text{a}_2+\Big(\frac{\text{n}-1}{2}\Big)\text{d} _2}=\frac{2\text{n}+5}{3\text{n}+4}$
Let, $\frac{\text{n}-1}{2}=\text{m}$
$\Rightarrow\text{n}-1=2\text{m}$
$\Rightarrow\text{n}=2\text{m}-1$
Replacing $\frac{\text{n}-1}{2}$ by m on both side of equation (2), we get
$\frac{\text{a}_1+\text{md}_1}{\text{a}_2\text{md}_2}=\frac{2(2\text{m}-1)+5}{3(2\text{m}-1)+4}$
$=\frac{4\text{m}-2+5}{6\text{m}-3+4}$
$=\frac{4\text{m}+3}{6\text{m}+1}$
$\therefore$ Ratio of $m^{th}$ terms $=\frac{4\text{m}+3}{6\text{m}+1}$
View full question & answer→MCQ 101 Mark
If in an $A.P., S_n = n^2p$ and $S_m = m^2p$, where $S_r$ denotes the sum of $r$ terms of the $A.P.$, then $S_p$ is equal to:
- A
$\frac{1}{2}\text{p}^3$
- B
$mn p$
- ✓
$P^3$
- D
$(m + n)P^2$
AnswerGiven: $\text{S}_\text{n}=\text{n}^2\text{p}$
$\Rightarrow\frac{\text{n}}{2}\{2\text{a}+(\text{n}-1)\text{d}\}=\text{n}^2\text{p}$
$\Rightarrow2\text{a}+(\text{n}-1)\text{d}=2\text{np}$
$\Rightarrow2\text{a}+(2\text{np}-(\text{n}-\text{1}\text{d}))\ .....(1)$
$\text{S}_\text{m}=\text{m}^2\text{p}$
$\Rightarrow\frac{\text{m}}{2}\{2\text{a}+(\text{m}-1)\text{d}\}=\text{m}^2\text{p}$
$\Rightarrow2\text{a}+(\text{m}-1)\text{d}=2\text{mp}$
$\Rightarrow2\text{a}=2\text{mp}-(\text{m}-1)\text{d}\ .....(2)$
From $(1)$ and $(2)$
$2\text{np}-(\text{n}-1)\text{d}=2\text{mp}-(\text{m}-1)\text{d}$
$\Rightarrow2\text{p}(\text{n}-\text{m})=\text{d}(\text{n}-1-\text{m}+1)$
$\Rightarrow2\text{p}=\text{d}$
Substituting $\text{d}=2\text{p}$ in equation $(1),$ we get:
$\text{a}=\text{p}$
Sum of $p$ terms of the $A.P.$ is given by:
$\frac{\text{p}}{2}\{2\text{a}+(\text{p}-1)\text{d}\}$
$=\frac{\text{p}}{2}\{2\text{p}+(\text{p}-1)2\text{p}\}$
$=\text{p}^3$
View full question & answer→Question 111 Mark
If the sums of n terms of two AP.'s are in the ratio (3n + 2) : (2n + 3), then find the ratio of their 12th terms.
AnswerLet the first terms of the two A.P.'s be a and a'; and their common difference be d and d'.
Now,
$\frac{\text{s}_\text{n}}{\text{s}_\text{n}}=\frac{(3\text{n}+2)}{(2\text{n}+3)}$
$\Rightarrow\frac{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]}{\frac{\text{n}}{2}[2\text{a}'+(\text{n}-1)\text{d}]}=\frac{(3\text{n}+2)}{(2\text{n}+3)}$
$\Rightarrow\frac{[2\text{a}+(\text{n}-1)]\text{d}}{[2\text{a}'+(\text{n}-1)\text{d}']}=\frac{(3\text{n}+2)}{(2\text{n}+3)}$
let n = 23
$\Rightarrow\frac{2\text{a}+(23-1)\text{d}}{2\text{a}'+(23-1)\text{d}'}=\frac{3\times23+2}{2\times23+3}$
$\Rightarrow\frac{2\text{a}+22\text{d}}{2\text{a}+22\text{d}}=\frac{69+2}{46+3}$
$\Rightarrow\frac{2(\text{a}+11\text{d)}}{2(\text{a}'11\text{d}')}=\frac{71}{49}$
$\therefore\frac{\text{a}_{12}}{\text{a}_{12}}=\frac{71}{49}$
So, the ratio of there 12th terms is 71 : 49.
View full question & answer→MCQ 121 Mark
If $7^{th}$ and $13^{th}$ terms of an $A.P$. be $34$ and $64$ respectively, then its $18^{th}$ term is:
Answer$a_7=34$
$\Rightarrow a+6 d=34 \ldots . . .(1)$
$\text { Also, } a_{13}=64$
$\Rightarrow a+12 d=64 \ldots . . .(2)$
Solving equations $(1)$ and $(2)$. we get:
$a=4 \text { and } d=5$
$\therefore a_{18}=a+17 d$
$=4+17(5)$
$=89$
View full question & answer→MCQ 131 Mark
Let $S_n$ denote the sum of $n$ terms of an $A.P.$ whose first term is a. If the common difference $d$ is given by $d=S_n-k$ $S_{n-1}+S_{n-2}$, then $k=$
AnswerLet the $A.P.$ be $\text{a},\ \text{a}+\text{d},\ \text{a}+\text{d},\ \text{a}+3\text{d}\ ...$
Given:
$\text{d}=\text{S}_\text{n}-\text{k}\text{S}_{\text{n}-1}+\text{S}_{\text{n}-2}$
For $\text{n}=3,$ we have
$\text{d}=(3\text{a}+3\text{d})-\text{k}(2\text{a}+\text{d})+\text{a}$
$\Rightarrow4\text{a}+2\text{d}=\text{k}(2\text{a}+\text{d})=0$
$\Rightarrow2(2\text{a}+\text{d})=\text{k}(2\text{a}+\text{d})$
$\Rightarrow2=\text{k}$
View full question & answer→Question 141 Mark
If $m^{th}$ term of an A.P. is n and $n^{th}$ term is m, then write its $p^{th}$ term.
AnswerWe have,
$a_m=a+(m-1) d=n$
$\Rightarrow a+(m-1) d=n$
$\text { and, } a_n=a+(n-1) d=m$
$\Rightarrow a+(n-1) d=m$
Subtracting equation (2) from equation (1), we get
$a+(m-1) d-a-(n-1) d=n-m$
$\Rightarrow(m-1) d-(n-1) d=n-m$
$\Rightarrow d(m-1-n+1)=n-m$
$\Rightarrow d(m-n)=n-m$
$\Rightarrow d=\frac{-(m-n)}{(m-n)}$
$\Rightarrow d=-1$
Putting $d=-1$ in equation (1), we get
$a+(m-1)(-1)=n$
$\Rightarrow a-m+1=n$
$\Rightarrow a=n+m-1$
Now,
$a_p=a+(p-1) d$
$=n+m-1+(p+1)(-1)$
$=n+m-1-p+1$
$=n+m-p$
$\therefore a_p=m+n-p$
Hence, $p ^{\text {th }}$ term is $m + n - p$.
View full question & answer→Question 151 Mark
Write the sum of first n odd natural numbers.
AnswerWe have,
$1 + 3 + 5 + ...$ up to n terms
$\therefore\text{S}_\text{n}=\frac{\text{n}}{2}[2\times1+(\text{n}-1)\times2]$
$=\frac{\text{n}}{2}[2+2\text{n}-2]$
$=\frac{\text{n}}{2}\times2\text{n}$
$=\text{n}^2$
$\therefore1+3+5+\ ...$ up to n terms $= n^2$
Hence, sum of first n odd natural numbers is $n^2$.
View full question & answer→Question 161 Mark
Write the value of n for which nth terms of the A.P.$s 3, 10, 17, ... $ and $63, 65, 67, ....$ are equal.
AnswerWe have,
$3 + 10 + 17 + ...$
$\Rightarrow a_n = 3 + (n - 1)2$
and, $63, 65, 67$
$\Rightarrow a_n =63 + (n - 1)2$
It is given that
$a_n = a_n$
$\Rightarrow 3 + (n - 1)7 = 63 + (n - 1)2$
$\Rightarrow 3 + 7n - 7 = 63 + 2n - 2$
$\Rightarrow 7n -2n = 61 + 4$
$\Rightarrow 5n = 65$
$\Rightarrow n = 13$
View full question & answer→Question 171 Mark
Write the common difference of an A.P. the sum of whose first n terms is $\frac{\text{p}}{2}\text{n}^2+\text{Qn}.$
AnswerWe have,
$\text{S}_\text{n}=\frac{\text{p}}{2}\text{n}^2+\text{Qn}$
$\therefore\text{S}_2=\frac{\text{p}}{2}(2)^2+\text{Qx}^2$
$\Rightarrow\text{S}_2=2\text{p}+2\text{Q}$
$\text{S}_3=\frac{\text{p}}{2}(3)^2+3\text{Q}$
$\Rightarrow\text{S}_3=\frac{\text{qp}}{2}+3\text{Q}$
and,
$\text{S}_4=\frac{\text{P}}{2}(4)^2+4\text{Q}$
$\Rightarrow\text{S}_4=8\text{P}+4\text{Q}$
Now,
$\text{T}_4=\text{S}_4-\text{S}_3=8\text{p}+4\text{Q}-\frac{\text{pq}}{2}-3\text{Q}$
$\Rightarrow\text{T}_4=\frac{7\text{P}}{2}+\text{Q}$
and,
$\text{T}_3=\text{S}_3-\text{S}_2=\frac{\text{pq}}{2}+3\text{Q}-2\text{P}-2\text{Q}$
$=\frac{5\text{P}}{2}+\text{Q}$
$\therefore\text{T}_4-\text{T}_3=\frac{7\text{p}}{2}+\text{Q}-\frac{5\text{p}}{2}-\text{Q}$
$=\frac{7\text{P}}{2}-\frac{5\text{P}}{2}$
$=\frac{2\text{P}}{2}$
$=\text{P}$
$\therefore\text{T}_4-\text{T}_3=\text{P}$
$\therefore$ Common difference = P.
View full question & answer→MCQ 181 Mark
The 10th common term between the $A.P.s\ 3, 7, 11, 15, ...$ and $1, 6, 11, 16, ...$ is:
AnswerAs, the common difference of the $A.P.\ 3,7,11,15, \ldots=7-3=4$ and the common difference of the $A.P.$ $1,6,11,16, \ldots$
$=6-1=5$ And, the common terms of both the $\text{A.P.S}$ will be in $\text{A.P.}$
So, the common difference of the $\text{A.P.}$ of the common terms,
$d=\operatorname{LCM}(4,5)=4 \times 5=$ its first common term, $a=11$
Now, the tenth common term, $a_{10}=a+(10-1) d$
$=11+9 \times 20$
$=11+180$
$=191$
Hence, the correct alternative is option $(a).$
View full question & answer→MCQ 191 Mark
If the sum of $n$ terms of an $A.P.$ is $2n^2 + 5 n$, then its $n^{th}$ term is
- A
$4n − 3$
- B
$3n − 4$
- ✓
$4n + 3$
- D
$3n + 4$
AnswerCorrect option: C. $4n + 3$
$\text{S}_\text{n}=2\text{n}^2+5\text{n}$
$\text{S}_1=2.1^2+5.1=8$
$\therefore\text{a}_1=7$
$\text{S}_2=2.2^2+5.2=18$
$\therefore\text{a}_1+\text{a}_2=18$
$\Rightarrow\text{a}_2=11$
Common difference, $\text{d}=11 -7 =4$
$\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$=7+(\text{n}-1)4$
$=4\text{n}+3$
View full question & answer→MCQ 201 Mark
In n A.M.'s are introduced between $3$ and $17$ such that the ratio of the last mean to the first mean is $3 : 1$, then the value of n is:
AnswerLet $A_1, A_2, A_3, A_4 .... A_n$ be the $n$ arithmetic means between $3$ and $17$.
Let d be the common difference of the $A.P. 3, A_1, A_2, A_3, A_4 .... A_n$ and $17$. Then, we have:
$\text{d}=\frac{17-3}{\text{n}+1}=\frac{14}{\text{n}+1}$
Now, $\text{A}_1=3+\text{d}=3+\frac{14}{\text{n}+1}=\frac{3\text{n}+17}{\text{n}+1}$
And, $\text{A}_\text{n}=3+\text{nd}=3+\text{n}\Big(\frac{14}{\text{n}+1}\Big)=\frac{17\text{n}+3}{\text{n}+1}$
$\therefore\frac{\text{A}_\text{n}}{\text{A}_1}=\frac{\text{3}}{1}$
$\Rightarrow\frac{\Big(\frac{17\text{n}+3}{\text{n}+1}\Big)}{\Big(\frac{3\text{n}+17}{\text{n}+1}\Big)}=\frac{3}{1}$
$\Rightarrow\frac{17\text{n}+3}{3\text{n}+17}=\frac{3}{1}$
$\Rightarrow17\text{n}+3=9\text{n}+51$
$\Rightarrow8\text{n}=48$
$\Rightarrow\text{n}=6$
View full question & answer→Question 211 Mark
If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times the least, then the numbers are:
- 5, 10, 15, 20
- 4, 10, 16, 22
- 3, 7, 11, 15
- None of these
Answer
- 5, 10, 15, 20
Solution:
Let the four numbers in A.P. be as follows:
$\text{a}-2\text{d},\ \text{a}-\text{d},\ \text{a},\text{a}+\text{d}$
Their sum = 50 (Given)
$\Rightarrow(\text{a}-2\text{d})+(\text{a}-\text{d})+\text{a}+(\text{a}+\text{d}=50$
$\Rightarrow2\text{a}-\text{d}=25\ .....(1)$
Also, $(\text{a}+\text{d})=4(\text{a}-2\text{d})$
$\Rightarrow\text{a}+\text{d}=4\text{a}-8\text{d}$
$\Rightarrow3\text{d}=\text{a}\ .....(2)$
From equations (1) and (2),
$\text{d}=5,\ \text{a}=15$
Hence, the numbers are 15 - 10, 15 - 5, 15, 15 + 5, i.e. 5, 10, 15, 20. View full question & answer→MCQ 221 Mark
If $a_1, a_2, a_3, .... a_n$ are in $A.P.$ with common difference d, then the sum of the series sin d,
$[\text{cosec}\ \text{a}_1\ \text{cosec}\ \text{a}_2\ \text{cosec}\ \text{a}_1\ \text{cosec}\ \text{a}_3\\+\ .....\ +\text{cosec}\ \text{a}_{\text{n}-1}\text{cosec}\ \text{a}_\text{n}]$ is:
- A
$\sec\text{a}_1-\sec\text{a}_\text{n}$
- B
$\text{cosec}\ \text{a}_1-\text{cosec}\ \text{a}_\text{n}$
- ✓
$\cot\text{a}_1-\cot\text{a}_\text{n}$
- D
$\tan\text{a}_1-\tan\text{a}_\text{n}$
AnswerCorrect option: C. $\cot\text{a}_1-\cot\text{a}_\text{n}$
We have,
$\sin\text{d}(\text{cosec}\ \text{a}_1\ \text{cosec}\ \text{a}_2+\ \text{coses}\ \text{a}_2\ \text{cosec}\ \text{a}_3\\+\ .....\ +\ \text{cosec}\ \text{a}_{\text{n}-1}\ \text{cosec}\ \text{a}_\text{n})$
$=\frac{\sin\text{d}}{\sin\text{a}_1\sin\text{a}_2}+\frac{\sin\text{d}}{\sin\text{a}_2\sin\text{a}_3}+\ .....\ +\frac{\sin\text{d}}{\sin\text{a}_{\text{n}-1}\sin\text{a}_\text{n}}$
$=\frac{\sin(\text{a}_2-\text{a}_1)}{\sin\text{a}_1\sin\text{a}_2}+\frac{\sin(\text{a}_3-\text{a}_2)}{\sin\text{a}_2\sin\text{a}_3}+\ .....\ +\frac{\sin(\text{a}_\text{n}-\text{a}_{\text{n}-1})}{\sin\text{a}_{\text{n}-1}\sin\text{a}_n}$
$=\frac{\sin\text{a}_2\cos\text{a}_1-\cos\text{a}_2\sin\text{a}_1}{\sin\text{a}_1\sin\text{a}_2}+\frac{\sin\text{a}3\cos\text{a}_2-\cos\text{a}_3\sin\text{a}_2}{\sin\text{a}_1\sin\text{a}_2}\\+\ .....\ +\frac{\sin\text{a}_2\cos\text{a}_1-\cos\text{a}_2\sin\text{a}_1}{\sin\text{a}_1\sin\text{a}_2}$
$=(\cot\text{a}_1-\cot\text{a}_2)+(\cot\text{a}_2-\cot\text{a}_3)\\+\ .....\ +(\cot\text{a}_{\text{n}-1}-\cot\text{a}_\text{n})$
$=\cot\text{a}_1-\cot\text{a}_\text{n}$
View full question & answer→Question 231 Mark
If in an A.P., the pth term is q and (p + q)th term is zero, then the qth term is:
- -p
- p
- p + q
- p - q
Answer
- p
Solution:
As, $\text{a}_\text{p}=\text{q}$
$\Rightarrow(\text{p}-1)\text{d}=\text{q}\ .....(1)$
Also, $\text{a}_{(\text{p}+\text{q})}=0$
$\Rightarrow\text{a}+(\text{p}+\text{q}-1)\text{d}=0\ .....(2)$
Subtracting (1) from (2), we get
$\text{a}+(\text{p}+\text{q}-1)\text{d}-\text{a}-(\text{p}-1)\text{d}=0-\text{q}$
$\Rightarrow(\text{p}+\text{q}-1)\text{d}-\text{a}-(\text{p}-1)\text{d}=0-\text{q}$
$\Rightarrow\text{pd}=-\text{q}$
$\Rightarrow\text{d}=\frac{-\text{q}}{\text{q}}$
$\Rightarrow\text{d}=-1$
Substituting $\text{d}=-1$ in (1), we get
$\text{a}+(\text{p}-1)\times(-1)=\text{q}$
$\Rightarrow\text{a}-\text{p}+\text{1}=\text{q}$
$\Rightarrow\text{a}=\text{p}+\text{q}-1$
Now,
$\text{a}_\text{q}=\text{a}+(\text{q}-1)\text{d}$
$=\text{p}+\text{q}-1+(\text{q}-1)\times(-1)$
$=\text{p}+\text{q}-1-\text{q}+1$
$=\text{p}$
Hence, the correct alternative is option (b). View full question & answer→MCQ 241 Mark
Let $S_n$ denote the sum of first $n$ terms of an $A.P.$ If $S_{2n} = 3 S_n$, then $S_{3n} : S_n$ is equal to:
AnswerAs, $\text{S}_{2\text{n}}=3\text{S}_\text{n}$
$\Rightarrow\frac{2\text{a}}{2}[2\text{a}+(2\text{n}-1)\text{d}]=\frac{3\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow2[2\text{a}+(2\text{n}-1)\text{d}]=3[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow4\text{a}+2(2\text{n}-1)\text{d}=6\text{a}+3(\text{n}-1)\text{d}$
$\Rightarrow4\text{a}+\text{nd}4-2\text{d}=6\text{a}+3\text{nd}-3\text{d}$
$\Rightarrow6\text{a}-4\text{a}+3\text{nd}-\text{3d}-4\text{nd}+2\text{d}=0$
$\Rightarrow2\text{a}-\text{nd}-\text{d}=0$
$\Rightarrow2\text{a}-(\text{n}+1)\text{d}=0$
$\Rightarrow2\text{a}=(\text{n}+1)\text{d}\ .....{(1)}$
Now,
$\frac{\text{S}_{3\text{n}}}{\text{S}_\text{n}}=\frac{\frac{3\text{n}}{2}[2\text{a}+(3\text{n}-1)\text{d}]}{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]}$
$=\frac{3[ (\text{n}+1)\text{d}+(3\text{n}-1)\text{d}]}{[(\text{n}+1)+(\text{n}-1)\text{d}]} [$Using $(1)]$
$=\frac{3[\text{nd}+\text{d}+3\text{nd}-\text{d}]}{[\text{nd}+\text{d}+\text{nd}-\text{d}]}$
$=\frac{3[4\text{nd}]}{[2\text{nd}]}$
$=3\times2$
$=6$
View full question & answer→MCQ 251 Mark
If the sum of $n$ terms of an $A.P.$ be $3n^2 − n$ and its common difference is $6$, then its first term is:
Answer$\text{S}_\text{n}=3\text{n}^2-\text{n}$
$\Rightarrow\text{S}_1=3(1)^2-1$
$\Rightarrow\text{S}_1=2$
$\therefore\text{a}_1=2$
View full question & answer→Question 261 Mark
Write the common difference of an A.P. whose nth term is $xn + y.$
AnswerWe have,
$a_n = xn + y$
Putting $n =1$, we get
$a_1 = x + y$
Putting $n = 2$, get
$a_2 = 2x + y$
$\therefore$ $a_2 - a_1 = 2x + y - x - y$
$= x$
$\Rightarrow a_2 - a_1 = x$
$\therefore$ Common difference $= x$
View full question & answer→MCQ 271 Mark
If $S_n$ denotes the sum of first $n$ terms of an $A.P. < a_n >$ such that $\frac{\text{S}_\text{m}}{\text{S}_\text{n}}=\frac{\text{m}^2}{\text{n}^2},$ then $\frac{\text{a}_\text{m}}{\text{a}_\text{n}}=$
- A
$\frac{2\text{m}+1}{2\text{n}+1}$
- ✓
$\frac{1\text{m}-1}{2\text{n}-1}$
- C
$\frac{\text{m}-1}{\text{n}-1}$
- D
$\frac{\text{m}+1}{\text{n}-1}$
AnswerCorrect option: B. $\frac{1\text{m}-1}{2\text{n}-1}$
$\frac{\text{S}_\text{m}}{\text{S}_\text{n}}=\frac{\text{m}^2}{\text{n}^2}$
$\Rightarrow\frac{\frac{\text{m}}{2}\{2\text{a}+(\text{m}-1\text{d})\}}{\frac{\text{n}}{2}\{2\text{a}+(\text{n}-1)\text{d}\}}=\frac{\text{m}^2}{\text{n}^2}$
$\Rightarrow\frac{\{2\text{a}+(\text{m}-1)\text{d}\}}{\{2\text{a}+(\text{n}-1)\text{d}\}}=\frac{\text{m}}{\text{n}}$
$\Rightarrow2\text{an}+\text{ndm}-\text{nd}=2\text{nmd}-\text{md}$
$\Rightarrow2\text{an}-2\text{am}-\text{nd}+\text{md}=0$
$\Rightarrow2\text{a}(\text{n}-\text{m})-\text{d}(\text{n}-\text{m})=0$
$\Rightarrow2\text{a}(\text{n}-\text{m})=\text{d}(\text{n}-\text{m})$
$\Rightarrow\text{d}=2\text{a}\ .....(1)$
Ratio of $\frac{\text{a}_m}{\text{a}_n}=\frac{\text{as}+(\text{m}-1)\text{d}}{\text{a}+(\text{n}-\text{1})\text{d}}$
$\Rightarrow\frac{\text{a}_\text{m}}{\text{a}_n}=\frac{\text{a}+(\text{m}-)2\text{a}}{\text{a}+(\text{n}-)2\text{a}}$ From $(1)$
$=\frac{\text{a}+2\text{am}-2\text{a}}{\text{a}+2\text{an}-2\text{a}}$
$=\frac{2\text{am}-\text{a}}{2\text{an}-\text{a}}$
$=\frac{2\text{m}-1}{2\text{n}-1}$
View full question & answer→Question 281 Mark
Find the sum of the following arithmetic progression:
50, 46, 42, ... to 10 terms.
Answer50, 46, 42, ... to 10 terms
$\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\text{s}=\frac{10}{2}[2\times50+(10-1)(-4)]$
$=320$
View full question & answer→Question 291 Mark
The first and last terms of an A.P. are 1 and 11. If the sum of its terms is 36, then the number of terms will be:
- 5
- 6
- 7
- 8
Answer
- 6
Solution:
$\text{a}=1,\ \text{a}_\text{n}=11,\ \text{S}_\text{s}=36$
$\because\text{a}_\text{n}=11$
$\Rightarrow\text{a}+(\text{n}-1)\text{d}=11$
$\Rightarrow1+(\text{n}-1)\text{d}=11$
$\Rightarrow(\text{n}-1)\text{d}=10\ .....(1)$
Also, $\text{S}_\text{n}=36$
$\Rightarrow\frac{\text{n}}{2}\{2\text{a}+(\text{n}-1)\text{d}\}=36$
$\Rightarrow\frac{\text{n}}{2}\{2\text{a}+(\text{n}-1)\text{d}\}=36$
$\Rightarrow\text{n}\{2+10\}=72$ (Uising(1))
$\Rightarrow\text{n}=6$ View full question & answer→Question 301 Mark
If $\log 2, \log \left(2^x-1\right)$ and $\log \left(2^x+3\right)$ are in A.P., write the value of $x$.
AnswerSince $\log \left(2^x-1\right)$ and $\log \left(2^x+3\right)$ are in A.P, therefore
$\log \left(2^x-1\right)-\log 2=\log \left(2^x+3\right)-\log \left(2^x-1\right)$
$2 \log \left(2^x-1\right)=\log \left(2^x+3\right)+\log 2$
$\left(2^x-1\right)^2=2\left(2^x+3\right)$
$2^{2 x}-2 \times 2^x+1=2.2^x+6$
$2^{2 x}-4 \times 2^x-5=0$
$2^{2 x}+2^x-5 \times 2^x-5=0$
$2^x\left(2^x+1\right)-5\left(2^x+1\right)=0$
$\left(2^x+1\right)\left(2^x-5\right)=0$
$2^x-5=0$
$2^x=5$
$x \log 2=\log 5$
$x=\frac{\log 5}{\log 2}$
$=\log _2 5$
View full question & answer→MCQ 311 Mark
If $a_1, a_2, a_3, .... a_n$ are in $A.P.$ with common difference d, then the sum of the series $\sin\text{d}\ [\text{sec}\ \text{a}_1\ \text{sec}\ \text{a}_2\ \text{sec}\ \text{a}_2\ \text{sec}\ \text{a}_3+\ .....\ +\text{sec}\ \text{a}_{\text{n}-1}\text{sec}\ \text{a}_\text{n}]$ is:
- A
$\sec\text{a}_1-\sec\text{a}_\text{n}$
- B
$\text{cosec}\ \text{a}_1-\text{cosec}\ \text{a}_\text{n}$
- C
$\cot\text{a}_1-\cot\text{a}_\text{n}$
- ✓
$\tan\text{a}_1-\tan\text{a}_\text{n}$
AnswerCorrect option: D. $\tan\text{a}_1-\tan\text{a}_\text{n}$
We have,
$\sin\text{d}(\text{sec}\ \text{a}_1\ \text{sec}\ \text{a}_2+\ \text{ses}\ \text{a}_2\ \text{sec}\ \text{a}_3\\+\ .....\ +\ \text{sec}\ \text{a}_{\text{n}-1}\ \text{sec}\ \text{a}_\text{n})$
$=\frac{\sin\text{d}}{\cos\text{a}_1\cos\text{a}_2}+\frac{\sin\text{d}}{\cos\text{a}_2\cos\text{a}_3}+\ .....\ +\frac{\sin\text{d}}{\cos\text{a}_{\text{n}-1}\cos\text{a}_\text{n}}$
$=\frac{\sin(\text{a}_2-\text{a}_1)}{\cos\text{a}_1\cos\text{a}_2}+\frac{\sin(\text{a}_3-\text{a}_2)}{\cos\text{a}_2\cos\text{a}_3}+\ .....\ +\frac{\sin(\text{a}_\text{n}-\text{a}_{\text{n}-1})}{\cos\text{a}_{\text{n}-1}\cos\text{a}_n}$
$=\frac{\sin\text{a}_2\cos\text{a}_1-\cos\text{a}_2\sin\text{a}_1}{\cos\text{a}_1\cos\text{a}_2}+\frac{\sin{\text{a}_3}\cos\text{a}_2-\cos\text{a}_3\sin\text{a}_2}{\cos\text{a}_1\cos\text{a}_2}\\+\ .....\ +\frac{\sin\text{a}_2\cos\text{a}_1-\cos\text{a}_2\sin\text{a}_1}{\cos\text{a}_1\cos\text{a}_2}$
$=(\tan\text{a}_1-\tan\text{a}_2)+(\tan\text{a}_2-\tan\text{a}_3)\\+\ .....\ +(\tan\text{a}_{\text{n}-1}-\tan\text{a}_\text{n})$
$=\tan\text{a}_1-\tan\text{a}_\text{n}$
View full question & answer→Question 321 Mark
If in an A.P. $\text{S}_\text{n}=\text{n}^2\text{q}$ and $\text{S}_\text{m}=\text{m}^2\text{q},$ where Sr denotes the sum of r terms of the A.P., then Sq equals:
- $\frac{\text{q}^3}{2}$
- $\text{mnq}$
- $\text{q}^3$
- $(\text{m}^2+\text{n}^2)$
Answer
- $\text{q}^3$
Solution:
As, $\text{S}_\text{n}=\text{n}^2\text{q}$
$\Rightarrow\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]=\text{n}^2\text{q}$
$\Rightarrow2\text{a}+(\text{n}-1)\text{d}=\frac{\text{n}^2\text{q}\times2}{\text{n}}$
$\Rightarrow2\text{a}+(\text{n}-1)\text{d}=2\text{nq}\ .....(1)$
Also, $\text{S}_\text{m}=\text{m}^2\text{q}$
$\Rightarrow\frac{\text{m}}{2}[2\text{a}+(\text{m}-1)\text{d}]=\text{m}^2\text{q}$
$\Rightarrow2\text{a}+(\text{m}-1)\text{d}=\frac{\text{m}^2\text{q}\times2}{\text{m}}$
$\Rightarrow2\text{a}+(\text{m}-1)\text{d}=2\text{mq}\ .....(2)$
Subtracting (1) from (2), we get
$2\text{a}+(\text{n}-1)\text{d}-2\text{a}-(\text{m}-1)\text{d}=2\text{nq}-2\text{mq}$
$\Rightarrow(\text{n}-1-\text{m}+1)\text{d}=2\text{nq}-2\text{mq}$
$\Rightarrow(\text{n}-\text{m})\text{d}=2\text{q}(\text{n}-\text{m})$
$\Rightarrow\text{d}=\frac{2\text{q}(\text{n}-\text{m})}{(\text{n}-\text{m})}$
$\Rightarrow\text{d}=2\text{q}$
Substituting $\text{d}=2\text{q}$ in (2), we get
$2\text{a}+(\text{m}-1)2\text{q}=2\text{mq}$
$\Rightarrow2\text{a}+2\text{mq}-2\text{q}=2\text{mp}$
$\Rightarrow2\text{a}=2\text{q}$
$\Rightarrow\text{a}=\text{q}$
Now,
$\text{S}_\text{q}=\frac{\text{q}}{2}[2\text{a}+(\text{q}-1)\text{d}]$
$=\frac{\text{q}}{2}[2\text{q}+(\text{q}-1)2\text{q}]$
$=\frac{\text{q}}{2}[2\text{q}+2\text{q}^2-2\text{q}]$
$=\frac{\text{q}}{2}[2\text{q}^2]$
$=\text{q}^3$
Hence, the correct alternative is option (c). View full question & answer→Question 331 Mark
If $\frac{3+5+7+\ ...\ +\text{Upto n terms}}{5+8+11+\ ...\ \text{Upto 10 terms}}=7,$ then find the value of n.
AnswerWe have,
$\frac{3+5+7+\ ...\ +\text{Upto n terms}}{5+8+11+\ ...\ \text{Upto 10 terms}}=7$
$\Rightarrow\frac{\frac{\text{n}}{2}[2\times3+(\text{n}-1)2]}{\frac{10}{2}[2\times5+(10-1)3]}=7$
$\Rightarrow\frac{\text{n}[6+2\text{n}-2]}{10[10+27]}=7$
$\Rightarrow\frac{\text{n}[4+2\text{n}]}{10\times37}=7$
$\Rightarrow4\text{n}+2\text{n}^2=7\times370$
$\Rightarrow2\text{n}+\text{n}^2=7\times185$
$\Rightarrow\text{n}^2+2\text{n}=1295$
$\Rightarrow\text{n}^2+2\text{n}-1295=0$
$\Rightarrow\text{n}^2+37\text{n}-35\text{n}-1295=0$
$\Rightarrow\text{n}(\text{n}+37)-35(\text{n}+37)=0$
$\Rightarrow(\text{n}=37)(\text{n}-35)=0$
$\Rightarrow\text{n}-35=0$
$\Rightarrow\text{n}=35$
View full question & answer→Question 341 Mark
If the sum of $n$ terms of an AP is $2n^2 + 3n$, then write its nth term.
AnswerWe have,
$S_n = 2n^2 + 3n$
$\Rightarrow S_{n-1} = 2 (n - 1)^2 + 3 (n - 1)$
$= 2 (n^2 + 1 - 2n) +3n - 3$
$=2n^2 + 2 -4n + 3n - 3$
$= 2n^2 - n - 1$
$\Rightarrow S_{n-1} = 2n^2- n - 1$
$\therefore$ $T_n = S_n - S_{n-1} = 2n^2 + 3n - (2n^2 - n - 1)$
$= 2n^2 + 3n - 2n^2 + n + 1$
$= 4n + 1$
Hence, $T_n= 4n + 1.$
View full question & answer→Question 351 Mark
Write the sum of first n even natural numbers.
AnswerWe have,
2 + 4 + 6 + ... up to n terms
$\therefore\text{S}_\text{n}=\frac{\text{n}}{2}[2\times2+(\text{n}-1)\times2]$
$=\frac{\text{n}}{2}[4+2\text{n}-2]$
$=\frac{\text{n}}{2}[2\text{n}+2]$
$=\frac{\text{n}}{2}\times2(\text{n}+1)$
$\text{n}(\text{n}+1)$
$\therefore2+4+6+\ ...$ up to n terms = n (n + 1)
Hence, the sum of first n even natural numbers is n (n + 1)
View full question & answer→MCQ 361 Mark
Sum of all two digit numbers which when divided by $4$ yield unity as remainder is:
AnswerCorrect option: B. $1210$
The given series is $13, 17, 21 ... 97.$
$a_1 = 13, a_2 = 17, a_n = 97$
$d = a_2 - a_1 = 7 - 3 = 4$
$a_n = 97$
$\Rightarrow a +(n - 1)d = 97$
$\Rightarrow 13 + (n - 1)4 = 97$
$\Rightarrow n = 22$
Sum of the above series:
$\text{S}_22=\frac{22}{2}\big\{2\times13+(22-1)4\big\}$
$=11\big\{26+84\big\}$
$=1210$
View full question & answer→Question 371 Mark
If the first, second and last term of an A.P are a, b and 2 a respectively, then its sum is:
- $\frac{2\text{n}}{\text{n}+1}$
- $\frac{\text{n}}{\text{n}+1}$
- $\frac{\text{n}+1}{2\text{n}}$
- $\frac{\text{n}+1}{\text{n}}$
Answer
- $\frac{\text{n}+1}{2\text{n}}$
Solution:
Let the A.P. be a, a + d, a + 2d ........ a + nd.
Here, let d be the common difference and n be the total number of terms.
Given:
$\text{a}_1=\text{a},$
$\text{a}_2=\text{b}$
$\Rightarrow\text{d}=\text{b}-\text{a}\ .....(1)$
$\text{a}_\text{n}=2\text{a}$
$\Rightarrow\text{a}+(\text{n}-1)\text{d}=2\text{a}$
$\Rightarrow(\text{n}-1)\text{d}=\text{a}$
$\Rightarrow\text{d}=\frac{\text{a}}{\text{n}-1}\ .....{(2)}$
From equations (1) and (2), we have
$\Rightarrow\frac{\text{a}}{\text{n}-1}=\text{b}-\text{a}$
$\Rightarrow\frac{\text{a}}{\text{b}-\text{a}}+1=\text{n}$
$\Rightarrow\frac{\text{a}+\text{b}-\text{a}}{\text{b}-\text{a}}=\text{n}$
$\Rightarrow\frac{\text{b}}{\text{b}-\text{a}}=\text{n}$
Now, sum of n terms of an A.P.
$\text{S}=\frac{\text{n}}{2}\{\text{a}+\text{a}_\text{n}\}$
$=\frac{\text{n}}{2}(3\text{a})$
$=\frac{3\text{ab}}{2(\text{b}-\text{a})}$ View full question & answer→Question 381 Mark
Find the sum of the following arithmetic progression:
1, 3, 5, 7, ... to 12 terms.
Answer1, 3, 5, 7, ... to 12 terms
$\text{s}_{12}=\frac{12}{2}[2\times1+(12-1)2]$
$=6\times24=144$
View full question & answer→