Question 12 Marks
Show that 107 is of the form 4q +3 for any integer q
Answer107 = 4 × 26 + 3.
This is of the form a = bq + r.
Hence it is proved.
View full question & answer→Question 22 Marks
A milk man has 175 litres of cow’s milk and 105 litres of buffalow’s milk. He wishes to sell the milk by filling the two types of milk in cans of equal capacity. Calculate the following :
Capacity of a can
Answer175 litres of cow’s milk.
105 litres of goat’s milk.
H.C.F of 175 & 105 by using Euclid’s division algorithm.
175 = 105 × 1 + 70, the remainder 70 ≠ 0
Again using division algorithm,
105 = 70 × 1 + 35, the remainder 35 ≠ 0
Again using division algorithm.
70 = 35 × 2 + 0, the remainder is 0.
∴ 35 is the H.C.F of 175 & 105.
∴ The milk man’s milk can’s capacity is 35 litres.
View full question & answer→Question 32 Marks
A milk man has 175 litres of cow’s milk and 105 litres of buffalow’s milk. He wishes to sell the milk by filling the two types of milk in cans of equal capacity. Calculate the following :
Number of cans of cow’s milk
Answer175 litres of cow's milk.
105 litres of goat's milk.
H.C.F of 175 \& 105 by using Euclid's division algorithm.
$
175=105 \times 1+70 \text {, the remainder } 70 \neq 0
$
Again using division algorithm,
$
105=70 \times 1+35 \text {, the remainder } 35 \neq 0
$
Again using division algorithm.
$70=35 \times 2+0$, the remainder is 0 .
$\therefore 35$ is the H.C.F of $175 \& 105$.
No. of cow's milk obtained $=\frac{175}{35}=5$ cans
View full question & answer→Question 42 Marks
A milk man has 175 litres of cow’s milk and 105 litres of buffalow’s milk. He wishes to sell the milk by filling the two types of milk in cans of equal capacity. Calculate the following :
Number of cans of buffalow’s milk
Answer175 litres of cow's milk.
105 litres of goat's milk.
H.C.F of 175 \& 105 by using Euclid's division algorithm.
$175=105 \times 1+70$, the remainder $70 \neq 0$
Again using division algorithm,
$105=70 \times 1+35$, the remainder $35 \neq 0$
Again using division algorithm.
$70=35 \times 2+0$, the remainder is 0 .
$\therefore 35$ is the H.C.F of $175 \& 105$.
No. of buffalow's milk obtained $=\frac{105}{35}=3$ cans
View full question & answer→Question 52 Marks
Find the sum of the series $\left(2^3-1^3\right)+\left(4^3-3^3\right)+\left(6^3-15^3\right)+\ldots$ to 8 terms
AnswerWhen $n =8$
$\text { Sum }=4 \times 8^3+3 \times 8^2$
$=2048+192=2240$
View full question & answer→Question 62 Marks
If $1^3+2^3+3^3+\ldots+k^3=44100$ then find $1+2+3+\ldots+k$
Answer$\begin{aligned} & \text { If } 1^3+2^3+3^3+\ldots+k^3=44100 \\ & 1+2+3+\ldots+k=\sqrt{44100} \\ & =210\end{aligned}$
View full question & answer→Question 72 Marks
If $1+2+3+\ldots+k=325$, then find $1^3+2^3+3^3+\ldots+k^3$
Answer$\begin{aligned} & 1+2+3+\ldots+k=325 \\ & 1^3+2^3+3^3+\ldots+k^3=\sum_1^n n^3 \\ & =\left(\frac{n(n+1)}{2}\right)^2 \\ & =\left(\sum_1^n n\right)^2 \\ & \text { If } 1+2+3+\ldots+k=325 \\ & 1^3+2^3+3^3+\ldots+k^3=(325)^2 \\ & =105625\end{aligned}$
View full question & answer→Question 82 Marks
Find the rational form of the number $0 . \overline{123}$
Answer$\begin{aligned} & \text { Let } x=0.123123123 \ldots \Rightarrow x=0 . \overline{123} \ldots(1) \\ & \text { Multiplying } 1000 \text { on both rides } \\ & 1000 x=123.123123 \ldots \Rightarrow 1000 x=123 . \overline{123} \ldots(2) \\ & (2)-(1)=1000 x-x 123 . \overline{123}-0 . \overline{123} \\ & \Rightarrow 999 x=123 \\ & \Rightarrow x=\frac{123}{999} \\ & \Rightarrow x=\frac{41}{333} \text { Rational number }\end{aligned}$
View full question & answer→Question 92 Marks
Find the sum to n terms of the series
3 + 33 + 333 + ………… to n terms
Answer$\begin{aligned} & S _{ n }=3+33+333+\ldots . \text { to } n \text { terms } \\ & =3[1+11+111+\ldots . \text { to } n \text { terms }] \\ & =\frac{3}{9}[9+99+999+\ldots n \text { terms }] \\ & =\frac{1}{3}[(10-1)+(100-1)+(1000-1)+\ldots n \text { terms }] \\ & =\frac{1}{3}[10+100+1000+\ldots n \text { terms }-(1+1+1 \ldots n \text { terms })] \\ & \left( a =10, r =10, S _{ n }=\frac{ a \left( r ^{ n }-1\right)}{ r -1}\right) \\ & =\frac{1}{3}\left[10 \cdot \frac{10^{ n }-1}{10-1}- n \right] \\ & =\frac{1}{3}\left[\frac{10}{9} \cdot\left(10^{ n }-1\right)- n \right] \\ & S _{ n }=\frac{10}{27}\left(10^{ n }-1\right)-\frac{ n }{3}\end{aligned}$
View full question & answer→Question 102 Marks
If the first term of an infinite G.P. is 8 and its sum to infinity is $\frac{32}{3}$ then find the common ratio
Answer$\begin{aligned} & a=8 \\ & S_{\infty}=\frac{32}{3} \\ & \Rightarrow \frac{a}{1-r}=\frac{32}{3} \\ & \frac{8}{1-r}=\frac{32}{3} \\ & 32(1-r)=24 \\ & 1-r=\frac{24}{32} \\ & =\frac{6}{8} \\ & =\frac{3}{4} \\ & -r=\frac{3}{4}-1 \\ & =\frac{3-4}{4} \\ & -r=\frac{-1}{4}\end{aligned}$
$\Rightarrow r=\frac{1}{4}$
View full question & answer→Question 112 Marks
Find the sum to infinity of 9 + 3 + 1 + ….
Answer$\begin{aligned} & 9+3+1+\ldots . \\ & a=9, r=\frac{3}{9}=\frac{1}{3}<1 \\ & S_{\infty}=\frac{a}{1-r} \\ & =\frac{9}{1-\frac{1}{3}} \\ & =\frac{\frac{9}{3-1}}{3} \\ & =\frac{\frac{9}{2}}{3} \\ & =9 \times \frac{3}{2} \\ & =\frac{27}{2}\end{aligned}$
View full question & answer→Question 122 Marks
Find the sum to infinity of $21+14+\frac{28}{3}+\ldots$
Answer$\begin{aligned} & 21+14+\frac{28}{3}+\ldots \\ & \text { Here } a=21, r=\frac{14}{21}=\frac{2}{3} \\ & S_{\infty}=\frac{a}{1-r} \\ & =\frac{21}{1-\frac{2}{3}} \\ & =\frac{21}{\frac{3-2}{3}} \\ & =\frac{21}{\frac{1}{3}} \\ & =21 \times 3=63 \\ & S_{\infty}=63\end{aligned}$
View full question & answer→Question 132 Marks
Find the first term of the G.P. whose common ratio 5 and whose sum to first 6 terms is 46872
Answer$\begin{aligned} & \text { Common ratio, } r=5 \\ & S _6=46872 \\ & \therefore \frac{ a \left( r ^6-1\right)}{ r -1}=46872 \\ & \Rightarrow S _6=\frac{ a \left(5^6-1\right)}{5-1}=46872 \\ & \Rightarrow a =\frac{46872 \times 4}{[25 \times 25 \times 25-1]} \\ & \Rightarrow a =\frac{187488}{15624}=12 \\ & \Rightarrow \text { first term }=12\end{aligned}$
View full question & answer→Question 142 Marks
Find the sum of first six terms of the G.P. 5, 15, 45, …
Answer$\begin{aligned} & \text { G.P. } 5,15,45 \\ & n=6, a=5, r=\frac{15}{5}=3>1 \\ & \therefore S_n=a \frac{\left(r^n-1\right)}{r-1} \\ & S_6=5\left(\frac{3^6-1}{3-1}\right) \\ & =5 \frac{\left(3^6-1\right)}{2} \\ & =\frac{5}{2}(729-1) \\ & =\frac{5}{2} \times 728 \\ & =5 \times 364 \\ & ={1820}\end{aligned}$
View full question & answer→Question 152 Marks
Find the sum of first $n$ terms of the G.P. $5,-3, \frac{9}{5},-\frac{27}{25}, \ldots$
Answer$\begin{aligned} & 5,-3, \frac{9}{5},-\frac{27}{25}, \ldots \\ & \text { Here } a =5, r =\frac{ t _2}{ t _1}=\frac{-3}{5}<1 \\ & S _{ n }= a \left[\frac{1- r ^{ n }}{1- r }\right] \\ & =5\left[\frac{1-\left(\frac{-3}{5}\right)^{ n }}{1-\left(\frac{-3}{5}\right)}\right] \\ & =5\left[\frac{1-\left(\frac{-3}{5}\right)^{ n }}{1-\frac{-3}{5}}\right] \\ & =5 \frac{\left[1-\left(\frac{-3}{5}\right)^{ n }\right]}{\frac{8}{5}} \\ & =5 \times \frac{5}{8}\left[1-\left(\frac{-3}{5}\right)^{ n }\right] \\ & \therefore S _{ n }=\frac{25}{8}\left[1-\left(\frac{-3}{5}\right)^{ n }\right]\end{aligned}$
View full question & answer→Question 162 Marks
Find the sum of first n terms of the G.P.
256, 64, 16, …
Answer$\begin{aligned} & 256,64,16, \ldots \\ & a=256 \\ & r=\frac{64}{256}=\frac{4}{16}=\frac{1}{4}<1 \\ & S_n=\frac{a\left(1-r^n\right)}{1-r} \\ & =256\left[\frac{1-\left(\frac{1}{4}\right)^n}{1-\frac{1}{4}}\right] \\ & =256 \frac{\left[1-\left(\frac{1}{4}\right)^n\right]}{\frac{3}{4}} \\ & =\frac{1024}{3}\left[1-\left(\frac{1}{4}\right)^n\right]\end{aligned}$
View full question & answer→Question 172 Marks
If $a, b, c$ is in A.P. then show that $3^a, 3^b, 3^c$ is in G.P.
AnswerIf $a, b, c$ are in A.P
$t_2-t_1=t_3-t_2$
$b-a=c-b$
$2 b=c+a$
To prove that $3^a, 3^b, 3^c$ is in G.P
$\Rightarrow 3^{2 b}=3^{c+a}+a \quad \ldots[\text { Raising the power both sides }]$
$\Rightarrow 3^b \cdot 3^b=3^c \cdot 3^a$
$\Rightarrow \frac{3^b}{3^a}=\frac{3^c}{3^b}$
$\Rightarrow \frac{t_2}{t_1}=\frac{t_3}{t_1}$
$\Rightarrow \text { Common ratio is same for } 3^a, 3^b, 3^c$
$\Rightarrow 3^a, 3^b, 3^c \text { forms a G.P }$
$\therefore$ Hence it is proved.
View full question & answer→Question 182 Marks
Find the $10^{\text {th }}$ term of a G.P. whose $8^{\text {th }}$ term is 768 and the common ratio is 2
Answer$t _8=768$
$=ar^7$
$r=2$
$t_{10}=ar^9$
$=ar^7 \times r \times r$
$=768 \times 2 \times 2$
$=3072$
View full question & answer→Question 192 Marks
Find the number of terms in the following G.P.
$\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots, \frac{1}{2187}$
Answer$\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots, \frac{1}{2187}$
| Hint: |
| 3 |
729 |
| 3 |
243 |
| 3 |
81 |
| 3 |
27 |
| 3 |
27 |
| 3 |
9 |
| 3 |
3 |
| |
|
$\text { Here } a=\frac{1}{3}, r=\frac{\frac{1}{9}}{\frac{1}{3}}$
$=\frac{1}{9} \times \frac{3}{1}$
$=\frac{1}{3}$
$t_n=a^{n-1}$
$\frac{1}{3} \times\left(\frac{1}{3}\right)^{ n -1}$
$=\frac{1}{2817}$
$\left(\frac{1}{3}\right)^{ n -1}=\frac{1}{2817} \times 3$
$=\frac{1}{729}{ }^{\left(\frac{1}{3}\right)^{ n -1}}=\frac{1}{3^6}$
$=\left(\frac{1}{3}\right)^6$
$n-1=6$
$n=6+1=7$
$\therefore$ No. of terms $=7$
$1$
$\text { Here } a=\frac{1}{3}, r=\frac{\frac{1}{9}}{\frac{1}{3}}$
$=\frac{1}{9} \times \frac{3}{1}$
$=\frac{1}{3}$
$t_n=a^{n-1}$
$\frac{1}{3} \times\left(\frac{1}{3}\right)^{ n -1}$
$=\frac{1}{2817}$
$\left(\frac{1}{3}\right)^{ n -1}=\frac{1}{2817} \times 3$
$=\frac{1}{729}{ }^{\left(\frac{1}{3}\right)^{ n -1}}=\frac{1}{3^6}$
$=\left(\frac{1}{3}\right)^6$
$n-1=6$
$n=6+1=7$
$\therefore$ No. of terms $=7$ View full question & answer→Question 202 Marks
Find $x$ so that $x + 6, x + 12$ and $x + 15$ are consecutive terms of a Geometric Progression
Answer$\text { G.P. }=x+6, x+12, x+15$
$ \text { In G.P. } r=\frac{t_2}{t_1}=\frac{t_3}{t_2}$
$ \frac{x+12}{x+6}=\frac{x+15}{x+12}$
$ (x+12)^2=(x+6)(x+5)$
$ x^2+24 x+144=x^2+6 x+15 x+90$
$ 24 x-21 x=90-144$
$ 3 x=-54$
$ x=\frac{-54}{3}=-18$
$ x=-18$
View full question & answer→Question 212 Marks
In a G.P. $729, 243, 81, …$ find $t_7$
Answer$\text { G.P }=729,243,81, \ldots .$
$ t _7=?$
$ t _{ n }= ar ^{ n -1}, \text { here } a =729, r =\frac{ t _2}{ t _1}$
$ r =\frac{243}{729}=\frac{1}{3}$
$ \therefore t _7=729\left(\frac{1}{3}\right)^{7-1}=729 \times\left(\frac{1}{3}\right)^6$
$ =729 \times \frac{1}{729}$
$ =1$
View full question & answer→Question 222 Marks
Write the first three terms of the G.P. whose first term and the common ratio are given below
a = 6, r = 3
Answer$t_n=a r^{n-1}$
$t_1=a r^{1-1}=a r^0=a=6$
$t_2=a r^{2-1}=a r^1=6 \times 3=18$
$t_3=a r^{3-1}=a r^2=6 \times 3^2=54$
$\therefore \text { The } 3 \text { terms are } 6,18,54, \ldots$
View full question & answer→Question 232 Marks
Write the first three terms of the G.P. whose first term and the common ratio are given below
$a=\sqrt{2}, r=\sqrt{2}$
Answer$a=\sqrt{2}, r=\sqrt{2} $
$t_n=a^{n-1} $
$t_1=\operatorname{ar}^{1-1}=a^0=\sqrt{2} \times 1=\sqrt{2} $
$t_2=\operatorname{ar}^{2-1}=\operatorname{ar}^1=\sqrt{2} \times \sqrt{2}=2 $
$t_3=\operatorname{ar}^{3-1}=\operatorname{ar}^2=\sqrt{2} \times(\sqrt{2})^2 $
$=\sqrt{2} \times 2=2 \sqrt{2}$
$\therefore$ The 3 terms are $\sqrt{2}, 2,2 \sqrt{2}, \ldots$
View full question & answer→Question 242 Marks
Write the first three terms of the G.P. whose first term and the common ratio are given below
$a=1000, r=\frac{2}{5}$
Answer$a=1000, r=\frac{2}{5} $
$t_n=a r^{n-1} $
$t_1=a r^{1-1}=a r^0=1000 \times 1=1000 $
$t_2=\operatorname{ar}^{2-1}=a r=1000 \times \frac{2}{5}=400 $
$t_3=a^{3-1}=\operatorname{ar}^2=1000\left(\frac{2}{5}\right)^2 $
$=1000 \times \frac{4}{25} $
$=160$
$\therefore$ The $3$ terms are $1000,400,160, \ldots \overline{ A }$
View full question & answer→Question 252 Marks
Identify the following sequence is in G.P.?
$1, −5, 25, −125, ...$
Answer$1,-5,25,-125, \ldots $
$r=\frac{t_2}{t_1}=\frac{-5}{1}=-5 $
$r=\frac{t_3}{t_2}=\frac{25}{-5}=-5 $
$-5=-5$
$\therefore$ It is a G.P.
View full question & answer→Question 262 Marks
Identify the following sequence is in G.P.?
120, 60, 30, 18, …
Answer$
120,60,30,18, \ldots
$
$
r=\frac{t_2}{t_1}=\frac{t_3}{t_2}=\frac{t_4}{t_3}
$
Here $r$ is not equal i.e. $\frac{60}{120}=\frac{30}{60} \neq \frac{18}{30}$
$\therefore$ It is not a G.P.
View full question & answer→Question 272 Marks
Identify the following sequence is in G.P.?
$16,4,1, \frac{1}{4}, \ldots$
Answer$16,4,1, \frac{1}{4}, \ldots $
$r =\frac{ t _2}{ t _1}=\frac{4}{16}=\frac{1}{4} $
$r =\frac{ t _3}{ t _2}=\frac{1}{4} $
$r =\frac{1}{4}=\frac{1}{4}$
$\therefore$ It is a G.P.
View full question & answer→Question 282 Marks
Identify the following sequence is in G.P.?
$\frac{1}{3}, \frac{1}{6}, \frac{1}{12}, \ldots$
Answer$ \frac{1}{3}, \frac{1}{6}, \frac{1}{12}, \ldots$
$ r=\frac{ t _2}{ t _1}$
$ =\frac{\frac{1}{6}}{\frac{1}{3}}$
$ =\frac{1}{6} \times \frac{3}{1}=\frac{1}{2}$
$ r=\frac{ t _3}{ t _2}$
$ =\frac{\frac{1}{12}}{\frac{1}{6}}$
$ =\frac{1}{12} \times \frac{6}{1}=\frac{1}{2}$
$ r=\frac{1}{2}=\frac{1}{2} $
$\therefore$ It is a G.P.
View full question & answer→Question 292 Marks
Identify the following sequence is in G.P.?
3, 9, 27, 81, …
Answer$
3,9,27,81, \ldots
$
$r=$ Common ratio
$
r=\frac{t_2}{t_1}=\frac{t_3}{t_2} \text { in G.P. }
$
Here $\frac{t_2}{t_1}=\frac{9}{3}=3$
$
\frac{ t _3}{ t _2}=\frac{27}{9}=3
$
$\therefore$ It is a G.P.
View full question & answer→Question 302 Marks
Identify the following sequence is in G.P.?
$4, 44, 444, 4444, .....$
Answer$4,44,444,4444, \ldots \ldots$
$r=\frac{t_2}{t_1}=\frac{44}{4}=11 $
$r=\frac{t_3}{t_2}=\frac{444}{44}=\frac{111}{11} $
$11 \neq \frac{111}{11}$
$\therefore$ It is not a G.P.
View full question & answer→Question 312 Marks
Identify the following sequence is in G.P.?
$0.5, 0.05, 0.005, …$
Answer$0.5,0.05,0.005, \ldots $
$r=\frac{t_2}{t_1} $
$=\frac{0.05}{0.5} $
$=\frac{0.05 \times 100}{0.5 \times 100} $
$=\frac{5}{50}=\frac{1}{10} $
$r=\frac{t_3}{t_2} $
$=\frac{0.005}{0.05} $
$=\frac{0.005 \times 1000}{0.05 \times 1000} $
$=\frac{5}{50}=\frac{1}{10} $
$r=\frac{1}{10}=\frac{1}{10} $
$\therefore$ It is a G.P.
View full question & answer→Question 322 Marks
Raghu wish to buy a laptop. He can buy it by paying $₹ 40,000$ cash or by giving it in 10 installments as $₹ 4800$ in the first month, $₹ 4750$ in the second month, $₹ 4700$ in the third month and so on. If he pays the money in this fashion, find total amount paid in $10$ installments
Answer$4800+4750+4700+\ldots 10 \text { terms }$
Here $a =4800$
$d=t_2-t_1=4750-4800=-50 $
$n=10 $
$S_n=\frac{n}{2}(2 a+(n-1) d) $
$S_{10}=\frac{10}{2}(2 \times 4800+9 \times-50) $
$=5(9600-450) $
$=5 \times 9150=45750$
Total amount paid in 10 installments $=₹ 45750$.
View full question & answer→Question 332 Marks
Raghu wish to buy a laptop. He can buy it by paying $₹ 40,000$ cash or by giving it in $10$ installments as $₹ 4800$ in the first month, $₹ 4750$ in the second month, $₹ 4700$ in the third month and so on. If he pays the money in this fashion, find how much extra amount that he has to pay than the cost?
Answer$4800+4750+4700+\ldots 10 \text { terms }$
Here $a=4800$
$d=t_2-t_1=4750-4800=-50$
$n=10 $
$S_n=\frac{n}{2}(2 a+(n-1) d) $
$S_{10}=\frac{10}{2}(2 \times 4800+9 \times-50) $
$=5(9600-450) $
$=5 \times 9150 $
$=45750$
Total amount paid in $10$ installments $=₹ 45750.$
The extra amount paid $=$ amount paid in $10$ installment - cost of the laptop
$=₹ 45750-₹ 40,000$
$=₹ 5750$
View full question & answer→Question 342 Marks
Find the sum of the following
$3, 7, 11, …$ up to $40$ terms
Answer$3,7,11, \ldots$ upto $40$ terms.
$a=3, d=t_2-t_1=7-3=4 $
$n=40 $
$S_n=\frac{n}{2}(2 a+(n-1) d) $
$S_{40}=\frac{20}{2}(2 \times 3+39 d) $
$=20(6+39 \times 4) $
$=20(6+156) $
$=20 \times 162 $
$=3240$
View full question & answer→Question 352 Marks
Find the sum of the following
$102, 97, 92, …$ up to $27$ terms.
Answer$\text { Here } a =102, d =97-102=-5$
$ n =27 \\ S _{ n }=\frac{ n }{2}[2 a +( n -1) d ]$
$ S _{27}=\frac{27}{2}[2(102)+26(-5)]$
$ =\frac{27}{2}[204-130]$
$ =\frac{27}{2} \times 74$
$ =27 \times 37=999$
$ S _{27}=999$
View full question & answer→Question 362 Marks
Find the sum of the following
$6 + 13 + 20 + .... + 97$
Answer$6+13+20+\ldots .+97$
$ a=6, d=7, I=97$
$ n=\frac{1-a}{d}+1$
$ =\frac{97-6}{7}+1=\frac{91}{7}+1$
$ =\frac{91+7}{7}=\frac{98}{7}=14$
$ S_n=\frac{n}{2}(a+1)$
$ S_{14}=\frac{14}{2}(6+97)$
$ =7 \times 103$
$ =721$
View full question & answer→Question 372 Marks
A brick staircase has a total of 30 steps. The bottom step requires 100 bricks. Each successive step requires two bricks less than the previous step
How many bricks are required for the top most step?
Answer100 + 98 + 96 + 94 + … 30 steps.
Here
a = 100
d = – 2
n = 30
$t_n = a + (n – 1)d$
$t_{30} = a + (n – 1)d$
= 100 + 29 × -2
= 100 – 58
= 42
No. of bricks required for the top step are 42.
View full question & answer→Question 382 Marks
A brick staircase has a total of $30$ steps. The bottom step requires $100$ bricks. Each successive step requires two bricks less than the previous step
How many bricks are required to build the stair case?
Answer$100 + 98 + 96 + 94 + … 30$ steps.
Here
$a = 100$
$d = – 2$
$n = 30$
$\therefore S _{ n }=\frac{ n }{2}(2 a +( n -1) d )$
$S _{30}=\frac{30}{2}(2 \times 100+29 \times-2)$
$= 15(200 – 58)$
$= 15 \times 142$
$= 2130$
No. of bricks required to build the stair case are $2130.$
View full question & answer→Question 392 Marks
If 3 + k, 18 – k, 5k + 1 are in A.P. then find k
Answer$3+k, 18-k, 5 k+1 \text { are in A.P }$
$\Rightarrow 2 b=a+c \text { if a, b, c are in A.P }$
$\therefore \underbrace{3+k}_a, \underbrace{18-k}_b, \underbrace{5 k+1}_c$
$2 b=a+c$
$\Rightarrow 2(18-k)=3+k+5 k+1$
$36-2 k=4+6 k$
$6 k+2 k=36-4$
$8 k=32$
$k=\frac{32}{8}=4$
View full question & answer→Question 402 Marks
First term a and common difference d are given below. Find the corresponding A.P.
a = 5, d = 6
Question 412 Marks
First term a and common difference d are given below. Find the corresponding A.P.
a = 7, d = – 5
Answera = 7, d = – 5
A.P. = a, a + d, a + 2d, ...
= 7, 7 + (– 5), 7 + 2(– 5), ...
= 7, 2, – 3, ...
View full question & answer→Question 422 Marks
First term a and common difference $d$ are given below. Find the corresponding A.P.
$
a =\frac{3}{4}, d =\frac{1}{2}
$
Answer$
a=\frac{3}{4}, d=\frac{1}{2}
$
The general form of the A.P. is $a, a+d, a+2 d, a+3 d \ldots .$.
$
\frac{3}{4}, \frac{3}{4}+\frac{1}{2}, \frac{3}{4}+2\left(\frac{1}{2}\right), \frac{3}{4}+3\left(\frac{1}{2}\right)
$
The A.P. $\frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \ldots \ldots$
View full question & answer→Question 432 Marks
Check whether the following sequence is in A.P.
1, –1, 1, –1, 1, –1, …
Answer1, –1, 1, –1, 1, –1, …
$t_2-t_1=-1-1=-2$
$t_3-t_2=1-(-1)=1+1=2$
$t_4-t_3=-1-(1)=-1-1=-2$
$t_5-t_4=1-(-1)=1+1=2$
Common difference are not equal
∴ The sequence is not an A.P.
View full question & answer→Question 442 Marks
Check whether the following sequence is in A.P.
$\frac{-1}{3}, 0, \frac{1}{3}, \frac{2}{3}, \ldots$
Answer$\frac{-1}{3}, 0, \frac{1}{3}, \frac{2}{3}, \ldots $
$t_2-t_1=0-\left(-\frac{1}{3}\right) $
$=0+\frac{1}{3}=\frac{1}{3} $
$t_3-t_2=\frac{1}{3}-0=\frac{1}{3} $
$t_2-t_1=t_3-t_2$
The sequence is in A.P.
View full question & answer→Question 452 Marks
Check whether the following sequence is in A.P.
a – 3, a – 5, a – 7, …
Answer$a-3, a-5, a-7 \ldots$
$t_2-t_1=a-5-(a-3)$
$=a-5-a+3$
$=-2$
$t_3-t_2=a-7-(a-5)$
$=a-7-a+5$
$=-2$
$t_2-t_1=t_3-t_2 \ldots \ldots \text { (common difference is same) }$
View full question & answer→Question 462 Marks
Check whether the following sequence is in A.P.
$\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots$
Answer$\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots $
$t_2-t_1=\frac{1}{3}-\frac{1}{2}=\frac{2-3}{6}=\frac{-1}{6} $
$t_3-t_2=\frac{1}{4}-\frac{1}{3}=\frac{3-4}{12}=\frac{-1}{12} $
$t_2-t_1 \neq t_3-t_2$
The sequence is not in A.P.
View full question & answer→Question 472 Marks
Check whether the following sequence is in A.P.
9, 13, 17, 21, 25, ...
Answer$9,13,17,21,25, \ldots$
$t_2-t_1=13-9=4$
$t_3-t_2=17-13=4$
$t_4-t_3=21-17=4$
$t_5-t_4=25-21=4$
Common difference are equal
$\therefore$ The sequence is in A.P.
View full question & answer→Question 482 Marks
Find the indicated terms of the sequences whose $n ^{\text {th }}$ terms are given by
$a_n=\frac{5 n}{n+2} ; a_6$ and $a_{13}$
Answer$a_n=\frac{5 n}{n+2}$
$ a_6=\frac{5(6)}{6+2}=\frac{30}{8}=\frac{15}{4}$
$ a_{13}=\frac{5(13)}{13+2}=\frac{5 \times 13}{15}=\frac{13}{3}$
$ a_6=\frac{15}{4}, a_{13}=\frac{13}{3}$
View full question & answer→Question 492 Marks
Find the indicated terms of the sequences whose $n ^{\text {th }}$ terms are given by $a_n=-\left(n^2-4\right) ; a_4$ and $a_{11}$
Answer$a_n=-\left(n^2-4\right) ; a_4$ and $a_{11}$
$a_4=-\left(4^2-4\right)$
$=-(16-4)$
$=-12$
$a_{11}=-\left(11^2-4\right)$
$=-(121-4)$
$=-117$
$a_4=-12 \text { and } a_{11}=-117$
View full question & answer→Question 502 Marks
Find the $n ^{\text {th }}$ term of the following sequence $2,5,10,17, \ldots$
Answer$2,5,10,17, \ldots$
$=1^2+1,2^2+1,3^2+1,4^2+1 \ldots$
$\therefore n^{\text {th }} \text { term is } n^2+1$
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