Exercise 2.5 - Chapter 2 Numbers and Sequences 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\operatorname{Ex} 2.5$
Question 1.
Check whether the following sequences are in A.P.
(i) $\mathrm{a}-3, \mathrm{a}-5, \mathrm{a}-7, \ldots \ldots \ldots$
(ii) $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{4}$,
(iii) $9,13,17,21,25, \ldots \ldots$.
(iv) $\frac{-1}{3}, 0, \frac{1}{3}, \frac{2}{3}, \ldots \ldots \ldots \ldots$
(v) $1,-1,1,-1,1,-1, \ldots$
Solution:
To prove it is an A.P, we have to show $\mathrm{d}=\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{t}_{3}-\mathrm{t}_{2}$.
(i) $a-3, a-5, a-7 \ldots \ldots \ldots$
$t_{1}, t_{2}, t_{3}$
$\mathrm{d}=\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{a}-5-(\mathrm{a}-3)=\mathrm{a}-5-\mathrm{a}+3=-2$
$\therefore \mathrm{d}=-2 \quad \therefore$ It is an A.P.
$\mathrm{d}=\mathrm{t}_{3}-\mathrm{t}_{2}=\mathrm{a}-7-(\mathrm{a}-5)=\mathrm{a}-7-\mathrm{a}+5=-2$

(ii) $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{4}, \ldots \ldots \ldots .$
$d=t_{2}-t_{1} \quad d=t_{3}-t_{2}$
$\Rightarrow \frac{1}{3}-\frac{1}{2} \quad \frac{1}{4}-\frac{1}{3}$
$=\frac{2-3}{6}=\frac{3-4}{12}=\frac{-1}{12}$
$=\frac{-1}{6}$
$\frac{-1}{6} \neq \frac{-1}{12}$
$\Rightarrow t_{2}-t_{1} \neq t_{3}-t_{2} \quad \therefore$ It is not an A.P.
(iii) $9,13,17,21,25, \ldots$
$\begin{aligned}
&d=t_{2}-t_{1}=13-9=4 \\
&d=t_{3}-t_{2}=17-13=4
\end{aligned}$
$\therefore$ It is an A.P.
(iv) $\frac{-1}{3}, 0, \frac{1}{3}, \frac{2}{3}, \ldots$
$\begin{aligned}
d=t_{2}-t_{1} &=0-\left(-\frac{1}{3}\right)=\frac{1}{3} \\
d=t_{3}-t_{2} &=\frac{1}{3}-0=\frac{1}{3} \\
\frac{1}{3} &=\frac{1}{3}
\end{aligned}$

$\therefore$ It is an A.P.
(v) $1,-1,1,-1,1,-1$,
$\mathrm{d}=\mathrm{t}_{2}-\mathrm{t}_{1}=-1-1=-2$
$\mathrm{d}=\mathrm{t}_{3}-\mathrm{t}_{2}=1-(-1)=2$
$-2 \neq 2 \therefore$ It is not an A.P.

Question $2 .$
First term a and common difference $d$ are given below. Find the corresponding A.P.
(i) $a=5, d=6$
(ii) $a=7, d=5$
(iii) $\mathrm{a}=\frac{3}{4}, \mathrm{~d}=\frac{1}{2}$
Solution:
(i) $a=5, d=6$
A.P a, a $+\mathrm{d}, \mathrm{a}+2 \mathrm{~d}, \ldots \ldots \ldots$
$=5,5+6,5+2 \times 6, \ldots \ldots . .$
$=5,11,17, \ldots$
(ii) $\mathrm{a}-7, \mathrm{~d}--5$
A.P. $=\mathrm{a}, \mathrm{a}+\mathrm{d}, \mathrm{a}+2 \mathrm{~d}, \ldots$
$=7,7+(-5), 7+2(-5), \ldots \ldots \ldots .$
$=7,2,-3$,
(iii) $\mathrm{a}=\frac{3}{4}, \mathrm{~d}=\frac{1}{2}$
A.P $=a, a+d, a+2 d, \ldots$
$=\frac{3}{4}, \frac{3}{4}+\frac{1}{2}, \frac{3}{4}+z\left(\frac{1}{z}\right), \ldots$

$\begin{aligned}
&=\frac{3}{4}, \frac{3+2}{4}, \frac{3+4}{4}, \ldots \\
\text { A.P } &=\frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \ldots .
\end{aligned}$

Question $3 .$
Find the first term and common difference of the Arithmetic Progressions whose nthterms are given below
(i) $t_{n}=-3+2 n$
Answer:
$t_{n}=-3+2 n$
$\begin{aligned}
&t_{1}=-3+2(1)=-3+2 \\
&=-1 \\
&t_{2}=-3+2(2)=-3+4
\end{aligned}$
$=1$
First term (a) $=-1$ and
Common difference
(d) $=1-(-1)=1+1=2$
(ii) $\mathrm{t}_{\mathrm{n}}=4-7 \mathrm{n}$
Answer:
$\mathrm{t}_{\mathrm{n}}=4-7 \mathrm{n}$

$\mathrm{t}_{\mathrm{n}}=4-7 \mathrm{n}$ $\mathrm{t}_{1}=4-7(1)$ $=4-7=-3$ $\mathrm{t}_{2}=4-7(2)$ $=4-14=-10$ First term $(\mathrm{a})=-3$ and Common difference $(\mathrm{d})=10-(-3)$ $=-10+3$ $=-7$
Common difference $(d)=10-(-3)$
$=-10+3$
$=-7$
 

Question $4 .$
Find the 19 th term of an A.P. $-11,-15,-19, \ldots . \ldots . .$.
Solution:
A.P $=-11,-15,-19, \ldots . . .$
$a=-11$
$\mathrm{d}=\mathrm{t}_{3}-\mathrm{t}_{1}=-15-(-11)$
$=-15+11$
$=-4$
$\mathrm{n}=19$
$\therefore \mathrm{t}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}$
$t_{19}=-11+(19-1)(-4)$
$=-11+18 \times-4$
$=-11-72$
$=-83$

Question 5.
Which term of an A.P. $16,11,6,1, \ldots$ is $-54$ ?
Answer:
First term (a) $=16$
Common difference $(d)=11-16=-5$
$\mathrm{t}_{\mathrm{n}}=-54$
$a+(n-1) d=-54$
$16+(n-1)(-5)=-54$
$54+21=-54$
$51+21=5 n$
$75=5 n$
$\mathrm{n}=\frac{75}{5}=15$
The $15^{\text {th }}$ term is $-54$

Question $6 .$
Find the middle term(s) of an A.P. $9,15,21,27, \ldots \ldots .183 .$
Solution:
A.P $=9,15,21,27, \ldots .183$
No. of terms in an A.P. is
$\mathrm{n}=\frac{l-a}{d}+1$
$\mathrm{a}=9, \mathrm{l}=183, \mathrm{~d}=15-9=6$
$\therefore \mathrm{n}=\frac{183-9}{6}+1$
$=\frac{174}{6}+1$
$=29+1=30$
$\therefore$ No. of terms $=30$. The middle must be 15 th term and 16 th term.
$\therefore \mathrm{t}_{15}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}$
$=9+14 \times 6$
$=9+84$
$=93$
$\mathrm{t}_{16}=\mathrm{a}+15 \mathrm{~d}$
$=9+15 \times 6$
$=9+90=99$
$\therefore$ The middle terms are 93,99 .
 

Question $7 .$
If nine times the ninth term is equal to the fifteen times fifteenth term, Show that six times twenty fourth term is zero.
Answer:
$t_{n}=a+(n-1) d$
9 times $9^{\text {th }}$ term $=15$ times $15^{\text {th }}$ term

$\begin{aligned}
&9 t_{9}-15 t_{15} \\
&9[a+8 d]=15[a+14 d] \\
&9 a+72 d=15 a+210 d \\
&9 a-15 a+72 d-210 d=0 \\
&-6 a-138 d=0 \\
&6 a+138 d=0 \\
&6[a+23 d]=0 \\
&6[a+(24-1) d]=0 \\
&6 t_{24}=0
\end{aligned}$
$\therefore$ Six times 24th terms is 0 .
 

Question $8 .$
If $3+k, 18-k, 5 k+1$ are in A.P. then find $k$.
Solution:
$3+k, 18-k, 5 k+1$ are in A.P
$\Rightarrow 2 \mathrm{~b}=\mathrm{a}+\mathrm{c}$ if $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in A.P
$\therefore \underbrace{3+k}_{a}, \underbrace{18-k}_{b}, \underbrace{5 k+1}_{c}$
$2 b=a+c$
$\Rightarrow \quad 2(18-k)=3+k+5 k+1$
$36-2 k=4+6 k$.
$6 k+2 k=36-4$
$8 k=32$

$k-\frac{32}{8}=4$
 

Question 9.
Find $x, y$ and $z$ gave that the numbers $x$. $10, y, 24, z$ are in A.P.
Answer:
$x, 10, y, 24, z$ are in A.P
$\mathrm{t}_{2}-\mathrm{t}_{1}=10-\mathrm{x}$
$\mathrm{d}=10-\mathrm{x} \ldots . .(1)$
$t_{3}-t_{2}-y-10$
$\mathrm{d}=\mathrm{y}-10 \ldots . .(2)$
$t_{4}-t_{3}=24-y$
$\mathrm{d}=24-\mathrm{y} \ldots . .(3)$
$t_{5}-t_{4}=z-24$
$\mathrm{d}=\mathrm{z}-24 \ldots . .(4)$
From (2) and (3) we get
$y-10-24-y$
$2 y=24+10$
$2 y=34$
$y=17$
From (1) and (2) we get
$10-x=y-10$
$-x-y=-10-10$
$-x-y=-20$
$x+y=20$
$x+17=20(y=17)$
$x=20-17=3$
From (1) and (4) we get
$z-24=10-x$
$z-24=10-3(x=3)$

$\begin{aligned}
&z-24=7 \\
&z=7+24 \\
&z=31
\end{aligned}$
The value of $x=3, y=17$ and $z=31$
 

Question 10.
In a theatre, there are 20 seats in the front row and 30 rows were allotted. Each successive row contains two additional seats than its front row. How many seats are there in the last row?
Solution:
$t_{1}=a=20$ $t_{2}=a+2=22$ $t_{3}=a+2+2=24 \Rightarrow d=2$ $\therefore$ There are 30 rows. $t_{30}=a+29 d$ $=20+29 \times 2$ $=20+58$ $=78$
$\therefore$ There will be 78 seats in the last row.
 

Question $11 .$
The Sum of three consecutive terms that are in A.P. is 27 and their product is 288 . Find the three terms.
Answer:
Let the three consecutive terms be $a-d$, a and $a+d$
By the given first condition
$a-d+a+a+d=27$
$\begin{aligned}
&3 a=27 \\
&a=\frac{27}{3}=9
\end{aligned}$
Again by the second condition
$(a-d)(a)(a+d)=288$
$\begin{aligned}
&\mathrm{a}\left(\mathrm{a}^{2}-\mathrm{d}^{2}\right)=288 \\
&9\left(81-\mathrm{d}^{2}\right)=288(\mathrm{a}=9) \\
&81-\mathrm{d}^{2}=\frac{288}{9} \\
&81-\mathrm{d}^{2}=32 \\
&\therefore \mathrm{d}^{2}=81-32 \\
&=49 \\
&\mathrm{~d}=\sqrt{49}=\pm 7
\end{aligned}$

When $\mathrm{a}=9, \mathrm{~d}=7$
$a+d=9+7=16$
$a=9$
$a-d=9-7=2$
When $\mathrm{a}=9, \mathrm{~d}=-7$
$a+d=9-7=2$
$a=9$
$a-d=9-(-7)=9+7=16$
The three terms are $2,9,16$ (or) $16,9,2$
 

Question $12 .$
The ratio of 6 th and $8^{\text {th }}$ term of an A.P is 7:9. Find the ratio of $9^{\text {th }}$ term to $13^{\text {th }}$ term.
Solution:
$\frac{t_{6}}{t_{9}}=\frac{7}{9}$
$\frac{a+5 d}{a+7 d}=\frac{7}{9}$
$9 a+45 d=7 a+49 d$
$9 a+45-7 d=7 a+49 d$
$9 a+45 d-7 a-49 d=0$
$2 a-4 d=0 \Rightarrow 2 a=4 d$
$a=2 d$
Substitue $a=2 d$ in
$\begin{aligned}
\frac{t_{9}}{t_{13}} &=\frac{a+8 d}{a+12 d} \\
&=\frac{2 d+8 d}{2 d+12 d}
\end{aligned}$

$\begin{aligned}
&=\frac{10 d}{14 d} \\
&=\frac{5}{7} \\
\therefore \quad t_{9}: t_{13} &=5: 7 .
\end{aligned}$

Question $13 .$
In a winter season let us take the temperature of Ooty from Monday to Friday to be in A.P. The sum of temperatures from Monday to Wednesday is $0^{\circ} \mathrm{C}$ and the sum of the temperatures from Wednesday to Friday is $18^{\circ} \mathrm{C}$. Find the temperature on each of the five days.
Solution:
Let the five days temperature be ( $a-d), a, a+d, a+2 d, a+3 d$.
The three days sum $=a-d+a+a+d=0$
$\begin{aligned}
&\Rightarrow 3 \mathrm{a}=0 \Rightarrow \mathrm{a}=0 \text {. (given) } \\
&\mathrm{a}+\mathrm{d}+\mathrm{a}+2 \mathrm{~d}+\mathrm{a}+3 \mathrm{~d}=18
\end{aligned}$
$3 \mathrm{a}+6 \mathrm{~d}=18$ $3(0)+6 \mathrm{~d}=18$ $6 \mathrm{~d}=18$ $\mathrm{~d}=\frac{18}{6}=3$ $\therefore$ The temperature of each five days is a $-\mathrm{d}, \mathrm{a}, \mathrm{a}+\mathrm{d}, \mathrm{a}+2 \mathrm{~d}, \mathrm{a}+3 \mathrm{~d}$ $0-3,0,0+3,0+2(3), 0+3(3)=-3^{\circ} \mathrm{C}, 0^{\circ} \mathrm{C}, 3^{\circ} \mathrm{C}, 6^{\circ} \mathrm{C}, 9^{\circ} \mathrm{C}$
$0-3,0,0+3,0+2(3), 0+3(3)=-3^{\circ} \mathrm{C}, 0^{\circ} \mathrm{C}, 3^{\circ} \mathrm{C}, 6^{\circ} \mathrm{C}, 9^{\circ} \mathrm{C}$

Question $14 .$
Priya earned $\square 15,000$ in the first month. Thereafter her salary increased by $\square 1500$ per year. Her expenses are $\square 13,000$ during the first year and the expenses increases by $\square 900$ per year. How long will it take for her to save $\square 20,000$ per month.
Solution:

We find that the yearly savings is in A.P with $\mathrm{a}_{1}=2000$ and $\mathrm{d}=600$.
We are required to find how many years are required to save 20,000 a year
$\begin{aligned}
&a_{n}=20,000 \\
&a_{n}=a+(n-1) d \\
&20000=2000+(n-1) 600 \\
&(n-1) 600=18000 \\
&n-1=\frac{18000}{600}=30 \\
&n=31 \text { years }
\end{aligned}$