Exercise 2.6 - Chapter 2 Numbers and Sequences 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $2.6$
Question $1 .$
Find the sum of the following
(i) $3,7,11, \ldots \ldots$ up to 40 terms.
(ii) $102,97,92, \ldots \ldots$ up to 27 terms.
(iii) $6+13+20+\ldots \ldots \ldots+97$
Solution:
(i) $3,7,11, \ldots$ upto 40 terms.
$a=3, d=t_{2}-t_{1}=7-3=4$
$\mathrm{n}=40$
$\mathrm{S}_{\mathrm{n}}=\frac{n}{2}(2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d})$
$S_{40}=\frac{20}{2}(2 \times 3+39 \mathrm{~d})$
$=20(6+39 \times 4)$
$=20(6+156)$
$=20 \times 162$
$=3240$
(ii) $102,97,952, \ldots$ up to 27 terms
$a=102$,
$\mathrm{d}=\mathrm{t}_{2}-\mathrm{t}_{1}$
$=97-102=-5$
$\mathrm{n}=27$

$\begin{aligned}
&\mathrm{S}_{n}=\frac{n}{2}(2 a+(n-1) d)\\
&S_{27}=\frac{27}{2}(2 \times 102+26 \times-5)\\
&=\frac{27}{2}(74)^{37}\\
&=27 \times 37=999 \text {. }\\
&\text { (iii) } 6+13+20+\ldots+97\\
&\text { a }=6, d=7,1=97\\
&n=\frac{l-a}{d}+1\\
&=\frac{97-6}{7}+1=\frac{91}{7}+1\\
&=\frac{91+7}{7}=\frac{98}{7}=14\\
&\therefore \quad \mathrm{S}_{n}=\frac{n}{2}(a+l)\\
&\mathrm{S}_{14}=\frac{14}{2}(6+97)\\
&=7 \times 103=721
\end{aligned}$

Question 2.
How many consecutive odd integers beginning with 5 will sum to 480 ?
Answer:
$\begin{aligned}
&5,7,9,11,13, \ldots \\
&\mathrm{S}_{\mathrm{n}}=480 \\
&\mathrm{a}=5, \mathrm{~d}=2, \mathrm{~S}_{\mathrm{n}}=480
\end{aligned}$
$\begin{aligned}
\mathrm{S}_{n} &=\frac{n}{2}(2 a+(n-1) d) \\
480 &=\frac{n}{2}[2 \times 5+(n-1) 2] \\
&=\frac{n}{2}[10+2 n-2] \\
480 &=\frac{n}{2}[8+2 n] \\
8 n+2 n^{2} &=960 \\
2 n^{2}+8 n-960 &=0
\end{aligned}$

$\begin{aligned}
&\Rightarrow n^{2}+4 n-480=0 \\
&\Rightarrow n^{2}+24 n-20 n-480=0 \\
&\Rightarrow n(n+24)-20(n+24)=0 \\
&\Rightarrow(n-20)(n+24)=0 \\
&\Rightarrow n=20,=24
\end{aligned}$
No. of terms cannot be -ve.
$\therefore$ No. of consecutive odd integers beginning with 5 will sum to 480 is 20 .
 

Question $3 .$
Find the sum of first 28 terms of an A.P. whose $n^{\text {th }}$ term is $4 n-3$.
Answer:
Number of terns $(n)=28$
$\begin{aligned}
&t_{n}=4 n-3 \\
&t_{1}=4(1)-3=4-3=1 \\
&t_{2}=4(2)-3=8-3=5 \\
&t_{3}=4(3)-3=12-3=9
\end{aligned}$
Here $a=1, d=5-1=4$
$\mathrm{S}_{28}=\frac{n}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$=\frac{28}{2}[2+(27)(4)]$
$=14[2+108]$
$=14 \times 110$
$=1540$
Sum of 28 terms $=1540$

Question $4 .$
The sum of first $n$ terms of a certain series is given as $2 \mathrm{n} 2-3 \mathrm{n}$. Show that the series is an A.P.
Solution:

Given $S_{n}=2 n^{2}-3 n$
$\begin{aligned}
&S_{1}=2(1)^{2}-3(1)=2-3=-1 \\
&\Rightarrow t_{1}=a=-1 \\
&S_{2}=2\left(2^{2}\right)-3(2)=8-6=2 \\
&t_{2}=S_{2}-S_{1}=2-(-1)=3 \\
&\therefore d=t_{2}-t_{1}=3-(-1)=4
\end{aligned}$
Consider $\mathrm{a}, \mathrm{a}+\mathrm{d}, \mathrm{a}+2 \mathrm{~d}, \ldots \ldots . \mathrm{c}$
$-1,-1+4,-1+2(4), \ldots \ldots . .$
$-1,3,7, \ldots$.
Clearly this is an A.P with $a=-1$, and $d=4$.

Question $5 .$
The $104^{\text {th }}$ term and 4 th term of an A.P are 125 and 0 . Find the sum of first 35 terms.
Solution:
$\mathrm{t}_{104}=125$
$t_{4}=0$

$\begin{aligned}
&\mathrm{a}+(\mathrm{n}-1) \mathrm{d}=\mathrm{t}_{\mathrm{n}}\\
&a+103 d=125\\
&\text { (-) } \underset{(-)}{a+3} 3=\underset{(-)}{0}\\
&(1)-(2) \Rightarrow 100 d=125\\
&d=\frac{125}{100}=\frac{5}{4}\\
&\text { Substitute } d=\frac{5}{4} \text { in (2) }\\
&a+3 \times \frac{5}{4}=0\\
&a+\frac{15}{4}=0 \Rightarrow a=-\frac{15}{4}\\
&\therefore \mathrm{S}_{n}=\frac{n}{2}(2 a+(n-1) d)\\
&S_{35}=\frac{35}{2}\left(22 \times \frac{-15}{A_{2}}+34^{17} \times \frac{5}{A_{2}}\right)\\
&=\frac{35}{2}\left(\frac{-15}{2}+\frac{85}{2}\right)\\
&=\frac{35}{2}\left(\frac{70}{2}\right)=\frac{35}{2} \times 35\\
&=\frac{1225}{2}\\
&=612.5
\end{aligned}$

Question $6 .$
Find the sum of all odd positive integers less than 450 .
Solution:
Sum of all odd positive integers less than 450 is given by
$1+3+5+\ldots+449$
$a=1$
$\mathrm{d}=2$
$1=419$
$\therefore n=\frac{l-a}{d}+1=\frac{449-1}{2}+1$
$=\frac{448}{2}+1$
$=224+1=225$
$\therefore \quad S_{n}=\frac{n}{2}(a+l)$
$\begin{aligned} \therefore \quad n=\frac{l-a}{d}+1 &=\frac{449-1}{2}+1 \\ &=\frac{448}{2}+1 \\ &=224+1=225 \\ \therefore \quad S_{n} &=\frac{n}{2}(a+l) \\ S_{225} &=\frac{225}{2}(1+449) \\ &=\frac{225}{2} \times 450{ }^{225} \\ \text { Se can use the formula } n^{2}=225^{2} \end{aligned}$
$\mathrm{S}_{225}=\frac{225}{2}(1+449)$
$=\frac{225}{2^{2}} \times 450^{225}$
$=225^{2}$
$=50625$
Another method:
Sum of all tve odd integers $=\mathrm{n}^{2}$.
We can use the formula $n^{2}=225^{2}$

$=50625$

Question $7 .$
Find the sum of all natural numbers between 602 and 902 which are not divisible by 4 .
Answer:
Natural numbers between 602 and 902
$\begin{aligned}
&603,604, \ldots, 901 \\
&\mathrm{a}=603, \mathrm{l}=901, \mathrm{~d}=1
\end{aligned}$
$\begin{aligned}
n=\frac{l-a}{d}+1 &=\frac{901-603}{1}+1 \\
&=298+1=299 \\
\mathrm{~S}_{n} &=\frac{n}{2}(a+l)
\end{aligned}$

$\begin{aligned}
\mathrm{S}_{299} &=\frac{299}{2}(603+901) \\
&=\frac{299}{2} \times 1504 \\
&=224848
\end{aligned}$
Sum of all natural numbers between 602 and 902 which are not divisible by 4 . = Sum of all natural numbers between 602 and 902
= Sum of all natural numbers between 602 and 902 which are divisible by 4 .
$1=902-2=900$
To make 602 divisible by 4 we have to add 2 to 602 .
$\therefore 602+2=604$ which is divisible by 4 .
To make 902 divisible by 4 , subtract 2 from 902 .
$\therefore 900$ is the last number divisible by 4 .

a=604, l=900, d=4, n=\frac{l-a}{d}+1

$\begin{aligned}
n=\frac{900-604}{4}+1 &=\frac{296}{4}+1 \\
&=74+1=75 \\
\mathrm{~S}_{n} &=\frac{n}{2}(a+l) \\
\mathrm{S}_{75} &=\frac{75}{2}(604+900) \\
&=\frac{75}{2}(1504) \\
&=56400
\end{aligned}$
Sum of all natural numbers between 602 and 902 which are not divisible 4 .
$\begin{aligned}
&=224848-56400 \\
&=168448
\end{aligned}$

Question 8.
Raghu wish to buy a Laptop. He can buy it by paying $\square 40,000$ cash or by making 10 installments as $\square 4800$ in the first month, $\square 4750$ in the second month, $\square 4700$ in the third month and so on. If he pays the money in this fashion. Find
(i) Total amount paid in 10 installments.
(ii) How much extra amount that he pay in installments.
Answer:
(i) Amount paid in 10 installments
$4800+4750+4700+\ldots \ldots \ldots \ldots \ldots .10$
Here $a=4800 ; d=-50 ; n=10$
$\begin{aligned}
&S_{\mathrm{n}}=\frac{n}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}] \\
&\mathrm{S}_{10}=\frac{10}{2}[2 \times 4800+9(-50)] \\
&=\frac{10}{2}[9600-450] \\
&=5[9150] \\
&=45750
\end{aligned}$
Amount paid in 10 installments

$=\square 45750$
(ii) Extra amount paid = amount paid in 10
installment - cost of the laptop
$=\square 45750-40,000$
$=\square 5750$
(i) Amount paid in 10 installments $=\square 45750$
(ii) Difference in payment $=\square 5750$
Question 9.
A man repays a loan of $\square 65,000$ by paying $\square 400$ in the first month and then increasing the payment by $\square 300$ every month. How long will it take for him to clear the loan?
Solution:
Loan Amount $=\square 65,000$
Repayment through installments
$400+700+1000+1300+\ldots$
$a=400$
$\mathrm{d}=300$
$\mathrm{S}_{\mathrm{n}}=65000$
$\mathrm{S}_{\mathrm{n}}=\frac{n}{2}(2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d})$
$=65000$
$\frac{n}{2}(2 \times 400+(n-1) 300)=65000$
$\mathrm{n}(800+300 \mathrm{n}-300)=130000$
$\mathrm{n}(500+300 \mathrm{n})=130000$
$500 n+300 n^{2}=130000$
$3 \emptyset \emptyset n^{2}+5 \emptyset \emptyset n=1300 \emptyset \emptyset$

$3 n^{2}+5 n-1300=0$
$(n-20)(3 n+65)=0$
$n=20, n=\frac{-65}{3}$
$\therefore n=20$

$\therefore n=20$
Number of terms should be (+ve) and cannot be (-ve) or fractional number.
$\therefore$ He will take 20 months to clear the loans.

Question $10 .$
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.
(i) How many bricks are required for the top most step?
(ii) How many bricks are required to build the stair case?
Answer:
Total number of steps $=30$
$\therefore \mathrm{n}=30$
Number of bricks for the bottom $=100$
$a=100$
2 bricks is less for each step
(i) Number of bricks required for the top most step
$\begin{aligned}
&t_{n}=a+(n-1) d \\
&t_{30}=100+29(-2) \\
&=100-58 \\
&=42
\end{aligned}$
(ii) Number of bricks required
$\begin{aligned}
&\mathrm{S}_{\mathrm{n}}=\frac{n}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}] \\
&\mathrm{S}_{30}=\frac{30}{2}[200+29(-2)] \\
&=15[200-58] \\
&=2130
\end{aligned}$

$=2130$
(i) Number of bricks required for the top most step $=42$ bricks
(ii) Number of bricks required $=2130$

Question $11 .$
If $S_{1}, S_{2}, S_{3}, \ldots ., S_{m}$ are the sums of $n$ terms of $m$ A.P.'s whose first terms are $1,2,3, \ldots, m$ and whose common differences are $1,3,5, \ldots,(2 \mathrm{~m}-1)$ respectively, then show that $\mathrm{S}_{1}+\mathrm{S}_{2}+\mathrm{S}_{3}+\ldots . \mathrm{Sm}=\frac{1}{2} \mathrm{mn}(\mathrm{mn}+1)$
Solution:

$\begin{aligned}
\mathrm{S}_{2} &=\frac{n}{2}(4+3 n-3)=\frac{n}{2}(3 n+1) \\
\mathrm{S}_{3} &=\frac{n}{2}(6+5 n-5)=\frac{n}{2}(5 n+1) \\
\mathrm{S}_{m} &=\frac{n}{2}[2 m+2 m n-2 m-n+1] \\
&=\frac{n}{2}(n(2 m-1)+1) \\
\therefore \mathrm{S}_{1} &+\mathrm{S}_{2}+\mathrm{S}_{3}+\ldots+\mathrm{S}_{m} \\
&=\frac{n}{2}[n+3 n+5 n+\ldots(2 m-1) n+m \times 1] \\
&=\frac{n}{2}[n(1+3+5+\ldots+(2 m-1)+m]\\
&=\frac{n}{2}\left[n \times \frac{m}{2}(2 m-1+\not)+m\right] \\
&=\frac{n}{2}\left[m^{2} n+m\right] \\
&=\frac{1}{2} m n(m n+1) \\
-12 n c e &=r o v e d .
\end{aligned}$
Hence proved.

Question $12 .$
Find the sum
$\left[\frac{a-b}{a+b}+\frac{3 a-2 b}{a+b}+\frac{5 a-3 b}{a+b}+\ldots \text { to } 12 \text { terms }\right]$

Solution:
$\begin{aligned}
=\frac{1}{a+b}[(a-b)+(3 a-2 b)+(5 a-3 b)+\ldots\\
\text { Here } a &=\frac{a-b}{a+b}, d=t_{2}-t_{1} \\
&=\frac{3 a-2 b}{a+b}-\frac{a-b}{a+b} \\
d &=\frac{2 a-b}{a+b} \\
\mathrm{~S}_{n} &=\frac{n}{2}(2 a+(n-1) d) \\
\mathrm{S}_{12} &=\frac{12}{2}\left[2\left(\frac{a-b}{a+b}\right)+11 \times\left(\frac{2 a-b}{a+b}\right)\right] \\
&=6\left[\frac{2 a-2 b+22 a-11 b]}{a+b}\right] \\
&=6\left[\frac{24 a-13 b}{a+b}\right]
\end{aligned}$