3 Marks Question

Question 513 Marks

The taxi fare after each km when the fare is ₹ $15$ for the first km and ₹ $8$ for each additional km. Is this situation make an arithmetic progression and why?

Answer

Taxi fare for $1 km = Rs 15 = a_1$_
Taxi Fare for 2 kms
$= R s 15 + R s 8 = R s 23 = a _ { 2 }$
Taxi fare for 3 km s
$= R s 23 + R s 8 = R s 31 = a _ { 3 }$
Taxi fare for 4 kms
$= R S 31 + R s 8 = R S 39 = a _ { 4 }$
and so on
$a _ { 2 } - a _ { 1 } = R s .23 - R s .15 = R s 8$
$a _ { 3 } - a _ { 2 } = R s 31 - R s \cdot 23 = R s 8$
$a _ { 4 } - a _ { 3 } = R s 39 - R s 31 = R s 8$
So, the arithmetic progression formed is:-
i.e.$,a_{k+1}-a_k$_ is the same every time.
So, this list of numbers form an arithmetic
Progression with the first term $a = Rs 15$ and
the common difference $d = Rs 8.$

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Question 523 Marks

A sum of ₹ $1000$ is invested at $8\ %$ simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of $30$ years making use of this fact.

Answer

Let P be the principle, R rate of interest and $I_n$_ be the interest at the end of n year
We know that
$I_n = \frac { P R n } { 100 }$
$\left[ \text { Using : Interest } = \frac { P R T } { 100 } \right]$
A sum of ₹1000 is invested at $8\ %$ simple interest per annum.
Here, we have
$P = ₹1000,$ and $R = 8\ % $per annum
$\therefore$ I_n = ₹$\left( \frac { 1000 \times 8 \times n } { 100 } \right)$= ₹ 80n
Putting $n = 1,2,3,...,$ we have
$l_n = 80n$
$I_1 = 80 \times 1 = ₹80$
$I_2 = 80 \times 2 = ₹160$
$I_3 = 80 \times 3 = ₹240$
$I_4 = 80 \times 4 = ₹320$ and so on.
Since, $I_n$_ is a linear expression in n.
Therefore, the sequence of interest forms an A.P. with common difference 80.
Hence, the sequence of interests is an A.P.
Also, Interest at the end of 30 years $= I_{30}= 80n = ₹ (80 \times 30) = ₹ 2400$

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Question 533 Marks

Check whether 301 is a term of the given list of numbers: $5, 11, 17, 23,...?$

Answer

We have :
$a_2-a_1=11-5=6, a_3-a_2=17-11=6, a_4-a_3=23-17=6$
As $a_{k+1}-a_k$ is the same for $k=1,2,3$, etc., so the given list of numbers are in AP.
Now, $a=5$ and $d=6$.
Let 301 be a term, say, the $n ^{\text {th }}$ term of this AP.
We know that
$a_n=a+(n-1) d$
So, $301=5+(n-1) \times 6$
i.e., $301=6 n-1$
So, $n=\frac{302}{6}=\frac{151}{3}$, since n is in the form of fraction , thus 301 is not the term of given AP

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Question 543 Marks

Determine the $AP$ whose 3rd term is $5$ and the $7th$ term is $9$.

Answer

We have
$a_3 = a + (3 – 1) d = a + 2d = 5 ....(i)$
and $a_7 = a + (7 – 1) d = a + 6d = 9 .....(ii)$
Solution by substitution method: Now from equation (i), value of$ a = 5 - 2d .....(iii)$
put value of a from equation (iii) in equation (ii), we get
$5 - 2d + 6d = 9$
$4d = 9 - 5$
$4d = 4$
$d = 1$
now put value of d in equation (iii), we get
$a = 5 - 2\times1$
$a = 3$
Hence, the required AP is $3, 4, 5, 6, 7,...$

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Question 553 Marks

Which term of the A.P$ 21, 18, 15, . . .$ is $– 81$? Also, is any term $0$? Give reason for your answer.

Answer

Here, $a = 21, d = 18 – 21 = – 3$ and $a_n = – 81$, and we have to find n.
As $a_n = a + ( n – 1) d$,
we have$ – 81 = 21 + (n – 1)(– 3)$
$– 81 = 24 – 3n$
$– 105 = – 3n$
$So, n = 35$
Therefore, the 35th term of the given $A.P$ is $– 81.$
Next, we want to know if there is any n for which an $= 0$. If such an n is there, then
$21 + (n – 1) (–3) = 0,$
$i.e., 3(n – 1) = 21$
$i.e., n = 8$
So, the eighth term is $0.$

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Question 563 Marks

Elpis Technology is a TV manufacturer company. It produces smart TV sets not only for the Indian market but also exports them to many foreign countries. Their TV sets have been in demand every time but due to the Covid-19 pandemic, they are not getting sufficient spare parts especially chips to accelerate the production. They have to work in a limited capacity due to the lack of raw material.

They produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find:

the production in the 1st year (2)
the production in the 10th year (1)
the total production in first 7 years (1)

Answer

Self Learning

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Question 573 Marks

If the sum of the first $14$ terms of an $A.P.$ is $1050$ and its first term is $10$ find its $20^{th}$ term.

Answer

Given, $a = 10,$ and $S_{14} = 1050$
Let the common difference of the A.P. be d
$\therefore \quad S _ { n } = \frac { n } { 2 } [ 2 a + ( n - 1 ) d ]$
$\therefore \quad S _ { 14 } = \frac { 14 } { 2 } [ 2 \times 10 + ( 14 - 1 ) d ]$
1050 = $7 (20 + 13 d )$
$20 + 13d$ = $\frac{1050}{7}$
$20 + 13d = 150$
$13d = 150 - 20$
$13d = 130$
$d$ = $\frac{130}{13} = 10$
$a_{20} = a + (n - 1)d$
$= 10 + (20 - 1) 10$
$= 10 + 19 $\times$ 10$
$= 10 + 190$
$= 200$
Hence, $a_{20} = 200$

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Maths 3 Marks Question