2 Marks Questions

Question 12 Marks

The taxi fare after each km , when the fare is Rs. $15$ for the first km and Rs. $8$ for each additional km , does not form an $AP$ as the total fare (in Rs.) after each km is $15,8,8,8.........$ Is the statement true? Give reasons.

Answer

$15, 8, 8, 8, .....$. are not the total fare for$ 1, 2, 3, 4$, km respectively.
Total fare for $1km = Rs. 15$
Total fare for $2km = Rs. 15 + Rs. 8 = Rs. 23$
Total fare for$ 3km = Rs. 23 + Rs. 8 = Rs. 31$
Total fare for $4km = Rs. 31 + Rs. 8 = Rs. 39$
Total fare for $1km, 2Km, 3km, 4km, ........$are $15, 23, 31, 39, ....$.respectively.
Now, $d_1 = 23 - 15 = 8d_2 = 31 - 23 = 8$
$d_3 = 39 - 31 = 8$
Hence, the total fare form an$ A.P. as 15, 23, 31, 39, ......$ But,
fare for each km does not form $A.P.$ as $15, 8, 8, 8, ......$

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Question 22 Marks

Justify whether it is true to say that the following are the nth terms of an $AP: 3n^2 + 5$

Answer

$a_n= 3n^2 + 5$
$\therefore$ $a_1 = 3(1)^2+ 5 = 3 \times 1 + 5$
$a_1 = 3 + 5 = 8$
$a_2 = 3(2)^2 + 5 = 3 \times 4 + 5$
$a_2 = 12 + 5 = 17$
$a_3 = 3(3)^2 + 5 = 3 \times 9 + 5(5)$
$a_3= 27 + 5 = 32$
$a_4 = 3(4)^2 + 5 = 3 \times 16 + 5$
$a_4= 48 + 5 = 53$
$a_5 = 3(5)^2 + 5 = 3 \times 25 + 5$
$a_5= 75 + 5 = 80$
$\therefore$ $d_1 = a_2 - a_1 = 17 - 8 = 9$
$d_2 = a_3- a_2 = 32 - 17 = 15$
$d_3 = a_4 - a_3 = 53 - 32 = 21$
$d_4 = a_5 - a_4 = 80 - 53 = 27$
Since $\text{d}_{1}\neq\text{d}_{2},$ so the list of numbers $8, 17, 32, 53, ...$.is not in $A.P.$

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Question 32 Marks

Justify whether it is truw to say that $-1,\frac{-3}{2},-2,\frac{5}{2},......$ froms an A.P. as $a_2 - a_1 = a_3 - a_2$

Answer

Main concept used: A form of numbers will form an A.P.if $d_1 = d_2 = d_3= ......d_n = d.$
Given form of numbers will form an A.P. if $d_1 = d_2 = d_3 = d$ otherewise not.
$\text{So}, \text{ d}_{1}=\text{a}_{2}-\text{a}_{1}$
$\text{d}_{1}=\frac{-3}{2}-\big(-1\big)=\frac{-3}{2}+1$
$\text{d}_{1}=\frac{-3+2}{2}=\frac{-1}{2}$
$\text{d}_{2}=\text{a}_{3}-\text{a}_{2}$
$\text{d}_{2}=-2-\bigg(\frac{-3}{2}\bigg)=-2+\frac{3}{2}$
$\text{d}_{2}=\frac{-4+3}{2}=\frac{-1}{2}$
$\text{d}_{3}=\text{a}_{4}-\text{a}_{3}$
$\text{d}_{3}=\frac{5}{2}-\big(-2\big)=\frac{5}{2}+2$
$\text{d}_{3}=\frac{5+4}{2}=\frac{9}{2}$
$\therefore\text{d}_{1}={\text{d}_{2}}\neq\text{d}_{3}$
$\text{Although}\text{ a}_{2}-\text{a}_{1}=\text{a}_{3}-\text{a}_{2}=\frac{-1}{2}$
$\text{but}\text{ a}_{4}-\text{a}_{3}\neq\frac{-1}{2}$
So, the given form of numbers will not form an A.P. Hence, the given statement is false.

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