Solution

Given A.P. is $121,117,113 \ldots$

$a=121$

$d=117-121=-4$

$a_n=a+(n-1) d$ $=121+(n-1)(-4)$

$=121-4 n+4$

$=125-4 n$

We have to find the first negative term of this A.P.

Therefore, $a_n<0$

$125-4 n<0$

$125<4 n$

$n>\dfrac{125}{4}$

$n>31.25$

Therefore, $32^{\text{nd }}$ term will be the first negative term of this A.P.

Solution

We know that,

$a_n=a+(n-1) d$

$a_3=a+(3-1) d$

$a_3=a+2 d$

Similarly, $a_7=a+6 d$

Given that, $a_3+a_7=6$

$(a+2 d)+(a+6 d)=6$

$2 a+8 d=6$

$a+4 d=3$

$a=3-4 d(i)$

Also, it is given that $(a_3) \times(a_7)=8$

$(a+2 d) \times(a+6 d)=8$

From equation ( $i$ ),

$ \begin{aligned} & (3-4 d+2 d) \times(3-4 d+6 d)=8 \\ & (3-2 d) \times(3+2 d)=8 \\ & 9-4 d^{2}=8 \\ & 4 d^{2}=9-8=1 \\ & d^{2}=\dfrac{1}{4} \\ & d= \pm \dfrac{1}{2} \\ & d=\dfrac{1}{2} \text{ or }-\dfrac{1}{2} \end{aligned} $

From equation (i),

$ \begin{aligned} & (\text{ When } d \text{ is } \dfrac{1}{2}) \\ & a=3-4 d \\ & a=3-4(\dfrac{1}{2}) \\ & =3-2=1 \end{aligned} $

(When $d$ is $.-\dfrac{1}{2})$

$a=3-4(-\dfrac{1}{2})$

$a=3+2=5$

$S_n=\dfrac{n}{2}[2 a(n-1) d]$

(When $a$ is 1 and $d$ is $.\dfrac{1}{2})$

$S _{16}=\dfrac{16}{2}[2(1)+(16-1)(\dfrac{1}{2})]$

$=8[2+\dfrac{15}{2}]$

$=4(19)=76$

(When $a$ is 5 and $d$ is $.-\dfrac{1}{2})$

$ \begin{aligned} & S _{16}=\dfrac{16}{2}[2(5)+(16-1)(-\dfrac{1}{2})] \\ & =8[10+(15)(-\dfrac{1}{2})] \\ & =8(\dfrac{5}{2}) \\ & =20 \end{aligned} $

Solution

It is given that the rungs are $25 cm$ apart and the top and bottom rungs are $2 \dfrac{1}{2} m$ apart.

$\therefore$ Total number of rungs

$ =\dfrac{2 \dfrac{1}{2} \times 100}{25}+1=\dfrac{250}{25}+1=11 $

Now, as the lengths of the rungs decrease uniformly, they will be in an A.P.

The length of the wood required for the rungs equals the sum of all the terms of this A.P.

First term, $a=45$

Last term, $I=25$

$n=11$

$S_n=\dfrac{n}{2}(a+l)$

$\therefore S _{10}=\dfrac{11}{2}(45+25)=\dfrac{11}{2}(70)=385 cm$

Therefore, the length of the wood required for the rungs is $385 cm$.

Solution

The number of houses was

$1,2,3 \ldots 49$

It can be observed that the number of houses are in an A.P. having $a$ as 1 and $d$ also as 1 .

Let us assume that the number of $x^{\text{th }}$ house was like this.

We know that,

Sum of $n$ terms in an A.P.

$ =\dfrac{n}{2}[2 a+(n-1) d] $

Sum of number of houses preceding $x^{\text{th }}$ house $=S _{x-1}$

$ \begin{aligned} & =\dfrac{(x-1)}{2}[2 a+(x-1-1) d] \\ & =\dfrac{x-1}{2}[2(1)+(x-2)(1)] \\ & =\dfrac{x-1}{2}[2+x-2] \\ & =\dfrac{(x)(x-1)}{2} \end{aligned} $

Sum of number of houses following $x^{\text{th }}$ house $=S _{49}-S_x$

$ \begin{aligned} & =\dfrac{49}{2}[2(1)+(49-1)(1)]-\dfrac{x}{2}[2(1)+(x-1)(1)] \\ & =\dfrac{49}{2}(2+49-1)-\dfrac{x}{2}(2+x-1) \\ & =(\dfrac{49}{2})(50)-\dfrac{x}{2}(x+1) \\ & =25(49)-\dfrac{x(x+1)}{2} \end{aligned} $

It is given that these sums are equal to each other.

$\dfrac{x(x-1)}{2}=25(49)-x(\dfrac{x+1}{2})$

$\dfrac{x^{2}}{2}-\dfrac{x}{2}=1225-\dfrac{x^{2}}{2}-\dfrac{x}{2}$

$x^{2}=1225$

$x= \pm 35$

However, the house numbers are positive integers.

The value of $x$ will be 35 only.

Therefore, house number 35 is such that the sum of the numbers of houses preceding the house numbered 35 is equal to the sum of the numbers of the houses following it.

Solution

From the figure, it can be observed that

$1^{\text{st }}$ step is $\dfrac{1}{2} m$ wide,

$2^{\text{nd }}$ step is $1 m$ wide,

$3^{\text{rd }}$ step is $\dfrac{3}{2} m$ wide.

Therefore, the width of each step is increasing by $\dfrac{1}{2} m$ each time whereas their height $\dfrac{1}{4} m$ and length $50 m$ remains the same.

Therefore, the widths of these steps are

$\dfrac{1}{2}, 1, \dfrac{3}{2}, 2, \ldots$

Volume of concrete in $1^{\text{st }}$ step $=\dfrac{1}{4} \times \dfrac{1}{2} \times 50=\dfrac{25}{4}$

Volume of concrete in $2^{\text{nd }}$ step $=\dfrac{1}{4} \times 1 \times 50=\dfrac{25}{2}$

Volume of concrete in $3^{\text{rd }}$ step $=\dfrac{1}{4} \times \dfrac{3}{2} \times 50=\dfrac{75}{4}$

It can be observed that the volumes of concrete in these steps are in an A.P.

$ \begin{aligned} & \dfrac{25}{4}, \dfrac{25}{2}, \dfrac{75}{4}, \ldots \\ & a=\dfrac{25}{4} \\ & d=\dfrac{25}{2}-\dfrac{25}{4}=\dfrac{25}{4} \\ & \text{ and } S_n=\dfrac{n}{2}[2 a+(n-1) d] \\ & S _{15}=\dfrac{15}{2}[2(\dfrac{25}{4})+(15-1) \dfrac{25}{4}] \\ & =\dfrac{15}{2}[\dfrac{25}{2}+\dfrac{(14) 25}{4}] \\ & =\dfrac{15}{2}[\dfrac{25}{2}+\dfrac{175}{2}] \\ & =\dfrac{15}{2}(100)=750 \end{aligned} $

Volume of concrete required to build the terrace is $750 m^{3}$.