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Q : 5     A small terrace at a football ground comprises of  $\small 15$ steps each of which is $\small 50$ m long and   built  of solid concrete. Each step has a rise of   $\small \frac{1}{4}\: m$   and a tread of   $\small \frac{1}{2}\: m$ . (see Fig. $\small 5.8$).  Calculate the total volume of concrete required to build the terrace.
[Hint : Volume of concrete required to build the first step  $\small =\frac{1}{4}\times \frac{1}{2}\times 50\: m^3$  ]

It is given that  football ground comprises of   steps each of which is m long and Each step has a rise of      and a tread of    Now, Volume required to make first step =  Similarly, Volume required to make 2nd step =  And Volume required to make 3rd step =  And so on We can clearly see that this is an AP with  Now,  total volume of concrete required to build the terrace of 15 such...

Q : 4    The houses of a row are numbered consecutively from $\small 1$ to $\small 49$. Show that there is a value of  $\small x$ such that the sum of the numbers of the houses preceding the house numbered $\small x$ is equal to the  sum of the numbers of the houses following it. Find this value of $\small x$.                                                              [Hint : $\small S_{x-1}=S_4_9-S_x$

It is given that the sum of the numbers of the houses preceding the house numbered is equal to the  sum of the numbers of the houses following it And 1,2,3,.....,49 form an AP with a = 1 and d = 1 Now, we know that Suppose their exist an n term such that ( n < 49) Now, according to given conditions  Sum of first n - 1 terms of AP = Sum of terms following the nth term Sum of first n - 1...

Q : 3    A ladder has rungs $\small 25$ cm apart. (see Fig. $\small 5.7$). The rungs decrease uniformly in length from $\small 45$ cm at the bottom to $\small 25$ cm at the top. If the top and the bottom rungs are  $\small 2\frac{1}{2}$  m apart, what is  the length of the wood required for the rungs?

[Hint : Number of rungs  $=\frac{250}{25}+1]$

It is given that Total distance between top and bottom rung  Distance between any two rungs = 25 cm Total number of rungs =   And it is also given that bottom-most rungs is of 45 cm length and topmost is of 25 cm length.As it is given that the length of rungs decrease uniformly, it will form an AP with  Now, we know that  Now, total  length of the wood required for the rungs is equal...

Q : 2    The sum of the third and the seventh terms of an AP is $\small 6$ and their product is $\small 8$. Find the sum of first sixteen terms of the AP.

It is given that  sum of the third and the seventh terms of an AP is and their product is Now, And  put value from equation (i) in (ii) we will get  Now, case (i)    Then, case (ii)   Then,

Q : 1     Which term of the AP :  $\small 121,117,113,...,$ is its first negative term?  [Hint : Find $n$ for $a_n<0$ ]

Given AP is  Here  Let suppose nth term of the AP is first negative term  Then, If nth term is negative then  Therefore, first negative term must be 32nd term
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