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**Q.1. Which term of the AP: 121, 117, 113, . . ., is its first negative term? [Hint: Find n for a _{n} < 0]**

Given the AP series is 121, 117, 113, . . .,

Thus, first term, a = 121

Common difference, d = 117-121= -4

By the nth term formula,

a

Therefore,

a

= 121-4n+4

=125-4n

To find the first negative term of the series, a

Therefore,

125-4n < 0

125 < 4n

n>125/4

n>31.25

Therefore, the first negative term of the series is 32

From the given statements, we can write,

a

And

a

By the nth term formula,

a

Third term, a

a

And Seventh term, a7= a+(7-1)d

a

From equation (iii) and (iv), putting in equation(i), we get,

a+2d +a+6d = 6

2a+8d = 6

a+4d=3

or

a = 3–4d …………………………………(v)

Again putting the eq.(iii) and (iv), in eq. (ii), we get,

(a+2d)×(a+6d) = 8

Putting the value of a from equation (v), we get,

(3–4d +2d)×(3–4d+6d) = 8

(3 –2d)×(3+2d) = 8

3

9 – 4d

4d

d = 1/2 or -1/2

Now, by putting both the values of d, we get,

a = 3 – 4d = 3 – 4(1/2) = 3 – 2 = 1, when d = 1/2

a = 3 – 4d = 3 – 4(-1/2) = 3+2 = 5, when d = -1/2

We know, the sum of nth term of AP is;

S

So, when a = 1 and d=1/2

Then, the sum of first 16 terms are;

S

And when a = 5 and d= -1/2

Then, the sum of first 16 terms are;

S

Given,

Distance between the rungs of the ladder is 25cm.

Distance between the top rung and bottom rung of the ladder is =

= 5/2 ×100cm

= 250cm

Therefore, total number of rungs = 250/25 + 1 = 11

As we can see from the figure, the ladder has rungs in decreasing order from top to bottom. Thus, we can conclude now, that the rungs are decreasing in an order of AP.

And the length of the wood required for the rungs will be equal to the sum of the terms of AP series formed.

So,

First term, a = 45

Last term,

Number of terms, n = 11

Now, as we know, sum of nth terms is equal to,

S

S

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

Given,

Row houses are numbers from 1,2,3,4,5…….49.

Thus we can see the houses numbered in a row are in the form of AP.

So,

First term, a = 1

Common difference, d=1

Let us say the number of x

Sum of nth term of AP = n/2[2a+(n-1)d]

Sum of number of houses beyond x house = S

= (x-1)/2[2.1+(x-1-1)1]

= (x-1)/2 [2+x-2]

= x(x-1)/2 ………………………………………(i)

By the given condition, we can write,

S

= 25(49) – x(x + 1)/2 ………………………………….(ii)

As per the given condition, eq.(i) and eq(ii) are equal to each other;

Therefore,

x(x-1)/2 = 25(49) – x(x-1)/2

x = ±35

As we know, the number of houses cannot be a negative number. Hence, the value of x is 35.

As we can see from the given figure, the first step is ½ m wide, 2

½ , 1, 3/2, 2, ……..

Volume of steps = Volume of Cuboid

= Length × Breadth Height

Now,

Volume of concrete required to build the first step = ¼ ×1/2 ×50 = 25/4

Volume of concrete required to build the second step =¼ ×1/×50 = 25/2

Volume of concrete required to build the second step = ¼ ×3/2 ×50 = 75/2

Now, we can see the volumes of concrete required to build the steps, are in AP series;

25/4 , 25/2 , 75/2 …..

Thus, applying the AP series concept,

First term, a = 25/4

Common difference, d = 25/2 – 25/4 = 25/4

As we know, the sum of n terms is;

S

Upon solving, we get,

S

S

Hence, the total volume of concrete required to build the terrace is 750 m

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