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**Ques 1: Which term of the AP: 121, 117, 113, ..., is its first negative term?****[Hint: Find n for a _{n} < 0]Sol:** We have the A.P. having a = 121 and d = 117 - 121 = - 4

âˆ´ a

= 121 + (n - 1) Ã— (- 4)

= 121 - 4n + 4

= 125 - 4n

For the first negative term, we have

a

â‡’ (125 - 4n) < 0

â‡’ 125 < 4n

â‡’

â‡’

or

Thus, the first negative term is 32nd term.

Let the first term = a and the common difference = d

âˆ´ T

âˆµ T

âˆ´ (a + 2d) + (a + 6d) = 6

â‡’ 2a + 8d = 6

â‡’ a + 4d = 3 ...(1)

Again T

âˆ´ (a + 2d) Ã— (a + 6d) = 8

â‡’ (a + 4d - 2d) Ã— (a + 4d + 2d) = 8

â‡’ [(a + 4d) - 2d] Ã— [(a + 4d) + 2d] = 8

â‡’ [(3) - 2d] Ã— [(3) + 2d] = 8 [From (1)]

â‡’ 3

â‡’ 9 - 4d

â‡’ - 4d

â‡’

â‡’

When d = 1/2.

From (1), we have:

â‡’ a + 2 = 3 or a = 3 âˆ’ 2 = 1

Now, Using

i.e., the sum of first 16 terms = 76

From (1), we have:

â‡’ a âˆ’ 2 = 3 â‡’ a = 5

Again, the sum of first sixteen terms

i.e., the sum of first 16 terms = 20

=

Distance between two consecutive rungs = 25 cm

âˆ´ Number of rungs

Length of the 1st rung (bottom rung) = 45 cm

Length of the 11th rung (top rung) = 25 cm

Let the length of each successive rung decrease by x cm

âˆ´ Total length of the rungs

= 45 cm + (45 - x) cm + (45 - 2x) cm + ..... + 25 cm

Here, the numbers 45, (45 - x), (45 - 2x), ....., 25 are in an A.P. such that

First term â€˜aâ€™ = 45

Last term â€˜lâ€™ = 25

Number of terms â€˜nâ€™ = 11

âˆ´ Using,

S

S

â‡’

â‡’ S

âˆ´ Total length of 11 rungs = 385 cm

i.e., Length of wood required for the rungs is 385 cm.

1, 2, 3, 4, 5, ....., 49.

These numbers are in an A.P., such that

a = 1

d = 2 - 1 = 1

n = 49

Let one of the houses be numbered as x

âˆ´ Number of houses preceding it = x - 1

Number of houses following it = 49 - x

Now, the sum of the house-numbers preceding x is given by:

Using S

The houses beyond x are numbered as

(x + 1), (x + 2), (x + 3), ....., 49

âˆ´ For these house numbers (which are in an A.P.),

First term (a) = x + 1

Last term (l) = 49

âˆ´ Using S

According to the question,

[Sum of house numbers preceding x] = [Sum of house numbers following x]

i.e., S

â‡’

â‡’

â‡’

â‡’ x^{2} = (49 Ã— 25)

â‡’

â‡’ x = Â± (7 Ã— 5) = Â± 35

But x cannot be taken as -ve

âˆ´ x = 35**Ques ****5: A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete.****Each step has a rise of 1/4m and a tread of 1/2m. (see Fig.). Calculate the total volume of concrete required to build the terrace.****[Hint: Volume of concrete required to build the first step ****Sol: ****For 1st step:**

Length = 50 m, Breadth = 1/2m, Height = 1/4m

âˆ´ Volume of concrete required to build the 1st step

= Volume of the cuboidal step

= Length Ã— Breadth Ã— height

= **For 2nd step:**

Length = 50 m, Breadth = 1/2m, Height =

âˆ´ Volume of concrete required to build the 2nd step**For 3rd step:**

Length = 50 m, Breadth = 1/2m, Height =

âˆ´ Volume of concrete required to build the 3rd step

........... ........... ........... ........... ...........

Thus, the volumes (in m^{3}) of concrete required to build the various steps are:

obviously, these numbers form an A.P. such that

a = 25/4

d =

Here, total number of steps n = 15

Total volume of concrete required to build 15 steps is given by the sum of their individual volumes.

âˆ´ Using,

Thus, the required volume of concrete is 750 m^{3}.

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