Arithmetic Progression
Mathematics Class Tenth
NCERT Exercise 5.4
Question. (1) Which term of the AP: 121, 117, 113, ...., is its first negative term? [Hint: Find `n` for `a_n<0`]
Solution:
Here, given, First term, `a=121`
Common difference, `d = 117-121=-4`
First negative term =?
We know that first negative term will be less than zero.
Here, it can be observed that, when a term is divided by `4` it gives a remainder of 1.
So, it is clear that last positive term will be 4+1=5
So, let us calculate, that 5 is which term of the given Arithmetic Progression (AP)
We know that, `a_n = a+(n-1)d`
`:. a_n = 121+(n-1)(-4)`
`=> 5 = 121-4n+4`
`=>5=125-4n`
`=>-4n = 5-125=-120`
`:. n = 120/4=30`
This means that 5 is the `30^(th)` term
Therefore, `31^(st)` term = `a_30+d`
`=>a_31=5+(-4)=1`
Thus, `a_32 = a_31+d`
`=>a_32 = 1+(-4) = -3`
Thus, First negative term `32^(nd)` term. Answer
Question. (2) The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
Solution:
Let, First term `=a` and Common difference `=d`
Given, `a_3+a_7=6`
And, `a_3xxa_7=8`
Therefore, Sum of first 16the term, `S_16=?`
We know that, `a_n = a+(n-1)d`
`:. a_3 = a+(3-1)d`
`=>a_3 = a+2d` ------(i)
And, `a_7=a+(7-1)d`
`=>a_7=a+6d` ----(ii)
According to question, `a_3+a_7=6`
From equation (i) and (ii), we get
`a+2d+a+6d=6`
`=>2a+8d=6`
`=>2(a+4d)=6`
`=>a+4d=6/2=3`
`=>a=3-4d` -----(iii)
Now, according to question, `a_3xxa_7=8`
From equation (i) and (ii), we get
`(a+2d)(a+6d)=8`
`=>a^2+6ad+2ad+12d^2=8`
`=a^2+8ad+12d^2=8`
Now, after substituting the value of `a` from equation (iii), we get
`(3-4d)^2+8(3-4d)d+12d^2=8`
`=>3^2+(4d)^2-2xx3xx4d` `+24d-32d^2+12d^2=8`
`=>9+16d^2-24d+24d` `-32d^2+12d^2=8`
`=>9+16d^2-32d^2` `+12d^2=8`
`=>9-4d^2=8`
`=>-4d^2=8-9`
`=>-4d^2 = -1`
`=>4d^2=1`
`=>(2d)^2=1`
`=>2d=sqrt1`
`=>2d=1`
`=>d = 1/2`
By substituting the value of `d=1/2` in quation (iii), we get
`a=3-4xx1/2`
`=>a=3-2=1`
Now, we have, `a=1 and d=1/2`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
∴ Sum of first 16 terms,
`S_16=16/2[2xx1+(16-1)1/2]`
`=8[2+15xx1/2]`
`=8[(4+15)/2]`
`=8xx19/2`
`=4xx19`
`=>S_16=76`
Thus, sum of first sixteen terms for given Arithmetic Progression (AP) = 76. Answer
Question. (3) A ladder has rungs 25cm apart. (see fig). The rungs decrease uniformly in length from 45cm at the bottom to 25cm at the top. If the top and the bottom rungs are `2\1/2m` apart, what is the length of the wood required for the rungs? [Hind: Number of rungs `=250/25+1`]
Solution:
Given distance between top and bottom rungs
`= 2.5m = 2.5xx100cm=250cm`
Distance between each rungs = 25cm
∴ Total no. of rungs `=250/25+1=11`
Now, we have, First term, `a = 25`
Total number of terms, `n = 11`
Last term `a_11=45`
∴ Sum of first 11 terms `S_11=?`
We know that, `a_n = a+(n-1)d`
`:. a_11 =25+(11-1)d`
`=>45=25+10d`
`=> 10d = 45-25=20`
`:. d = 20/10=2`
Now, we know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_11 = 11/2[2xx25+(11-1)2]`
`=11/2xx2[25+10]`
[After taking 2 as common]
`=11xx35`
`=>S_11=385`
Thus, lenght of wood required = 385 cm Answer
Question. (4) The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of `x` such that the sum of the numbers of the houses preceding the house numbered `x` is equal to the sum of the numbers of the houses following it. Find this value of `x`. [Hint: `S_(x-1)=S_49-S_x`]
Solution:
Here, given, First term, `a=1` [∵ number starts from 1]
Common difference `d = 1` [∵ house numbered from 1 to 49 consecutively]
Total terms, `n = 49`
And, `a_49=49`
Now, According to question
`S_(x-1)=S_49-S_x` ------(i)
Now, we know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_x=x/2[2xx1+(x-1)1]`
`=x/2[2+x-1]`
`=x/2[x+1]`
`=>S_x=(x^2+x)/2` ----(ii)
And, `S_(x-1)=(x-1)/2[2xx1+(x-1-1)1]`
`=(x-1)/2[2+x-2]`
`=((x-1)x)/2`
`S_(x-1)=(x^2-x)/2` ----(iii)
And, `S_49=49/2[2xx1+(49-1)1]`
`=49/2[2+48]`
`=49/2xx50=49xx25`
`=>S_49=1225` ------(iv)
After substituting the values of `S_(x-1), S_x and S_49` from equation (ii), (iii) and (iv) in equation (i), we get
`(x^2-x)/2=1225-(x^2+x)/2`
`=>1225=(x^2-x)/2+(x^2+x)/2`
`=> 1225=(x^2-x+x^2+x)/2`
`=>1225=(2x^2)/2`
`=>x^2=1225`
`=>x=sqrt1225`
`=>x=35`
Thus, value of `x` = 35 Answer
Question. (5) A small terrace at a football ground comprises of 15 steps each of which is 50m long and built of solid concrete. Each step has a rise of `1/4m` and a tread of `1/2m`. Calculate the total volume of comcrete required to built the terrace. [Hint: Volume of concrete required to build the first step `=1/4xx1/2xx50\ m^3` ]
Solution:
Given, lenght of step = 50m.
Width of step `= 1/2=0.5m`
Height of step `=1/4=0.25m`
Volume of concrete required to build one step = Volume of one step
Now, volume of 1st step = `50mxx0.5mxx0.25m` `=6.25m^3`
For Second step
Now, lenght of 2nd step = 50m
Width of 2nd step = 0.5m`
Height of 2nd step = height of 1st step + height of 2nd step
`=0.25m+0.25m= 0.5m`
∴ Volume of 2nd step `= 50mxx0.5mxx0.5m` `=12.5m^3`
For Third step
Now, lenght of 3rd step = 50m
Width of 3rd step = 0.5m`
Height of 3rd step = height of 2nd step + height of 3rd step
`=0.25m+0.25m+0.25m` `=0.75m`
∴ Volume of 3rd step `=50.mxx0.5mxx0.75m` `=18.75m^3`
Thus, here we have,
First term, `a= 6.25`
Common difference, `d=12.5-6.25=6.25`
Total no. of terms `n =15`
Therefore, Sum of first `n` terms, i.e. `S_15=?`
Now, we know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`S_15=15/2[2xx6.25+(15-1)6.25]`
`=15/2[12.5+14xx6.25]`
`=15/2[12.5+87.5]`
`=15/2xx100`
`=15xx50`
`=>S_15=750`
Thus, `750m^3` concrete will be required to build the terrace. Answer
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