Arithmetic Progression
Mathematics Class Tenth
NCERT Exercise 5.3 Q4-11
Question (4) How many terms of the AP: 9, 17, 25, .... must be taken to give a sum of 636?
Solution:
Here, given, First term, `a = 9`
Common difference, `d = 17-9=8`
`S_n = 636`
Therefore, `n = ?`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_n = n/2[2xx9+(n-1)8]`
`=> 636 = n/2[18+8n-8]`
`=>636=n/2[8n+10]`
`=>636=n/2xx2[4n+5]`
`=>636=n[4n+5]`
`=>636=4n^2+5n`
`=>4n^2+5n-636=0`
`=>4n^2-48n+53n-636=0`
`=>4n(n-12)+53(n-12)=0`
`=>(4n+53)(n-12)=0`
Now, If `4n+53=0`
`:. 4n = -53`
`=> n = -53/4`
And, if `n-12=0`
`:. n = 12`
After discarding the negative value of `n`, we have
`n = 12`
Thus, 12 terms must be taken for given Arithmetic Progression to give required sum.
Thus, Answer `=12`
Question. (5) The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the commond difference.
Solution:
Given, First term, `a = 5`.
Last term, `l = 45`
Sum of all terms, `S = 400.
Therefore, no. of terms , `n=?` and common difference `d=?`
We know that sum of all terms, `S= n/2(a+l)`
`:. 400 = n/2(5+45)`
`=> 400 = n/2xx50`
`=> 400 = 25n`
`:. n = 400/25=16`
Now, we know that, `a_n=a+(n-1)d`
`:. a_16 = 5+(16-1)d`
`=> 45 = 5+15d`
`=> 15d = 45-5=40`
`=> d = 40/15=8/3`
`=> d = 8/3`
Thus, n = 16 and `d=8/3` Answer
Question. (6) The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution:
Given, First term, `a = 17,
And, Last term, `l = 350`
Common difference, `d = 9`
Number of terms, `n =?`
And, `S =?`
We know that, `a_n=a+(n-1)d`
`:. 350 = 17+(n-1)9`
`=> (n-1)9 = 350-17=333`
`=>n-1=333/9=37`
`:. n = 37+1=38`
Now, we know that sum of all terms, `S= n/2(a+l)`
`:. S= 38/2(17+350)`
`=> S = 19xx367`
`=> S = 6973`
Thus, total number of terms, `n = 38` and their sum = 6973. Answer
Question. (7) Find the sum of first 22 terms of an AP in which `d=7` and 22nd term is 149.
Given, `a_22 = 149`
Common difference, `d = 7`
`:. S_n =?`
We know that, `a_n=a+(n-1)d`
`:. a_22 = a+(22-1)7`
`=>149=a+21xx7`
`=>149=a+147`
`=>a=149-147=2`
Now, we know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_22 = 22/2[2xx2+(22-1)7]`
`=11[4+21xx7]`
`=11(4+147)`
`=11(151)=1661`
`:. S_n = 1661`
Thus, required sum = 1661.Answer
Question. (8) Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
Given, Second term, `a_2=14`
And, Third term, `a_3 = 18`
And number of terms, `n=51`
Therefore, Sum of first 51 terms, `S_51=?`
We know that, `a_n=a+(n-1)d`
`:. a_2 = a+(2-1)d`
`=>14=a+d` -----(i)
And, `a_3 = a+(3-1)d`
`=>18=a+2d` -----(ii)
After subtracting equation (i) from equation (i), we get
`18-14=a+2d-(a+d)`
`=>4=a+2d-a-d`
`=>4=d`
`:. d = 4`
After putting the value of `d` in equation (i), we get
`14=a+4`
`=>a=14-4`
`=>a = 10`
Now, we know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_51 = 51/2[2xx10+(51-1)4]`
`=> S_51 = 51/2xx2[10+(50)2]`
`=> S_51=51[10+100]`
`=>S_51 = 51xx110`
`=>S_51 = 5610`
Thus, sum of first 51 terms `S_51= 5610`. Answer
(9) If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first `n` terms.
Solution:
Given, `S_7 = 49`
And, `S_17=289`
`:. S_n = ?`
Let, first term `=a` and Common difference `=d`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_7 = 7/2[2a+(7-1)d]`
`=>49=7/2[2a+6d]`
`=>49=7/2xx2(a+3d)`
`=>49/7=a+3d`
`=>7=a+3d` -------(i)
Again, `S_17=17/2[2a+(17-1)d]`
`=>289=17/2[2a+16d]`
`=>289=17/2xx2[a+8d]`
`=>289=17(a+8d)`
`=>289/17=a+8d`
`=>17=a+8d` -----(ii)
After subtracting equation (i) from (ii) we get
`=>17-7=a+8d-(a+3d)`
`=>10=a+8d-a-3d`
`=>10=5d`
`=>:. d = 10/5=2`
After putting the value of 'd' in equation (i) we get
`7=a+3xx2`
`=>a+6=7`
`=>a=7-6=1`
Now, sum of first `n` terms
`S_n=n/2[2xx1+(n-1)2]`
`=n/2xx2[1+n-1]`
`=n[n]=n^2`
`=>S_n = n^2`
Thus, sum of first `n` terms `=n^2`. Answer
Question. (10) Show that `a_1, a_2`,....,`a_n` ... form an AP where `a_n` is defined as below:
(i) `a_n = 3+4n` (ii) `a_n = 9-5n`
Also find the sum of the first 15 terms in each case.
Solution:
(i) `a_n = 3+4n`
Here, given, `a_n = 3+4n`
∴ First term, `a_1 = 3+4xx1`
`=>a_1 = 3+4 = 7`
Similarly, Second term, `a_2 = 3+4xx2`
`=>a_2 = 3+8=11`
And, Third term, `a_3 = 3+4xx3`
`=>a_3= 3+12=15`
And, Fourth term, `a_4 = 3+4xx4`
`=>a_4 = 3+16=19`
Here, `a_2-a_1 = a_3-a_2`=.....`=4=d`
Thus, list of number formed
`=7, 11, 15, 19`, .....
Here, `a_(k+1)-a_k=4` which is equal for all values of `k`. Thus, given list of number is in Arithmetic Progression(AP)
Here we have, `a= 7, and d = 4`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
∴ Sum of first 15 terms,
`:. S_15 = 15/2[2xx7+(15-1)4]`
`= 15/2xx2[7+14xx2]`
`= 15(7+28)=15xx35`
`=>S_15 = 525`
Thus, `a_1, a_2`,....,`a_n` ... form an AP. Proved
And, Sum of first 15 terms = 525. Answer
(ii) `a_n = 9-5n`
Solution:
Given, `a_n = 9-5n`
∴ First term, `a_1 = 9-5xx1`
`=>a_1 = 9-5=4`
Similarly, Second term, `a_2 = 9-5xx2`
`=>a_2=9-10=-1`
And, Third term, `a_3 = 9-5xx3`
`=>a_3 = 9-15=-6`
And, Fourth term, `a_4 = 9-5xx4`
`=>a_4 = 9-20=-11`
Thus, list of numbers formed
`= 4, -1, -6, -11`, .....
Here, `a_2-a_1=a_3-a_2`= ..... `=-5=d`
Here, `a_(k+1)-a_k=4` which is equal for all values of `k`. Thus, given list of number is in Arithmetic Progression(AP)
Here, `a=4 and d = -5`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
∴ Sum of first 15 terms,
`:. S_15 = 15/2[2xx4+(15-1)(-5)]`
`=15/2[8+14xx(-5)]`
`=15/2[8-70]`
`=15/2xx(-62)`
`=15xx(-31)`
`=> S_15 = -465`
Thus, `a_1, a_2`,....,`a_n` ... form an AP in the given AP. Proved
And, Sum of first 15 terms `= -465`. Answer
Question (11) The sum of the first `n` terms of an AP is `4n-n^2`, what is the first term (that is `S_1`)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and `n^(th)` terms.
Solution:
Given, Sum of first `n` terms, `S_n = 4n-n^2`
∴ First term `S_1 = ?`
Sum of first 2 terms `S_2 =?`
Second term ` a_2 = ?`
3rd term `a_3=?`
10th term `a_10=?`
`n^(th)` term `a_n =?`
Here, ∵ `S_n = 4n-n^2`
`:. S_1 = 4xx1-1^2`
`=4-1=3`
`=>S_1 = 3` ---(i)
Similarly, sum of first two terms,
`S_2 = 4xx2-2^2`
`=8-4=4`
`=>S_2 = 4` -----(ii)
[To find out third term `a_3` we need to find sum of first 3 terms]
Similarly, sum of first three terms,
`S_3 = 4xx3-3^2=12-9`
`=> S_3 =3` ------(iii)
[To find out 10th term `a_10` we need to find sum of first 9 terms `(S_9)` and sum of first 10 terms `(S_10)`]
Similarly, sum of first nine terms,
`S_9 = 4xx9-9^2=36-81`
`=>S_9 = -45` ------(iv)
And, `S_10 = 4xx10-10^2 = 40-100`
`=>S_10=-60` -----(v)
[To find out 10th term `a_n` we need to find sum of first `(n-1)` terms `(S_(n-1))` and sum of first n terms `(S_n)` which is already given in question]
`:. S_(n-1)= 4(n-1)-(n-1)^2`
`=4n-4-(n^2+1-2n)`
`=4n - 4 - n^2-1+2n`
`=>S_(n-1)=6n-5-n^2` ----(vi)
Now, Second term
`a_2 = S_2 - S_1`
By putting the values of `S_2 and S_1` from equation (i) and (ii), we get
`a_2 = 4-3`
`=>a_2 =1`
Similarly, Third term
`a_3 = S_3 -S_2`
By putting the values of `S_3 and S_2` from equation (iii) and (ii), we get
`a_3 = 3-4`
`a_3 = -1`
Similarly, tenth term
`a_10 = S_10-S_9`
After substituting the values of `S_10 and S_9` from equation (v) and (iv), we get
`a_10 = -60-(-45)=-60+45`
`=>a_10 = -15`
Similarly, `n^(th)` term
`a_n = S_n - S_(n-1)`
After substituting the value of `S_n` as given in question and value of `S_(n-1)` from equation (vi), we get
`a_n = 4n-n^2-(6n-5-n^2)`
`=>a_n=4n-n^2-6n+5+n^2`
`a_n=5+4n-6n-n^2+n^2`
`a_n = 5-2n`
Thus, `S_1 = 3, S_2 = 4, a_2 = 1,` `a_3 = -1, a_10=-15,` `a_n = 5-2n` Answer
Alternate Method: After finding the `d`
Given, Sum of first `n` terms, `S_n = 4n-n^2`
∴ First term `S_1 = ?`
Sum of first 2 terms `S_2 =?`
Second term ` a_2 = ?`
3rd term `a_3=?`
10th term `a_10=?`
`n^(th)` term `a_n =?`
Here, ∵ `S_n = 4n-n^2`
`:. S_1 = 4xx1-1^2`
`=4-1=3`
`=>S_1 = 3`
Similarly, sum of first two terms,
`S_2 = 4xx2-2^2`
`=8-4=4`
`=>S_2 = 4`
Now, ∵ Sum of first two terms, `S_2 = 4`.
And First term `S_1 = 3`
∴ Second term, `a_2=S_2-S_1`
`=4-3=1`
`=>a_2 = 1`
We know that, if first term `=a` and common difference `=d`
∴ Second term `a_2 = a+d`
`:. d = a_2 - a`
`=>d=1-3=-2`
Here, we have, `a=3 and d=-2`
We know that, `a_n=a+(n-1)d`
`:. a_3 = 3+(3-1)(-2)`
`=3+2xx(-2)=3-4`
`=>a_3 = -1`
Similarly, `a_10 = 3+(10-1)(-2)`
`=3+9xx(-2)=3-18`
`a_10 = -15`
And, `a_n = 3+(n-1)(-2)`
`= 3-2n+2`
`=>a_n = 5-2n`
Thus, `S_1 = 3, S_2 = 4, a_2 = 1,` `a_3 = -1, a_10=-15,` `a_n = 5-2n` Answer
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