Arithmetic Progression
Mathematics Class Tenth
NCERT Exercise 5.3
Question: (1) Find the sum of the following APs:
(i) 2, 7, 12, ....., to 10 terms
Solution:
Here, First term `a=2`
Common difference `(d) = 7-2= 5`
And no. of terms `n = 10`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_10 = 10/2[2xx2+(10-1)5]`
`=10/2 [4+9xx5]`
`=5[4+45]`
`=5xx49`
`=>S_10=245`
Thus, sum of first 10 terms of given AP = 245. Answer
(ii) `-37, -33, -29`, ...., to 12 terms.
Solution:
Given, First term `a= -37`
Common difference (d) `= -33-(-37)` `=-33+37 = 4`
Number of terms, `n = 12`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_12=12/2[2(-37)+(12-1)4]`
`=6[-74+44]`
`= 6xx(-30)`
`=>S_12=-180`
Thus, sum of first 12 terms for given AP `= -180` Answer
(iii) `0.6, 1.7, 2.8`, ...., to 100 terms
Solution:
Here, given, First term `(a) = 0.6`
Common difference `(d)=1.7-0.6=1.1
`n = 100`
Sum of first 100 terms, `S_100=?`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`S_100 = 100/2[2xx0.6+(100-1)1.1]`
`=50[1.2+99xx1.1]`
`=50[1.2+108.9]`
`=50[110.1]=5505`
`=>S_100=5505`
Thus, sum of first 100 terms for the given Arithmetic Progression (AP) = 5505. Answer
(iv) `1/15, 1/12, 1/10`, ..., to 11 terms.
Solution:
Here, given, First term, `a=1/15`
Common difference, `d=1/12-1/15`
`=>d=(5-4)/60`
`=>d=1/60`
And, `n =11`
∴ Sum of first 11 terms, `S_11=?`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_11=11/2[2xx1/15+(11-1)1/60]`
`=11/2[2/15+10/60]`
`=11/2[2/15+1/6]`
`=11/2[(4+5)/30]`
`=11/2xx9/30`
`=99/60=33/20`
`=>S_11=1\13/20`
Thus, sum of first 11 terms for the given Arithmetic Progression (AP)`=1\13/20` Answer
Question. (2) Find the sums given below:
(i) `7+10\1/2+14+...+84`
Solution:
Here, given, First term, `a=7`
Common difference, `d=10\1/2 -7`
`=>d=21/2-7=(21-14)/2`
`=>d=7/2`
`a_n=84`
`:. S_n=?`
Now, we know that, `a_n=a+(n-1)d`
`:. a_n=7+(n-1)7/2`
`=>84=7+(n-1)7/2`
`=>84-7=(n-1)7/2`
`=>(n-1)7/2=77`
`=>n-1=(77xx2)7`
`=>n-1=11xx2`
`=> n = 22+1`
`=> n = 23`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_23=23/2[2xx7+(23-1)xx7/2]`
`=>S_23=23/2[14+22xx7/2]`
`=23/2[14+77]`
`=23/2xx91`
`=2093/2=1046\1/2`
`=>S_23=1046\1/2`
Thus, Sum of the given Arithmetic Progression(AP) `=1046\1/2` Answer
(ii)`34+32+30+...+10`
Solution:
Here, given, First term, `a=34`
Common difference, `d = 32-34=-2`
`n^(th)` term, `a_n = 10`
`:. S_n = ?`
Now, we know that, `a_n=a+(n-1)d`
`:. a_n = 34+(n-1)xx(-2)`
`=> 10 = 34 + (n-1)xx(-2)`
`=>(n-1)(-2)= 10-34`
`=>(n-1)(-2)=-24`
`=> n-1=(-24)/(-2)`
`=>n-1 = 12`
`=> n = 12+1=13`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_13=13/2[2xx34+(13-1)xx(-2)]`
`=> S_13=13/2[68+12xx(-2)]`
`=>S_13=13/2[68-24]`
`=>S_13=13/2xx44`
`=>S_13=13xx22`
`=>S_13=286`
Thus, sum of the given term in Arithmetic Progression (AP) = 286 Answer
(iii) `-5+(-8)+(-11)+...+(-230)`
Solution:
Here, First term, `a=-5`
Common difference, `d=-8-(-5)=-3`
nth term `a_n = -230`
`:. S_n =?`
Now, we know that, `a_n=a+(n-1)d`
`:. -230 = -5+(n-1)xx(-3)`
`=>-230 +5 =(n-1)xx(-3)`
`=>n-1=(-225)/(-3)`
`=>n-1 = 75`
`=>n = 75+1=76`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`S_76 = 76/2[2xx(-5)+(76-1)xx(-3)]`
`=38[-10+75xx(-3)]`
`=38[-10-225]`
`=38xx(-235)`
`=>S_76= -8930`
Thus, sum of the given terms for Arithmetic Progression `= -8930` Answer
Question. (3) In an AP:
(i) Given, `a=5, d=3, a_n =50` find `n and S_n`
Solution:
Now, we know that, `a_n=a+(n-1)d`
`:. 50 = 5+(n-1)3`
`=> 50-5 =(n-1)3`
`=>(n-1)3=45`
`=>n-1 = 45/3=15`
`=> n = 15+1=16`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_16 = 16/2[2xx5+(16-1)3]`
`=8[10+45]=8xx55`
`=>S_16 = 440`
Thus, `n=16 and S_n=440` Answer
(ii) given, `a=7, a_13=35,` find `d and S_13`
Solution:
We know that, `a_n=a+(n-1)d`
`:. a_13 = 7+(13-1)d`
`=>35 = 7+12d`
`=> 12d = 35-7=28`
`=> d = 28/12=7/3`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_13 = 13/2[2xx7+(13-1)7/3]`
`=13/2[14+12xx7/3]`
`=13/2[14+28]`
`=13/2xx42 = 13xx21`
`=>S_13=273`
Thus, `d = 7/3 and S_13 = 273` Answer
(iii) given, `a_12 = 37, d=3` find `a and S_12`
We know that, `a_n=a+(n-1)d`
`:. a_12 = a+(12-1)3`
`=>37=a+33`
`:. a = 37-33=4`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_12 = 12/2[2xx4+(12-1)3]`
`=6[8+11xx3]`
`=6(8+33)=6xx41`
`=>S_12 = 246`
Thus, `a=4 and S_12 = 246` Answer
(iv) given, `a_3 = 15, S_10 = 125` find `d and a_10`
Solution:
We know that, `a_n=a+(n-1)d`
`:. a_3 = a+(3-1)d`
`=>15=a+2d` ------(i)
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_10 = 10/2[2a+(10-1)d]`
`=>125=5[2a+9d]`
`=>125/5=2a+9d`
`=>25=2a+9d` ------(ii)
Now, after multiplying equation (i) by 2 we get
`30=2a+4d` -----(iii)
Now, after subtracting equation (ii) from equation (iii) we get
`30-25=2a+4d-(2a+9d)`
`=>5 = 2a+4d-2a-9d`
`=>5 = -5d`
`:. d = -5/5 = -1`
Now, By putting the value of d in equation (i) we get
`15 = a + 2xx(-1)`
`=>15=a-2`
`:. a = 15+2 = 17`
Now, by using `a_n = a+(n-1)d` we get
`a_10 = 17+(10-1)(-1)`
`=>a_10 = 17+9(-1)=17-9`
`=>a_10 = 8`
Thus, `d = -1 and a_10 = 8` Answer
(v) given, `d=5, s_9 = 75`, find `a and a_9`
Solution:
Given, `d=5, s_9 = 75`
`:. a = ? and a_9=?`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_9 = 9/2[2a+(9-1)5]`
`=>75 = 9/2[2a+8xx5]`
`=>75 = 9/2[2a+40]`
`=>75 = 9/2xx2[a+20]`
`=>75 = 9[a+20]`
`=>75=9a+180`
`=>9a=75-180=-105`
`=>a=-105/9`
`=> a = -35/3`
We know that, `a_n=a+(n-1)d`
`:. a_9 = -35/3+(9-1)5`
`=-35/3+8xx5`
`=-35/3+40`
`=(-35+120)3`
`=>a_9=85/3`
Thus, `a = -35/3 and a_9 = 85/3` Answer
(vi) Given, `a=2, d=8, S_n = 90`, find `n and a_n`
Solution:
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_n = n/2[2xx2+(n-1)8]`
`=> 90 = n/2[4+8n-8]`
`=>90xx2=n[8n-4]`
`=>180=8n^2-4n`
`=>8n^2-4n-180=0`
`=>4(2n^2-n-45)=0`
`=>2n^2 -n -45=0`
`=>2n^2 -10n +9n-45=0`
`=>2n(n-5)+9(n-5)`
`=>(2n+9)(n-5)=0`
Now, if `2n+9=0`
`:. 2n = -9`
`=> n = -9/2`
And, if `n-5 = 0`
`:. n = 5`
After discarding the negative value of n, we have
`n = 5`
We know that, `a_n=a+(n-1)d`
`:. a_5 = 2+(5-1)8`
`=>a_5 = 2+4xx8=2+32`
`=>a_5 = 34`
Thus, `n = 5 and a_n = 34` Answer
(vii) Given, `a=8, a_n = 62, S_n =210`, find `n and d`
Solution:
Here, given, `a=8, a_n = 62, S_n =210`
`:. n = ? and d =?`
We know that, `a_n=a+(n-1)d`
`:. a_n = 8+(n-1)d`
`=>62 = 8+(n-1)d`
`=>(n-1)d = 62-8`
`=>(n-1)d = 54` ----(i)
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_n = n/2[2xx8+(n-1)d]`
`=>210 = n/2[16+(n-1)d]`
`=>(210xx2)/n= 16+(n-1)d`
`=>420/n = 16+(n-1)d`
After substituting the value of `(n-1)d` from equation (i), we get
`=>420/n = 16+54`
`=>420/n = 70`
`:. n = 420/70`
`=> n = 6`
Now, by putting the value of `n=6` in equation (i), we get
`(6-1)d= 54`
`=>5d= 54`
`:. d = 54/5`
Thus, `n = 6 and d = 54/5` Answer
(viii) Given, `a_n = 4, d = 2, S_n = -14`, find `n and a`
Solution:
We know that, `a_n=a+(n-1)d`
`:. 4 = a+(n-1)2`
`=>4=a+2n-2`
`=>a+2n = 4+2`
`=>a+2n = 6`
`=>a = 6-2n` ----(i)
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. -14 = n/2[2a+(n-1)2]`
`=>-14 = n/2xx2[a+(n-1)]`
`=>-14 = n [a+n-1]`
After putting the value of `a` from equation (i), we get
`-14 = n [6-2n +n-1]`
`=>-14 = n [5-n]`
`=>-14=5n - n^2`
`=>-n^2+5n+14=0`
`=>-(n^2-5n-14)=0`
`=>n^2 -5n -14 = 0`
`=> n^2 -7n+2n-14=0`
`=>n(n-7)+2 (n-7)=0`
`=>(n+2)(n-7)=0`
Now, if `n+2 = 0`
`:. n = -2`
And, if `n-7=0`
`:. n = 7`
After discarding the negative value of `n`, we have
`n = 7`
Now, by putting the value of `n=7` in equation (i), we get
`a=6-2xx7=6-14`
`=>a=-8`
Thus, `n = 7 and a = -8` Answer
(ix) Given, `a=3, n=8, S= 192`, find `d`
Solution:
Here, given, `a=3, n=8, and S= 192`
Therefore, `d=?`
We know that sum of all terms, `S= n/2(a+l)`
`:. 192 = 8/2(3+l)`
`=>192=4(3+l)`
`=>3+l = 192/4=48`
`=> l = 48-3=45`
This means 45 is the 8th term
Or, `a_8 = 45`
We know that, `a_n=a+(n-1)d`
`:. a_8 = 3+(8-1)d`
`=> 45 = 3+7d`
`=>7d = 45-3=42`
`:. d = 42/7`
`=> d = 6`
Thus, `d = 6` Answer
(x) Given, `l=28, S = 144` and there are total 9 terms, find `a`
Solution:
Here, given, `l=28, S = 144 and n=9`
We know that sum of all terms, `S= n/2(a+l)`
`=>S = 9/2(a+28)`
`=> 144 = 9/2 (a+28)`
`=>a+28 = (144xx2)/9`
`=>a+28 = 288/9`
`=> a +28 = 32`
`=> a = 32 - 28`
`=> a = 4`
Thus, `a = 4` Answer
Reference: