Arithmetic Progression
NCERT Exercise 5.3
Part 3
Question: 3 - In an AP:
(a) Given a = 5, d = 3, an = 50, find n and Sn.
Answer: Number of terms can be calculated as follows:
`a_n = a + (n – 1)d`
Or, `50 = 5 + (n – 1)3`
Or, `(n – 1)3 = 50 – 5 = 45`
Or, `n – 1 = 15`
Or, `n = 16`
Sum of n terms can be given as follows:
`S=n/2[2a+(n-1)d]`
`=(16)/(2)(2xx5+15xx3)`
`=8(10+45)`
`=8xx55=440`
Thus, n = 16 and sum = 440
(b) Given a = 7, a13 = 35, find d and S13.
Answer: Common difference can be calculated as follows:
`a_n = a + (n – 1)d`
Or, `35 = 7 + 12d`
Or, `12d = 35 – 7 = 28`
Or, `d = 7/3`
Sum of n terms can be given as follows:
`S=n/2[2a+(n-1)d]`
`=(13)/(2)(2xx7+12xx7/3)`
`=(13)/(2)(14+28)`
`=(13)/(2)xx42=273`
Thus, d = 7/3 and sum = 273
(c) Given a12 = 37, d = 3, find a and S12.
Answer: First term can be calculated as follows:
`a_n = a + (n – 1)d`
Or, `37 = a + 11 xx 3`
Or, `a = 37 – 33 = 4`
Sum of n terms can be given as follows:
`S=n/2[2a+(n-1)d]`
`=(12)/(2)(2xx4+11xx3)`
`=6(8+33)`
`=6xx41=246`
Thus, a = 4 and sum = 246
(d) Given a3 = 15, S10 = 125, find d and a10.
Answer: Sum of n terms can be given as follows:
`S=n/2[2a+(n-1)d]`
Or, `125=(10)/(2)(2a+9d)`
Or, `125=5(2qa+9d)``
Or, `2a+9d=25` -----------(1)
As per question; the 3rd term is 15, which means;
`a + 2d = 15` …… (2)
Subtracting equation (2) from equation (1), we get;
`2a + 9d – a – 2d = 25 – 15`
Or, a` + 7d = 10` ……. (3)
Subtracting equation (2) from equation (3), we get;
`a + 7d – a – 2d = 10 – 15`
Or, `5d = - 5`
Or, `d = - 1`
Substituting the value of d in equation (2), we get;
`a + 2(- 1) = 15`
Or, `a – 2 = 15`
Or, `a = 17`
Now, tenth term can be calculated as follows;
`a_(10) = a + 9d`
`= 17 – 9 = 8`
Thus, d = - 1 and 10th term = 8
(e) Given d = 5, S9 = 75, find a and a9.
Answer: Sum of n terms can be given as follows:
`S=n/2[2a+(n-1)d]` Or, `75=9/2(2a+8xx5)`
Or, `75=9/2(2a+40)`
Or, `2a+40=75xx2/9`
Or, `2a+40=(50)/(3)`
Or, `2a=(50)/(3)-40`
Or, `2a=(50-120)/(3)`
Or, `a=-(70)/(3)xx1/2`
Or, `a=-(35)/(3)`
Now, the 9th term can be calculated as follows:
`S=n/2[2a+(n-1)d]`
`a_9=a+8d`
`=-(35)/(3)+40=(85)/(3)`
(f) Given a = 2, d = 8, Sn = 90, find n and an.
Answer: Sum of n terms can be given as follows:
`S=n/2[2a+(n-1)d]`
Or, `90=n/2[2xx2+(n-1)8]`
Or, `90=n[2+(n-1)4]`
Or, `90=n(2+4n-4)`
Or, `90=n(4n-2)`
Or, `90=4n^2-2n`
Or, `4n^2-2n-90=0`
Or, `2n^2-n-45=0`
Or, `2n^2-10n+9n-45=0`
Or, `2n(n-5)+9(n-5)=0`
Or, `(2n+9)(n-5)=0`
Hence, `n=-9/2` and `n = 5`
Discarding the negative value, we have n = 5
Now, 5th term can be calculated as follows:
`a_5 = a + 4d`
`= 2 + 4 xx 8`
`= 2 + 32 = 34`
(g) Given a = 8, an = 62, Sn = 210, find n and d.
Answer: Sum of n terms can be given as follows:
`S=n/2[2a+(n-1)d]`
Or, `210=n/2[a+a+(n-1)d]`
Or, `420=n(8+62)`
Because `a + (n – 1)d = a_n`
Or, `420 = n xx 70`
Or,`n = 6`
Now, d can be calculated as follows:
`a_6 = a + 5d`
Or, `62 = 8 + 5d`
Or, `5d = 62 – 8 = 54`
Or, `d = (54)/(5)`
(h) Given an = 4, d = 2 Sn = -14, find n and a.
Answer: We know;
`S=n/2[2a+(n-1)d]`
Or, `-14=n/2[a+a+(n-1)d]`
Since, `a+(n-1)d=a_n`
Hence, `-28=n(a+4)`
Or, `n=(-28)/(a+4)` -------- (1)
We know;
`a_n = a + (n – 1)d`
Or, `4 = a + (n – 1)2`
Or, `4 = a + 2n – 2`
Or, `a + 2n = 6`
Or, `2n = 6 – a`
Or, `n = (6 – a)/(2)` ……. (2)
From equations (1) and (2);
`-(28)/(a+4)=(6-a)/(2)`
Or, `-56=(6-a)(a+4)`
Or, `24+6a-4a-a^2=-56`
Or, `24+2a-a^2=-56`
Or, `24+56+2a-a^2=0`
Or, `a^2-2a-80=0`
Or, `a^2-10a+8a-80=0`
Or, `a(a-10)+8(a-10)=0`
Or, `(a+8)(a-10)=0`
Hence, a = - 8 and a = 10
Since an is smaller than 10 and d has positive value, so we need to take a = - 8
Using this, we can find the number of terms as follows:
`a_n = a + (n – 1)d`
Or, `4 = - 8 + (n – 1)2`
Or, `(n – 1)2 = 4 + 8 = 12`
Or, `n – 1 = 6`
Or, `n = 7`
Thus, n = 7 and a = - 8
(i) Given a = 3, n = 8, S = 192, find d.
Answer: We know;
`S=n/2[2a+(n-1)d]`
Or, `192=8/2(2xx3+7d)`
Or, `192=4(6+7d)`
Or, `6+7d=48`
Or, `7d=42`
Or, `d=42÷7=6`
(j) Given l = 28, S = 144, and there are total 9 terms, find a.
Answer: We know;
`S=n/2[2a+(n-1)d]`
Or, `144=9/2(a+a_n)`
Or, `288=9(a+28)`
Or, `9a+252=288`
Or, `9a=288-252=36`
Or, `a=36÷9=4`