Arithmetic Progression NCERT Exercise 5.3 part three Class Ten Mathematics

Arithmetic Progression

Exercise 5.3 Part 3

Question: 3 - In an AP:

(a) Given a = 5, d = 3, an = 50, find n and Sn.

Solution: 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.

Solution: 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.

Solution: 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.

Solution: 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.

Solution: 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.

Solution: 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.

Solution: 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.

Solution: 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.

Solution: We know;

sum of n terms

`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.

Solution: 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`



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