Class 10 Maths

# 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.

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; 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`