Here are some questions on arithmetic progression and their solutions

**Question 1**: Which term of the A.P. 3,8,13 …is 78?

**Solution**: Here a_{n }= a + (n – 1) d = 78

a= 3, d = 8- 3 = 5

Therefore,

3 + (n -1) (5) = 78

(n-1) * 5 = 78 – 3 = 75

n – 1 = 75/5 = 15

n = 15 + 1 = 16

Hence a _{16 }(sixteenth term) is 78.

**Question 2**: Is – 150 a term of the series 11, 8, 5, 2,…?

**Solution**: Here, a = 11, d = 8-11 = -3. Let a_{n }= -150

Therefore,

a + (n-1) d = -150

11+ (n-1) (-3) = -150

-3 (n-1) = -150 – 11 = -161

(n-1) = + 161/3 = 53 2/3 which is not an integral number.

Since number of terms can never be a fraction

Hence, -150 is not a term of the given series.

**Question 3**: Find the 31^{st} term of an A.P. whose 11^{th} term is 38 and the 16^{th} term is 73.

**Solution:** Let a be the 1^{st} term and d the common difference.

Here a_{11 }= a + 10d = 38**….. (1)**

a_{16 }= a + 15d = 73…..** (2)**

Subtracting (2) from (1), we get

a + 10d – 1 – 15 d = 38 – 73

-5 = – 35

d = 7

Putting d= 7 in (1), we get

a + 10 * 7 = 38

a= 38 – 70 = – 32

a_{31}= a + 30 d

= -32 + 30 * 7

= -32 + 210 = 178

Hence, 31^{st} term is 178.

**Question 4**: Which term of the A.P. 3, 15, 27, 39 … will be 132 more than its 54^{th} term?

**Solution:** Given series is 3, 15, 27, 39…

Here, a =3, d= 15-3 = 12

Since a_{n }= a_{k }= (n-k) d

a_{n }– a_{54 }= (n-54) 12

132 = 12n – 54 * 12 …..(since a_{n }– a _{54 }= 132 given)

12 n = 132 + 54 * 12 = 12 (11+ 54)

n= 11 + 54 = 65

if the sum of the 8th and 9th terms of an arithmetic progression is 72 and the 4th term is -6, find the common difference.

first term a And 9th term a+8d sum of these terms is 2a+8d=24 ie a+4d=12 sum of the first 9 numbers is 9/2(2a+8d) ie 9/2(2(a+4d))ie 9/2(2.12)ie 108

@RAHUL

first n odd numbers are

1,3,5……….n

Sn=1+3+5……+n

a=1, d=2

Sn=n/2(2a+(n-1)d)

Sn=n/2(2*1+(n-1)2)

Sn=n/2(2+2n-2)

Sn=n/2(2n)

Sn=n*n

Sn=n square

@qwerty

100 and 200 are divisible of 5

x1=100

xn=200

d=5

n=(xn-x1/d)+1 = (200-100/5)+1 =(100/5)+1=20+1=21

sum= (1/2)n(xn+x1)=21/2(300)=(21)150=3150