Solved Problems in Arithmetic Progression 54


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 an = 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 an = -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 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.

Solution: Let a be the 1st term and d the common difference.

Here a11 = a + 10d = 38….. (1)

a16 = 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

a31= a + 30 d

= -32 + 30 * 7

= -32 + 210 = 178

Hence, 31st term is 178.

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

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

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

Since an = ak = (n-k) d

an – a54 = (n-54) 12

132 = 12n – 54 * 12    …..(since an – a 54 = 132 given)

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

n= 11 + 54 = 65


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54 thoughts on “Solved Problems in Arithmetic Progression

  • pavan

    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