HOW MANY TERMS TO BE ADDED TO GET THE GIVEN SUM OF AP

Problem 1 :

How many terms of the series 32 + 36 + 40 + … when added together will give 200 ?

Solution :

32 + 36 + 40 + ….....

Here, a = 32

d = 36 – 32

d = 4

S_n = n/2[2a + (n – 1)d]

200 = n/2[2 × 32 + (n – 1)4]

200 × 2 = n[64 + 4n – 4]

400 = n[60 + 4n]

400 = 60n + 4n²

4n² + 60n – 400 = 0

Dividing 4 on each sides.

n² + 15n – 100 = 0

(n – 5) (n + 20) = 0

n = 5 (or) n = -20

So, required numbers is added is 5.

Problem 2 :

How many terms of the series 550 + 575 + 600 + … when added together will give 4375 ?

Solution :

Given, 550 + 575 + 600 + …

Here, a = 550

d = 575 – 550

d = 25

S_n = (n/2)[2a + (n – 1)d]

4375 = (n/2)[2 × 550 + (n – 1)25]

4375 = n/2[1100 + 25n – 25]

8750 = n[1075 + 25n]

8750 = 1075n + 25n²

25n² + 1075n – 8750 = 0

Dividing 25 on each sides.

n² + 43n – 350 = 0

(n – 7) (n + 50) = 0

n = 7 (or) n = -50

So, required numbers is added is 7.

Problem 3 :

Find the number of terms of AP 20, 30, 40, … gives sum 3500.

Solution :

Given, AP is 20, 30, 40, …

Here, a = 20

d = 30 – 20 ==> d = 10

S_n = n/2[2a + (n – 1)d]

3500 = n/2[2 × 20 + (n – 1)10]

7000 = n[40 + 10n – 10]

7000 = n[30 + 10n]

7000 = 30n + 10n²

Dividing by 10, we get

n² + 3n – 700 = 0

(n – 25) (n + 28) = 0

n = 25 (or) n = -28

Problem 4 :

Find the number of terms of AP 177, 174, 171, … gives sum 5310.

Solution :

Here, a = 177

d = 174 – 177  ==> d = -3

S_n = n/2[2a + (n – 1)d]

5310 = n/2[2 × 177 + (n – 1) × -3]

10620 = n[354 – 3n + 3]

10620 = 354n – 3n² + 3n

10620 = 357n – 3n²

3n² – 357n + 10620 = 0

Dividing by 3, we get

n² – 119n + 3540 = 0

(n – 60) (n – 59) = 0

n = 60 (or) n = 59

Problem 5 :

Find the number of terms of AP 15, 13.5, 12, … gives sum 82.5.

Solution :

a = 15

d = 13.5 – 15  ==>  d = -1.5

S_n = n/2[2a + (n – 1)d]

82.5 = n/2[2 × 15 + (n – 1) × (-1.5)]

82.5 = n/2[30 + (-1.5n + 1.5]

165 = n[30 – 1.5n + 1.5]

165 = 30n – 1.5n² + 1.5n

165 = 31.5n – 1.5n²

1.5n² – 31.5n + 165 = 0

Dividing 10 on each sides.

15/10n² – 315/10n + 165 = 0

3/2n² – 63/2n + 165 = 0

Multiplying 2 on each sides.

3n² – 63n + 330 = 0

n² – 21n + 110 = 0

(n – 10) (n – 11) = 0

n – 10 = 0 (or) n – 11 = 0

n = 10 (or) n = 11

Problem 6 :

How many terms of the AP -15, -13, -11, … are needed to make the sum -55 ?

Solution :

Here, a = -15, d = -13 + 15 => d = 2

S_n = n/2[2a + (n – 1)d]

-55 = n/2[2 × (-15) + (n – 1) × 2]

-55 = n/2[-30 + 2n – 2]

-110 = n[-32 + 2n]

-110 = -32n + 2n²

2n² – 32n + 110 = 0

Dividing 2 on each sides.

n² – 16n + 55 = 0

(n – 5) (n – 11) = 0

n = 5 (or) n = 11 

So, 11 (or) 5 terms are needed to make the sum is -55.