HOW MANY TERMS ARE NEEDED TO YIELD A SUM

Question 1 :

How many terms of the AP

9, 17, 25,..........

must be taken to give a sum of 636?  

Solution :

S_n =  636 

a  =  9, d  =  17 - 9  =  6

S_n  =  (n/2) [2a + (n - 1)d]

(n/2) [2(9) + (n - 1)8]  =  636

(n/2) [18 + 8n - 8]  =  636

(n/2) [10 + 8n]  =  636

n[5 + 4n]  =  636

5n + 4n²  =  636

4n² + 5n - 636  =  0

4n² - 48n + 53n - 636  =  0

4n(n - 12) + 53(n - 12)  =  0

(n - 12) (4n + 53)  =  0

n  =  12

By solving other factor (4n + 53), we get negative value for n, which is not admissible. 

So, 12 terms to be added to get the sum 636.

Question 2 :

The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and common difference. 

Solution :

Given that :

First term (a)  =  5, last term (l)  =  45 and S_n  =  400

S_n  =  (n/2) [a + l]

(n/2) [5 + 45]  =  400

(n/2) [50]  =  400

25n  =  400

n  =  400/25

n  =  16

Hence the given series consist of 16 terms.

Question 3 :

In an arithmetic sequence

60, 56, 52, 48,.......

starting from the first term, how many terms are needed so that  their sum is 368 ?

Solution :

Given that :

S_n  =  368

60, 56, 52, 48,.......

a = 60, d = 56 - 60 = -6

(n/2)[2a+(n-1)d]  =  368

  (n/2)[2(60)+(n-1)(-4)]  =  368           

  (n/2) (120-4n+4)  =  368           

  (n/2) (124 - 4n)  =  368           

n(124-4n)  =  368(2)

124n - 4n²  =  736

  4n²-124n+736  =  0

÷ by 4 = > n²-31n+184  =  0

(n-23)(n-8)  =  0 

   n  =  23, 8

Sum of 8 or 23 terms of arithmetic sequence is  368.

Question 4 :

the sum of first n terms of an arithmetic progression is 3n² + 5n. The series is

a)  8, 14, 20, 26, ...........      b)  8, 22, 42, 68, ..........

c)  22, 68, 114, ...........         d)  8, 14, 28, 44, ...........

Solution :

Sn = 3n² + 5n

When applying n = 1, we get the first term

S₁ = 3(1)² + 5(1)

= 3 + 5

S₁ = 8

Applying n = 2, we get

S₂ = 3(2)² + 5(2)

= 12 + 10

S₂ = 22

Here S₂ is the sum of first two terms.

a + a + d = 22

Where a = 8,

2a + d = 22

2(8) + d = 22

16 + d = 22

d = 22 - 16

d = 6

First term is 8 and common difference is 6.

8, (8 + 6), 14 + 6, .........

8, 14, 20..........

Question 5 :

The number of terms of the series 5 + 7 + 9 + ........... must be taken so that the sum may be 480

a)  20     b)  10     c)  15     d)  25

Solution :

5 + 7 + 9 + ...........

a = 5, d = 7 - 5 ==> 2

Sum of n terms = 480

(n/2)[2a+(n-1)d]  =  480

(n/2)[2(5) + (n - 1)2] = 480

(n/2) [10 + 2n - 2] = 480

(n/2) [8 + 2n] = 480

8n + 2n² = 480(2)

Dividing by 2, we get

4n + n² = 480

n² + 4n - 480 = 0

(n - 20) (n + 24) = 0

n = 20 and n = -24

The value of n cannot be negative. Then, the value of n is 20.

Question 6 :

If the sum of five terms of AP is 75. Find the third term of the series.

Solution :

S_n = (n/2)[2a+(n-1)d]

S₅ = (5/2)[2a+(5-1)d] = 75

(5/2)[2a + 4d] = 75

2a + 4d = 75(2/5)

2a + 4d = 30

Dividing by 2, we get

a + 2d = 15

Writing third term in terms of a and d, we get a + 2d = 15

So, the answer is 15.

Question 7 :

If 2 + 6 + 10 + 14 + 18 + .............. + x = 882

then the value of x

a)  78   b)  80     c)  82    d)  86

Solution  :

2 + 6 + 10 + 14 + 18 + .............. + x = 882

a = 2, d = 4, Sn = 882

S_n = (n/2)[2a+(n-1)d]

882 = (n/2)[2(2) + (n - 1) 4]

882 = (n/2) [4 + 4n - 4]

882 = (n/2) [4n]

882 = 2n²

Dividing by 2, we get

n² = 441

n² = 21²

So, the required number of terms is 21.

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