HOW MANY TERMS ARE NEEDED TO YIELD A SUM

We have to consider the given sum as Sn. Instead of Sn, we may use one of the formulas given below.

Sn  =  (n/2) [a + l] (or)

Sn  =  (n/2) [2a + (n - 1)d]

a = first term, d = common difference and n = number of terms.

Question 1 :

How many terms of the AP

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

must be taken to give a sum of 636?  

Solution :

Sn =  636 

a  =  9, d  =  17 - 9  =  6

Sn  =  (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 Sn  =  400

Sn  =  (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 :

Sn  =  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.

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