HOW MANY TERMS OF THE ARITHMETIC SEQUENCE MUST BE ADDED

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. 

Hence, 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 :

The first and last term of an AP are 17 and 350 respectively.If the common difference is 9, how many terms are there and what is their sum?  

Solution :

Given that :

First term (a)  =  17, last term (l)  =  350 and common difference (d)  =  9 .

S_n  =  (n/2)[a + l]

By applying the values of a and l, we get

S_n  =  (n/2)[17 + 350]

S_n  =  (n/2)(367)  ----(1)

S_n  =  (n/2)[2a + (n - 1)d]

S_n  =  (n/2)[2(17) + (n - 1)(9)]

S_n  =  (n/2)[34 + 9n - 9]

S_n  =  (n/2)[25 + 9n]  -----(2)

(1)  =  (2)

(n/2)(367)  =  (n/2)[25 + 9n] 

25 + 9n  =  367

9n  =  367 - 25

9n  =  342

Divide each side by 9, we get

 n  =  342/9  =  38

So, there are 38 terms in the sequence. In order to find their sum, let us apply the value of n in (1).

S_n  =  (n/2)(367) 

S₃₈  =  (38/2)(367) 

  =  19(367)

S₃₈  =  6973

Hence, the required sum is 6973.

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