Question

how many terms of the AP : 9, 17,25,... must be taken to give sum of 636 ?​

Answer 1

Answer:

ap = 9,17,25

Step-by-step explanation:

a= 9

d= 17-9=8

d=8

an= 636

an= a+(n-1)d

636= 9+(n-1)8

636= 9+8n-8

636-1=8n

636=8n

n= 635/8

Answer 2

Given:-

• The AP , 9, 17 ,25 ...
• A sum of 636

To find:-

• Find the nth term..?

Solutions:-

• Let the be n term of the Ap.

=> a = 9

=> d = a2 - a1 = 17 - 9 = 8

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

=> 636 = n/2 [2 × 9 + (n - 1) 8]

=> 636 = n/2 [18 + 8n - 8]

=> 636 = n/2 [10 + 8n]

=> 636 = n/2 × 2 [5 + 4n]

=> 636 = n [5 + 4n]

=> 636 = 5n + 4n²

=> 4n² + 5n - 636 = 0

=> n² + 53n - 48n - 636 = 0

=> n(4n + 53) - 12(4n + 53) = 0

=> (n - 12) (4n + 53)

=> n - 12 = 0 or 4n + 53 = 0

=> n = 12 or n = -53/4

• n is cannot be negative and fraction.

=> n = 12

Hence, the nth term of Ap is 12.