Question

how many terms of the AP 9,17,25 e.t.c must be taken to give a sum of 636

Answer 1

Number Series : 9,17,25, ...

⇒ a1 = 9

⇒ d = 17 - 9 = 8

Find the last term (an):

an = a1 + (n - 1 )d

an = 9 + 8(n - 1)

an = 9 + 8n - 8

an = 8n + 1

Find the number of terms (n):

sn = n/2 (a1 + an)

636 = n/2 (9 + an)

1272 = n(9 + an)

Since we found earlier that an = 8n + 1

1272 = n(9 + 8n + 1)

1272 = 9n + 8n² + n

8n² + 10n - 1272 = 0

4n² + 5n - 636 = 0

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

n = 12 or n = -53/4 (Rejected, since n cannot be negative)

Answer:  Adding 12 terms of the AP will get 636

Answer 2

Here is your solution

Given :-

a = 7  

d = 17 - 9=8   

Sn = 636

We know that

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

636 = (n/2)[2(9) + (n-1) 8]

(636 x 2) = n[18 + 8 n - 8]

(636 x 2) = n[10 + 8 n]

636 = n[5 + 4n]

636 = 5n + 4 n²

4 n² + 5 n - 636 = 0

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

 n = -53/4     

n = 12(positive only)

Hence ,

12 terms are need to give a sum of 636.