Question

How many terms of the A.P 9, 17, 25... must be added together to get 636 as the sum

Answer 1

the AP 9, 17, 25 .....
Given:  a = 9 & d= 17-9 = 8  sum= 636
the formula to find the Sum of AP
  Sn = n/2(2a + (n-1)d)
  636 = n/2(2*9 + (n-1) 8)
 636 * 2 = n( 18 + 8n - 8)
1272 = n( 10 + 8n)
10n + 8n^2 = 1272
8n^2 +10n - 1272 =0
for finding n formula   = -b + sq (b^2 - 4ac)
                                        -------------------------
                                                    2a
                              n   =  [ -10 + sq(100 - 4*8*(-1272))]/2*8
                                   = [-10 + sq(100 + 40704)]/16
                                   = [-10 + sq(40804)] / 16
                                   = (-10 + 202)/16
                                   = (-10+202)/16    &  (-10-202)/16
                                   = 12  & -13.25

 -13.25 does not exist
So  n=12 is the solution

Answer 2

AP= 9,17 ,25..... Sn=636 ,a=9 ,d= 17-9 =8
as we know the sum formula in arithmetic progession,
Sn=n/2 [2a + [n-1] d]
putting the values ,we get
636 = n/2 [2 x 9 +(n-1) 8]
636 = n/2 [18 +8n -8]
⇒636= n/2 x 2 [9 +4n -4]
⇒636 =n[9+4n-4]
⇒636 =9n+4n²-4n
⇒636 =5n+4n²
⇒4n²+5n-636 =0