Question

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

Answer 1

Answer:

n=313/4

Step-by-step explanation:

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

a=9,d=17-9=8, Sn=636

Sn=2a+(n-1)d

636=2(9)+(n-1)8

636=18+8n-8

636=10+8n

636-10=8n

8n=626

n=626/8

n=313/4

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

AP - 9 , 17 , 25 , ......

★Here,

a = 9

d = 17 - 9

= 8

$\tt{\rightarrow S_{n}=636}$

$\tt{\rightarrow S_{n}=\dfrac{n}{2}[2\times 9+(n-1)8]}$

$\tt{\rightarrow 636=\dfrac{n}{2}[18+8n-8]}$

$\tt{\rightarrow 636=\dfrac{n}{2}[10+8n]}$

636 = n[5 + 4n]

636 = 4n² + 5n

4n² + 5n - 636 = 0

$\fbox{Splitting\;the\;middle\;term}$

4n² + 53n - 48n - 636 = 0

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

(n - 12) = 0

n = 12

(4n + 53) =0

$\tt{\rightarrow n=-\dfrac{53}{4}}$

★Rejection (-) value

★Therefore ,

$\large{\fbox{12th\;term\;needed\;to\;get\;sum\;636}}$