how many terms of the AP:9,17,25,.........must be taken to give a sum of 636?
Answers
Step-by-step explanation:
a=9,d=17-9=8
Sn=636
Sn=n/2[2a+(n-1) d]
n/2[2(9)+(n-1) 8]=636
n/2[18+18n-8]=636
n[4n+5]=636
4n2+5n-636=0
a=4,b=5,c=-636
D=(5)2-4×4×(-636)
=25+10176=10201
n=-b+-√D/2a
=-5+-√10201/2×4=-5+-101/8
=-106/8 ,96/8
-53/4 or 12
n=-53/4 rejected
n=12 answer
Answer:
Step-by-step explanation:
Solution :-
Given series, 9, 17, 25, .....
Here, a = 9, d = 8, S(n) = 636
Since, S(n) = n/2[2a + (n - 1)d]
⇒ 636 = n/2[18 + (n - 1)8]
⇒ 636 = n/2[9 + (n - 1)4]
⇒ 636 = n(9 + 4n - 4)
⇒ 636 = n(5 + 4n)
⇒ 636 = 5n + 4n²
⇒ 4n² + 5n - 636 = 0
⇒ 4n² + 53n - 48n - 636 = 0
⇒ 4n² - 48n + 53n - 636 = 0
⇒ 4n(n - 12) + 53(n - 12) = 0
⇒ (4n + 53) (n - 12) = 0
⇒ n = - 53/4, 12 (Neglecting negative once)
⇒ n = 12
n is a natural number, n = 12
Hence, 12 terms are required to give sum 636.