Q36 of 40 Page 1

If 4 times the 4th term of an AP is equal to 18 times the 18th term, then find the 22nd term.

OR


How many terms of the AP: 24, 21, 18, ….. must be taken so that their sum is 78?


Given 4a4=18a18, a22=7

2(a+3d)=9(a+17d)


2a+6d=9a+153d


7a+197d=0


7(a+21d)=0


a+21d=0


∴ a22=0


OR


Given: 24,21,18, ….sn=78


A=24, d=21-24=-3


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


78=n/2 [2(24)+(n-1)(-3)]


156=n[48-3x+3]


156=n[51-3x]


156=51n-3n2


3x2-51n+156=0


X2-17n+52=0


X2-13n-4n+52=0


n(n-13)-4(n-13)=0


(n-13)(n-4)=0


n=13, n=4


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