How many terms must added in an arithmetic sequence whose first term is 6 and whose common difference is 5 to obtain a sum of 5220?

1 Answer
Oct 10, 2017

#45# terms to be added to obtain a sum of #5220#

Explanation:

Sum formula of arithmatic sequenc is #s_n=n/2{2a+(n-1)d}#

When 1st term is #a=6# , common difference is #d=5# , Sum of

the sequence is #S_n=5220 ; n= ? :. #

#5220=n/2{2*6+(n-1)5}# or #5220=n/2{12+(n-1)5}#

multiplying by #2# on both sides we get

# n{12+(n-1)5} = 10440 or 12n+5n^2-5n -10440=0# or

# 5n^2+7n - 10440=0 ; a=5 ; b= 7 ; c= -10440#

#[ax^2+bx+c=0] :. n= (-b+- sqrt (b^2-4ac))/(2a)# or

# n= (-7+- sqrt (7^2-4*5*(-10440)))/(2*5)# or

# n= (-7+- sqrt (49+208800))/10# or

# n= (-7+- 457)/10 or n= -46.4 , n=45 ; n# can not be negative

#:. n=45 :. 45# terms to be added to obtain a sum of #5220# [Ans]