Question

the sum of n terms of the series 25,22,19,16... is 116. find the number of terms and the last term​

Answer 1

Answer:

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

a=25

d= -3

116=n/2(50+(n-1) (-3))

116*2=(50+(n-1) (-3))

232=(50+(n-1) (-3) ) n

232= 50n-3n(n-1)

232=50n-3n^2+3n

232=53n-3n^2

3n^2-53n+232=0

solve the equation.

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