Question

how many terms of ap 25,22,19... are needed to give the sum 116​

Answer 1

Answer:

Hope this helps:

Step-by-step explanation:

Here a=25,d=−3,S  

n

​  

=116  

we know

S  

n

​  

=  

2

n

​  

[2a+(n−1)d]

116=  

2

n

​  

[2×25+(n−1)(−3)]

⇒232=n(50−3n+3)

⇒232=n(53−3n)

⇒232=53n−3n  

2

 

⇒3n  

2

−53n+232=0

⇒3n  

2

−24n−29n+232=0

⇒3n(n−8)−29(n−8)=0

⇒(3n−29)(n−8)=0

∴n=8,n  



​  

=  

3

29

​  

 (fraction)

∴ Number of terms required = '8'

t  

n

​  

=a+(n−1)d

⇒t  

n

​  

=25−1(8−1)(−3)

⇒t  

n

​  

=25−21=4

Hence, Last term is '4'