Question

For an A.P, if t1=5,tn=45,Sn=125,find n

Answer 1

Step-by-step explanation:

we have know

first term of AP (a) =5

last term of AP (an) =45

sum of n terms (sn) =125

so

a+(n-1)d=an

5+(n-1)d=45

(n-1)d=45-5

(n-1)d=40

d=40/(n-1) ......(1)

and

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

125=n/2[(2×5)+(n-1)d]

125=n/2[10+(n-1)d]

125×2=n[10+(n-1)d]

250=n[10+(n-1)×40/(n-1)]

250=n[10+40]

250=50n

250/50=n

5=n