Question

for an A.p the first term is 4 and them is
31 the sum of All the term
last
is
420 find the value of n


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Answer 1

Answer:

Answer

Sn=2n{2a+(n−1)d}

a+(n−1)d=31⇒a=31−4(n−1)

∴136=2n{2[31−4(n−1)]+(n−1)d}

n[70−8n+4n−4]=272

n(66−4n)=272

4n2−66n+272=0

2n2−33n+136=0

2n(n−8)−17(n−8)=0

(2n−17)(n−8)=0

∴n=8