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Question

# If 3+14(3+P)+142(3+2P)+143(3+3P)+…=8, then the value of P is

A
9
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B
15
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C
21
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D
12
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Solution

## The correct option is A 9Given series is 3+14(3+P)+142(3+2P)+143(3+3P)+… It can be written as 1⋅3+14(3+P)+142(3+2P)+143(3+3P)+… So, the given series is AGP with a=3, r=14, d=p For |r|<1,n→∞ S∞=a1−r+dr(1−r)2 ⇒31−14+p⋅14(1−14)2=8 ⇒4+4p9=8 ⇒p=9 Alternate Solution: 3+14(3+P)+142(3+2P)+143(3+3P)+…=8 ⇒3(1+14+142+…)+P(14+242+…)=8 …(1) Now Let x=14+242+… ⇒x4=142+243+… Then x−x4=14+142+143+… ⇒x=43(14+142+143+…) From equation (1) 3(1+14+142+…)+P⋅43(14+142+143+…)=8 ⇒3⋅11−14+P⋅43⋅141−14=8 ⇒4+4P9=8 ⇒P=9

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