In this Question you have to find the value of p which makes three consecutive terms of an AP which is given.
This is the Important question based on Arithmetic progression Chapter of R.S Aggarwal book for ICSE & CBSE
Board.
This is the Question Number 3 Of Exercise 11 D of RS Aggarwal Solution.
Deepak BoraNewbie
For what value of p are (2p+1),13,(5p-3) three consecutive terms of an AP.
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If these terms are in AP then,
a₂ – a₁ = a₃ – a₂ = ………….
a₂ =13
a₁ = 2p + 1
a₃ = 5p – 3
⇒ 13 – (2p + 1) = (5p – 3) – 13
⇒ 13 – 2p – 1 = 5p – 3 – 13
⇒ – 2p – 5p = – 3 – 13 – 13 + 1
⇒ – 7p = – 29 + 1
⇒ – 7p = -28
⇒ 7p = 28
⇒ p = 28/7
⇒ p = 4
So, for the value of p = 4, these three consecutive terms are in AP