This is the Important question based on Arithmetic progression Chapter of R.S Aggarwal book for ICSE & CBSE

Board.

Here the 2 terms of AP is given. you have to find the first term, common difference and the AP

This is the Question Number 16 Exercise 11 A of RS Aggarwal Solution.

Deepak BoraNewbie

# The 7th term of an AP is -4 and its 13th term is -16 Find the AP.

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Let us say a be the first term and d be the common difference of an AP

a

_{n}= a + (n – 1)da

_{7}= a + (7 – 1)d= a + 6d = -4 …………(1)

And a

_{13}= a + 12d = -16 …..…..(2)Subtracting equation (1) from (2), we get

6d = -16 – (-4) = -12

From (1), a + 6d = -4

a + (-12) = -4

a = -4 + 12 = 8

a = 8, d = -2

AP will be 8, 6, 4, 2, 0, ……