An Important Question of RS Aggarwal book Based on Arithmetic Progression For ICSE & CBSE Board

Question Number 19 of Exercise 11A of RS Aggarwal Solution

Here the AP is given. you have to find which term of the given AP is more than its *n*th term

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# Which term of the AP 3,8,13,18, ….. will be 55 more than its 20th term?

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Given AP is 3, 8, 13, 18,…

First term = a = 3

Common difference = d = 8 – 3 = 5

And n = 20 and a_20 be the 20th term, then

a_20 = a + (n – 1)d

= 3 + (20 – 1) 5

= 3 + 95

= 98

The required term = 98 + 55 = 153

Now, 153 be the nth term, then

a_n = a + (n – 1)d

153 = 3 + (n – 1) x 5

153 – 3 = 5(n – 1)

150 = 5(n – 1)

n – 1 = 30

n = 31

Required term will be 31st term.