ICSE & CBSE Board Question Based on Arithmetic Progression of RS Aggarwal . Here the AP is given. you have to find which term of the given AP is more than its *n*th term. Question Number 20 of Exercise 11A of RS Aggarwal Solution

# Which term of the AP 5, 15, 25,… will be 130 more than its 31st term?

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AP is 5, 15, 25,…

First term = a = 5

Common difference = d = 15 – 5 = 10

Find 31st term:

a_31 = a + (n – 1)d

= 5 + (31 – 1) 10

= 5 + 30 x 10

= 305

Required term = 305 + 130 = 435

Now, say 435 be the nth term, then

a_n = a + (n – 1)d

435 = 5 + (n – 1)10

435 – 5 = (n – 1)10

n – 1 = 43

n = 44

The required term will be 44th term.