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 nth term. Question Number 20 of Exercise 11A of RS Aggarwal Solution
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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.