An important question from arithmetic progression chapter as it was already asked in previous year paper of 2012 in which we have been asked to find the nth term, if the 17th term of AP is 5 more than twice its 8 term and if the 11th term of the AP is 43.

Book – RS Aggarwal, class 10, chapter 5A, question no 37

we know that the nth term of the arithmetic progression is given by a+(n−1)d

Given that the 17th term is 5 more than twice the 8th term.

Therefore, 5+17thterm=2(8thterm)

⟹a+(17−1)d=2(a+(8−1)d)+5

⟹a+16d−5=2a+14d

⟹a−2d=−5 ——-(1)

Given that the 11th term is 43

Therefore, a+(11−1)d=43

⟹a+10d=43 ——(2)

subtracting (1) from (2) we get

(a+10d)−(a−2d)=43−(−5)

⟹12d=48

⟹d=4

⟹a=3

Therefore, the nth term is 3+(n−1)4=4n−1