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 theÂ nthÂ term of the arithmetic progression is given byÂ a+(nâˆ’1)d

Given that theÂ 17thÂ term isÂ 5Â more than twice theÂ 8thÂ term.

Therefore,Â 5+17thterm=2(8thterm)

âŸ¹a+(17âˆ’1)d=2(a+(8âˆ’1)d)+5

âŸ¹a+16dâˆ’5=2a+14d

âŸ¹aâˆ’2d=âˆ’5Â ——-(1)

Given that theÂ 11thÂ term isÂ 43

Therefore,Â a+(11âˆ’1)d=43

âŸ¹a+10d=43Â ——(2)

subtracting (1) from (2) we get

(a+10d)âˆ’(aâˆ’2d)=43âˆ’(âˆ’5)

âŸ¹12d=48

âŸ¹d=4

âŸ¹a=3

Therefore, theÂ nthÂ term isÂ 3+(nâˆ’1)4=4nâˆ’1