Question

Find the seventh term from the end of the series : $\sqrt{2},2,2\sqrt{2},....................,32.$

Answer 1

L= last term = 32
d= common difference= 2-√2
therefore, 7th term from the end = 32+(7-1)(-(2-√2))
[ because nth term from the end = l+(n-1)(-d)
= 32+6(-2+√2)
= 32-12+6√2
= 20+6√2