Q30 of 49 Page 1

The 24th term of an AP is twice its 10th term. Show that its 72nd term is 4 times its 15th term.

Let the nth term of AP is denoted by an.


According to question-


a24 = 2×a10


⇒ a+(24-1)d = 2[a+(10-1)d]


[∵ an = a+(n-1)d; where, a=first term and d=common difference]


⇒ a+23d = 2a+18d


⇒ a = 5d ....(1)


Now,


L.H.S = a72 = a+(72-1)d


= 5d+71d [from(1)]


= 76d


R.H.S = 4×a15 = 4[a+(15-1)d]


= 4[5d+14d] [from(1)]


= 4× 19d


= 76d


∴ L.H.S = R.H.S ....Proved


Thus, 72nd term of given A.P is 4 times its 15th term.


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