If (m + l)th term of an A.P. is twice the (n + l)th term. Prove that (3m + l)th term is twice the (m + n + l)th term.

Solution

Let a be the first and d be the common difference of an A.P. Since (m + 1)th term of an A.P. is twice the (n + 1)th term.

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or a + (m + 1 - 1)d = 2[a + (n + 1 - 1)d]
or a + md = 2a + 2nd
or a = (m - 2n)d
Now, $straight t subscript 3 straight m plus 1 end subscript space equals space straight a space plus space left parenthesis 3 straight m space plus space 1 minus 1 right parenthesis straight d space equals space straight a plus space 3 md space equals space left parenthesis straight m minus 2 straight n right parenthesis straight d space plus space 3 md space equals space left parenthesis straight m minus 2 straight n plus 3 straight m right parenthesis straight d$
 $equals left parenthesis 4 straight m minus 2 straight n right parenthesis straight d space equals space 2 left parenthesis 2 straight m minus straight n right parenthesis straight d$
$rightwards double arrow$ $straight t subscript 3 straight m plus 1 end subscript space equals space 2 left parenthesis 2 straight m minus straight n right parenthesis straight d$ ...(i)
Now, $straight t subscript straight m plus straight n plus 1 end subscript space equals space straight a plus left parenthesis straight m plus straight n plus 1 minus 1 right parenthesis straight d space equals space left parenthesis straight m minus 2 straight n right parenthesis straight d space plus space left parenthesis straight m plus straight n right parenthesis straight d$
 $equals space left parenthesis straight m minus 2 straight n space plus space straight m plus space straight n right parenthesis straight d$
or $space space space space space straight t subscript straight m plus straight n plus 1 end subscript space equals space left parenthesis 2 straight m space minus space straight n right parenthesis straight d$ ...(ii)