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Show that the sequence $space space open curly brackets straight t subscript straight n close curly brackets$ defined by $straight t subscript straight n space equals space 2 straight n squared plus straight n plus 1$ is not an A.P.

305 Views

12.

Which term of an A.P. 5, 2, -1, ......... is -22?

1001 Views

13.

Which term of the sequence 12 + 8i, 11 + 6i, 10+41i, ....... is

(i) purely real (ii) purely imaginary?

819 Views

14.

Which term of the sequence $space space 24 comma space 23 1 fourth comma space 22 1 half comma space 21 3 over 4 comma space space.....$ is the first negative term?

384 Views

15.

Determine the number of terms in an A.P. 3, 8, 13, .......253. Also, find 12th term from the end.

509 Views

16.

Find the 2nd term and the rth term of an A.P. whose 6th term is 12 and 8th term is 22.

484 Views

17.

Determine K, so that K + 2, 4K – 6 and 3K – 2 are three consecutive terms of an A.P.

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Long Answer Type

18.
In an A.P., the mth term is $1 over straight n$ and the nth term is $1 over straight m comma$ find the (mn)th term.

Let a be the first term and d be the c.d.

 $space space space space space space straight t subscript straight m space equals space 1 over straight n space space space space space space space space space rightwards double arrow space space space space space space space space space space space straight a space plus space left parenthesis straight m space minus space 1 right parenthesis straight d space equals space 1 over straight n$ ...(i)

 $space space straight t subscript straight n space equals space 1 over straight m$ $rightwards double arrow space space straight a space plus space left parenthesis straight n space minus space 1 right parenthesis straight d space equals space 1 over straight m$ ... (ii)

Subtracting (ii) from (i), we get

 (m - 1)d - (n - 1)d = $1 over straight n minus space 1 over straight m$ or (m - 1 - n + 1)d = $fraction numerator straight m minus straight n over denominator mn end fraction$

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Using this value of d in (i), we get

 $straight a plus left parenthesis straight m minus 1 right parenthesis 1 over mn space equals space 1 over straight n$ or $space space straight a plus straight m over mn minus 1 over mn space equals space 1 over straight n$

or $straight a plus 1 over straight n minus 1 over mn equals 1 over straight n$ or $straight a space equals space 1 over mn$

 $space space straight t subscript mn space equals space straight a plus space left parenthesis mn minus 1 right parenthesis straight d space equals space 1 over mn plus left parenthesis mn minus 1 right parenthesis 1 over mn space equals space fraction numerator 1 plus mn minus 1 over denominator mn end fraction space equals space mn over mn space equals space 1$