Question

in a h.p , if 4th term is 1/9 and 13th term is 1/27, then the first term is​

Answer 1

Answer:

Given:-

• ${4}^{th} term \: = \frac{1}{9}$
• ${13}^{th} \: term \: = \frac{1}{27}$

To find:-

• first term of A.p

Solution:-

here,

$t_{n} = a + (n - 1) \times d \\ \\t_{4} = a + (4 - 1) \times d \\ \\ \frac{1}{9} = a + 3d \: \: ...........(1) \\ \\ for \: 2nd \: \: condition \\ \\ t_{n} = a + (n - 1) \times d \\ \\ t_{13} = a + (13 - 1) \times d \\ \\ \frac{1}{27} = a + 12d \: \: \: ................(2) \\ \\ now \: \\ \\ substract \: the \: equation \: 1 \: and \: 2 \\ \\ \: \: \: \: \: \frac{1}{9} = a + 3d \\ - \: \: \frac{1}{27} = a + 12d \\ - - - - - - - - \\ - \frac{2}{27} = - 9d \\ \\ d = \frac{ 9 \times 27}{2} \\ \\ d = \frac{243}{2} \\ \\ put \: \\ d = \frac{243}{2} \: \: in \: \: equation \: 1 \\ \\ \frac{1}{9} = a + 3d \\ \\ \frac{1}{9} = a + 3 \times \frac{243}{2} \\ \\ \frac{1}{9} = a + \frac{729}{2} \\ \\ \frac{2}{9 \times 729} = a \\ \\ a = \frac{2}{6561}$

Answer 2

Answer:

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