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Question

Answers

A). 16.

B). 14.

C). ${27}/{14}\;$.

D). ${56}/{15}\;$.

Answer
Verified

Now the 7th term of HP is equal to 8.

As described in the hint the general term is given by $\dfrac{1}{a+\left( n-1 \right)d}\ =\ {{T}_{n}}$

Now for the beneath term $n=7$

$\therefore \ {{T}_{7}}=\dfrac{1}{a+6d}$

$\because \ {{T}_{7}}=\ 8$

$\Rightarrow \ \dfrac{1}{a+6d}\ =\ 8$

$\Rightarrow \ a+6d=\ \dfrac{1}{8}$ (1)

Now the 8th term of HP is equal to 7.

$\therefore \,{{T}_{8}}\ =\ 7\ =\dfrac{1}{a+\left( 8-7 \right)d}$

$7\ =\ \dfrac{1}{a+7d}$

$\therefore \ a+7d\ =\ \dfrac{1}{7}$ (2)

Now on subtracting $eq\ \left( 1 \right)\ \text{from}\ \text{eq}\ \left( 2 \right)$

$\left( a+7d \right)-\left( a+6d \right)\ =\ \dfrac{1}{7}-\dfrac{1}{8}$

$a+7d-a-6d=\dfrac{8-7}{56}$

$\therefore \ d\ =\ \dfrac{1}{56}$

Now since we know the value of d, we can compute the value of a by substituting in equation (1), we get

$a\ +\ 6\ \times \ \dfrac{1}{56}=\ \dfrac{1}{8}$

$a\ =\ \dfrac{1}{8}-\dfrac{6}{56}$

$a\ =\ \dfrac{7-6}{56}$

$a\ =\ \dfrac{1}{56}$

Now if we want to compute the 15th term we can substitute the value of $a,d\ and\ n=15$ in general terms.

$\therefore \ {{T}_{15\ }}\ =\ \dfrac{1}{\dfrac{1}{56}+\left( 15-1 \right)\times \dfrac{1}{56}}$

$\Rightarrow \ \ \ {{T}_{15}}=\ \ \dfrac{1}{\dfrac{1}{56}+\dfrac{14}{56}}$

$\Rightarrow \ \ \ {{T}_{15}}\ \ \ \ =\dfrac{1}{\begin{align}

& 15 \\

& \overline{56} \\

\end{align}}$

$\therefore \ {{T}_{15}}\ \ \ =\ \ \dfrac{56}{15}$

$\therefore $ The 15th term of HP is $\dfrac{56}{15}$

Therefore series of terms are a HP series when their reciprocal are in AP.