Name: n- Geometric Means Between Two Numbers
Uploaded: 2019-07-19
Duration: 3 min 7 s
Description: There are so many numbers between two numbers, but how do we find out n geometric means between two numbers? Watch this video to learn more about it.

Question

If $θ_{1}, θ_{2}, θ_{3}, \dots, θ_{n}$ are in A.P, whose common difference is $d$ , then show that $\sec θ_{1} \sec θ_{2} + \sec θ_{2} \sec θ_{3} + \dots + \sec θ_{n - 1} \sec θ_{n} = \frac{\tan θ_{n} - \tan θ_{1}}{\sin d}$ .

Accepted Answer

Given that, $θ_{1}, θ_{2}, θ_{3}, . . θ_{n}$ are in A.P.

Therefore, $θ_{2} - θ_{1} = θ_{3} - θ_{2} = θ_{4} - θ_{3} = \dots = θ_{n} - θ_{n - 1} = d$

Let $T_{1} = \sec θ_{1} . s e c θ_{2} = \frac{1}{\cos θ_{1} . \cos θ_{2}}$

$= \frac{\sin d}{\sin d \cos θ_{1} \cos θ_{2}}$ [ multiply and divide by $\sin d$ ]

$= \frac{\sin (θ_{2} - θ_{1})}{\sin d \cos θ_{1} \cos θ_{2}}$ $[∵ θ_{2} - θ_{1} = d]$

$= \frac{\sin θ_{2} . \cos θ_{1} - \cos θ_{2} . \sin θ_{1}}{\sin d \cos θ_{1} \cos θ_{2}} = \frac{1}{\sin d} [\tan θ_{2} - \tan θ_{1}]$ $\dots (i)$

Similarly, we can get

$T_{2} = \sec θ_{2} . \sec θ_{3} = \frac{1}{\sin d} [\tan θ_{3} - \tan θ_{2}]$

$T_{3} = \sec θ_{3} . \sec θ_{4} = \frac{1}{\sin d} [\tan θ_{4} - \tan θ_{3}]$
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$T_{n} = \sec θ_{n - 1} . \sec θ_{n} = \frac{1}{\sin d} [\tan θ_{n} - \tan θ_{n - 1}]$

Now, $T_{1} + T_{2} + T_{3} + \dots + T_{n} = \frac{1}{\sin d} [(\tan θ_{2} - \tan θ_{1}) + (\tan θ_{3} - \tan θ_{2}) + \dots + (\tan θ_{n} - \tan θ_{n - 1})$

$= \frac{1}{\sin d} [{tanθ}_{n} - \tan θ_{1}]$

Hence, proved.

If $θ_{1}, θ_{2}, θ_{3}, \dots, θ_{n}$ are in A.P, whose common difference is $d$ , then show that $\sec θ_{1} \sec θ_{2} + \sec θ_{2} \sec θ_{3} + \dots + \sec θ_{n - 1} \sec θ_{n} = \frac{\tan θ_{n} - \tan θ_{1}}{\sin d}$ .

Important Questions on Sequences and Series

Learn and Prepare for any exam you want !

If θ1, θ2, θ3,…,θn are in A.P, whose common difference is d, then show that secθ1secθ2+secθ2secθ3+…+secθn-1secθn=tanθn-tanθ1sind.

Important Questions on Sequences and Series

Learn and Prepare for any exam you want !

If $θ_{1}, θ_{2}, θ_{3}, \dots, θ_{n}$ are in A.P, whose common difference is $d$ , then show that $\sec θ_{1} \sec θ_{2} + \sec θ_{2} \sec θ_{3} + \dots + \sec θ_{n - 1} \sec θ_{n} = \frac{\tan θ_{n} - \tan θ_{1}}{\sin d}$ .