A sequence of numbers a1-a2-a3- satisfies the relation a2 n-1-an an-2-1n then find sum of digits of a3 if a1-2 and a2-5 -

Putting n=1 , in the given relation , we obtain a^2 _2=a_1a_3+( 1)^1 ⇔ nbsp; nbsp; 5^2=2a_3 1 nbsp; nbsp; nbsp; nbsp; ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Properties of AP

A sequence of...

Question

A sequence of numbers $a_{1}, a_{2}, a_{3}, . .$ satisfies the relation $a_{n + 1}^{2} = a_{n} a_{n + 2} + (- 1)^{n}$ then find sum of digits of $a_{3}$ if $a_{1} = 2$ and $a_{2} = 5$ ?

Open in App

Solution

Suggest Corrections

Similar questions

a_{1}, a_{2}, a_{3}, . . . . .

a_{n + 1} = a_{n} + a_{n - 1}

n \geq 2

a_{4}

a_{1} = a_{2} = 1

a_{r} > 0; \forall r, n \in N

a_{1}, a_{2}, a_{3}, . . . . . a_{2 n}

\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}} =

a_{r} > 0, r ϵ N

a_{1}

a_{2}

a_{3}

a_{2 n}

\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}}

a_{r} > 0, r \in N

a_{1}, a_{2}, a_{3}, . . ., a_{2 n}

\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}}

a_{1}, a_{2}, a_{3}, . . . . a_{2 n + 1}

\frac{a_{2 n + 1} - a_{1}}{a_{2 n + 1} + a_{1}} + \frac{a_{2 n} - a_{2}}{a_{2 n} + a_{2}} + . . . + \frac{a_{n + 2} - a_{n}}{a_{n + 2} + a_{n}}

Join BYJU'S Learning Program