For any increasing sequence of real numbers $a_{1}, a_{2}, a_{3}, ...$ let us denote A as $≺ ≺ a_{1}, a_{2}, a_{3}, a_{4}, ... ≻ ≻,$ and a sequence $Δ A$ is defined as $≺ ≺ a_{2} - a_{1}, a_{3} - a_{2}, a_{4} - a_{3}, ....... ≻ ≻ .$ Suppose that all of the terms of the sequence $Δ (Δ A)$ are 1 and that $a_{19} = a_{92} = 0.$ Then the sum of digit(s) of $a_{3}$ is M, then the value of $[\frac{M}{2}]$ is ____, where $[.]$ is greatest integer function.

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 5.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Suppose that the first term of the sequence $Δ A$ is d then
$Δ A = {d, d + 1, d + 2, ....... (d + (n - 1))}$
Hence $A = (a_{1}, a_{1} + d, a_{1} + d + (d + 1)),$
$a_{1} + d + (d + 1) + (d + 2) ...........)$
$a_{n} = a_{1} + (n - 1) d + \frac{1}{2} (n - 1) (n - 2)$
so $a_{n}$ is a quadratic polynomial in n.
so $a_{n} = \frac{(n - α) (n - β)}{2}$
Since $a_{19} = a_{92} = 0$ we must have $a_{n} = \frac{1}{2} (n - 19) (n - 92)$
So $a_{3} = \frac{1}{2} (3 - 19) (3 - 92) = 712$