5 votes

In a sequence of $12$ consecutive odd numbers, the sum of the first $5$ numbers is $425$. What is the sum of the last $5$ numbers in the sequence?

6 votes

Best answer

Let $a$ be the first odd number.

So the terms of sequence will be

$a, a+2, a+4, a+6, \dots, a+20, a+22$

Sum of first $5$ terms $= a + a+2 + a+4 + a+6 + a+8 = 5a+20 = 425$

We get, $5a = 405$

$\quad \implies a = 81$

Sum of last $5$ terms $= a+22 + a+20 + a+18 + a+16 + a+14$

$\quad \quad \quad = 5a + 90$

Now, we have $a = 81.$ Substituting it we get,

Answer as $5\times 81 + 90 = 405 + 90 = 495$

**Answer: 495**