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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?
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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 $5\times 81 + 90 = 405 + 90 = 495$

by Loyal (9.7k points)
edited by

x+x+2+x+4+x+6+x+8=425 so x=81

last 5 digits will be x+14+x+16+x+!8+x+20+x+22=425
by Active (1.3k points)

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