Arithmetic progression 2

Algebra Level 3

In a 24-term arithmetic progression , it is known that

a 1 + a 5 + a 10 + a 15 + a 20 + a 24 = 225 a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225

What is the sum of all 24 terms?


The answer is 900.

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1 solution

With the n n th term of the A.P. being a n = a 1 + ( n 1 ) d a_{n} = a_{1} + (n - 1)d , where d d is the common difference between successive terms, we see that a 1 + a 24 = a 5 + a 20 = a 10 + a 15 = 2 a 1 + 23 d a_{1} + a_{24} = a_{5} + a_{20} = a_{10} + a_{15} = 2a_{1} + 23d .

Given that a 1 + a 5 + a 10 + a 15 + a 20 + a 24 = 225 a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225 we then have that a 1 + a 24 = 225 3 = 75 a_{1} + a_{24} = \dfrac{225}{3} = 75 .

With the sum of the first n n terms of an A.P. being S n = n 2 ( a 1 + a n ) S_{n} = \dfrac{n}{2}(a_{1} + a_{n}) we have that

S 24 = 24 2 ( a 1 + a 24 ) = 12 75 = 900 S_{24} = \dfrac{24}{2}(a_{1} + a_{24}) = 12*75 = \boxed{900} .

True. The challenge is to understand that the sum of the terms equidistant from the end and beginning of the AP are equal. Nice solution (+1)

Ashish Menon - 5 years ago

Nice one..

Sparsh Sarode - 5 years ago

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