The number of terms of an Arithmetic Progression is even.
The sum of the terms in the odd places (First term + Third Term + Fifth Term + ...and so on) is 24;
The sum of the terms in the even places (Second Term + Fourth Term + Sixth Term + ... and so on) is 30; and
The last term exceeds the first by 21/2, then:
What is the arithmetic progression?
(In reply to
Puzzle Answer by K Sengupta)
Let the number of terms in the A.P. be 2n.
a= first term (say)
d= common difference (say)
Then, by condition (1), we have:
a+(a+2d)+........+{a+(2n-2)d}=24
=> n{a+(n-1)d} = 24 (upon simplification) .......(#)
Condition (2) gives:
(a+d)+(a+3d)+......+{a+(2n-1)d} =30
=> n(a+nd) = 30 (upon simplification) .........(##)
Condition (3) gives:
(2n-1)d = 21/2 ........(###)
Then, (##)-(#) gives:
nd=6
Then, (2n-1)d =21/2 gives
=> 2nd - d = 21/2
=> 2*12 - d = 21/2
=> d = 3/2, so that:
6
n = ------- = 4
(3/2)
Therefore, 4(a+ 3* 3/2 ) = 24
=> a+ 9/2 = 5
=> a= 3/2
Accordingly (a,d, 2n) = { 3/2, 3/2, 8}, and consequently the required arithmetic progression is given by:
{3/2, 3, 9/2, 6, 15/2, 9, 21/2, 12}
Edited on June 12, 2022, 11:25 pm
Explanation to Puzzle Answer
