This is the way i solved it ..
The question gives us a general formula that confirms that we are dealing with an A.P. , therefore we know that the nth term = a (n-1)d .. In this case d = x ...
First term of the AP = A ; Last term = L ..
we know that A + L = 14 ...........[I]
Now we are to find the sum of 6 terms , ie the first three and the last three ... Lets assume that this series has 6 numbers .. therefore Sum of 6 numbers can be written as
S6 = 6/2 [ A + L ]
= 3 x 14 ( From I)
= 42 (D)
As an alternative to this method , we could
PLUG IN values for X and solve for the sum of 6 terms of the AP as follows ------------->
Let us assume x =1 , therefore we would need to come up with a sequence whose first and last term adds to 14 , and which as at least 6 terms ...
let us use this sequence ...
3 , 4 , 5, 6, 7, 8, 9, 10, 11
The first and the last terms add to 14 ...
Each term is x greater than the one it succeeds ( in this case i have assumed x to be 01)
The first three and the last three terms are : 3 + 4 + 5 + 9 + 10 + 11 = 42 (D)
To further test the plug in method (not required to confirm the answer but just to test it out) ...
Now let us assume that x = 2
1 3 5 7 9 11 13
First three and last three terms add up to = 42 (D) ....
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