00:01
Hello everyone, in the first part of the problem we need to find the first term and the common difference.
00:06
So this is the general form of an ap.
00:08
So now let us take sn to be the sum of the first n terms.
00:12
So it would be sn as n by 2 of 2a plus n minus 1 of b.
00:23
So here we are given that the value of the sum of first 15 terms to be 225.
00:32
So equating that here in this formula, so it would be 225 will be equal to 15 by 2 of 2a plus of 14d.
00:50
So from this we get 2a plus 14d to be equal to 225 .2 divided by 15.
01:03
So it is also given that the sum of the next 15 terms is 705.
01:09
So if we add the sum 225 of the first 15 terms to the above sum which is of 705.
01:18
So the sum of first 30 terms is 930.
01:32
So again using the sum of nth terms formula, so we have 30 by 2 of 2a plus 29d to be equal to 930.
01:48
So from this we get 2a plus 29d to be equal to 62.
01:58
So here let us take this to be here it is 30.
02:02
Let us take this to be equation number 1 and this to be equation number 2.
02:06
Now solving these two equations 1 and 2, we have the value of d which is to be 32 divided by 15 and from the equation 1 we have a plus 70 to be equal to 15.
02:23
Now substituting the value of d and simplifying we have the value of a to be 1 by 15.
02:29
So with this we can say that the first term is 1 by 15 and its common difference d is 32 by 15.
02:44
So in the next part of the equation we are given with a gp.
02:48
So let us take that to be a, ar, ar square, ar cube and so on up to ar bar n where n belongs to n.
03:01
So the first sum of two terms is 6...