 Hello friends, and how are you all doing today? The question says find the sum of first 30 terms of an AP whose second term is 2 and Seventh term is 22 So let us start with our solution We are given That the second term of an AP that we see that A2 is given to us as 2 and The seventh term of an AP is given to us as 22 we can say that here A2 can be written as a plus t which is given to us as 2 Which can further implies that the value of a is equal to 2 minus d let this be the first equation Similarly for this we can write that a plus 6d is given to us as 22 This implies that the value of a is 22 minus 6d Let this be the second equation since the first term of an AP is same for the both the cases therefore on equating 1 and 2 we have 2 minus d equal to 22 minus 6d Which implies 2 minus 22 is equal to minus 6d plus d which further implies minus 20 is equal to minus 5d and The value of t comes out to be 4 now on substituting the value of d in any of these equation We can have the value of a which means the value of a is 2 minus 4 that is minus 2 now we need to find out the sum of first 30 Terms of an AP whose first term is minus 2 and the difference is 4 So we can find the sum that is SN where n is 30 equal to 1 by 2 n that is 30 2 a that is minus 2 Plus n that is 30 minus 1 into d Now we need to solve it where the formula was if n is equal to 1 by 2 n 2 a plus n minus 1 d in this formula we have substituted all these values And what we need to do next is we just need to simplify it using the concept of word mass We have 15 into minus 4 plus 1 1 6 will give us the value as 1 1 2 and their product comes out to be 1 6 8 0 So the answer to this session is 1 6 8 0 that is the sum of the first 30 terms of an arithmetic Progression whose second term is 2 and the seventh term is 20 second Right, so hope you understood it well do kitty do keep do take care of your calculations and have a nice day