for today's video we are going to talk about how to find the sum of arithmetic sequence so when we say sum of arithmetic sequence the sum S sub n if the end theorem of arithmetic series is S sub N equals n over 2 times the quantity of a sub 1 plus a sub N or S sub N equals n over 2 times the quantity of 2 times a sub 1 plus n minus 1 times V so this are the different formulas we are going to use in order for us to find the sum of arithmetic sequence where a sub 1 is the first term a sub n is the nth term n is the number of terms and the is the common difference so let's start and let's have an example on example number one we are going to find the sum of the first 10 terms letter a our given is a sub 1 which is 5 and a sub 10 is 68 so in this kind of given we are going to use the formula S sub N equals n over 2 times a sub 1 plus a sub n so the value of n is 10 because we are looking for the sum of the first 10 forms so that is 10 over 2 times a sub 1 which is 5 plus 68 okay then 10 divided by 2 is 5 times 5 plus 68 is 73 then 5 times 73 so that is 365 so the sum of the first 10 terms is 365 what about letter B a letter B we have a sub 1 that is 5 and the comment Frances negative seven so this time we are going is the second formula which is S sub N equals n over 2 times 2 a sub 1 plus n minus 1 times D because we don't have a value of a sub n that's why we use the second formula so our value of n which is 10 over 2 times 2 a sub 1 is 5 plus n which is 10 minus 1 times negative 7 then let us simplify then over 2 or 10 divided by 2 is 5 multiplied by 2 times 5 is 10 plus 9 times negative 7 then 5 times 10 plus 9 times negative 7 that is negative 63 then let us divide 5 so we have 10 plus negative 63 that is negative 53 negative 53 times 5 that is negative 265 so the sum of the first tantrum for now letter B is negative 265 what about for letter C she can see the key event is 3 5 7 and 11 so we are looking for the sum of the first 10 firms so we have a sub 1 that is 3 common difference that is 5 minus 3 is 2 and then n is 10 so since we don't have the value of a sub n let us use this formula S sub N equals n over 2 times 2 a sub 1 plus n minus one times d so we have n which is 10 over 2 times 2 times a sub 1 which is 3 plus 10 minus 1 the common difference is 2 then 10 divided by 2 is 5 times 2 times 3 6 plus 10 minus 1 is 9 times 2 that is 18 and then let us simplify we have 5 times 6 plus 18 that is 24 then 24 times 5 that is 120 so the sum of the first 10 terms is 120 what about letter D on letter D a sub 1 is negative 6 or 6 rather common difference that is 3 minus 6 to find a common difference simply subtract the second term to the first term 3 minus 6 is negative 3 and then n is 10 so let us use the same formula S sub N equals n over 2 times 2 a sub 1 plus n minus 1 times D so the value of n is 10 over 2 then 2 a sub 1 which is 6 plus 10 minus 1 the common difference is negative 3 then we have 10 divided by 2 is 5 go I'm six yes well las 10 minus 1 is 9 times negative 3 is negative 27 then we have 5 x 12 plus negative 27 we are adding different sign whenever you are different sign you subtract and follow the sign of the bigger number so we have negative 15 5 times negative 15 is negative 75 so the sum of the first 10 terms for letter e d is negative 75 let's have another example on this example we are going to find the first term last term and the sum of each finit arithmetic series for example number one there's a mention of 6 n when n starts from 1 up to 31 so let us find first that in the first term and the last term so the first term that is 6 n so since the value of n starts from 1 the value of n will be 1 so 6 times 1 so the first term is 6 times 1 is 6 so this will be our first term then a sub n that is 6 n since the summation is up to 31 so 6 times 31 so 6 multiplied by 31 is 186 so this will be the value of n now let us find a value of n so the value of n starts from 1 to 31 so any state 1 there are 31 terms so since we have a sub 1 + a sub and we are using the formula S sub N equals n over 2 times the quantity of a sub 1 plus a sub n so we have the value of n which is 31 over 2 times a sub 1 which is 6 plus e sub n that is 186 then we have 31 over to the let's add 6 plus 186 that is 192 and then let us divide 192 by 2 we have 31 times 96 192 divided by 2 is 96 and then let us multiply 31 times 96 that is 7 9 2 so the sum is 2976 what about for number 2 summation of 3 and minus 1 when n starts from 3 up to 20 so we have here a sub 1 that is 3 n minus 1 since the value of n starts from 3 so the value of n is 3 3 times 3 minus 1 so 3 times 3 is 9 minus 1 so a sub 1 is 8 and then a sub n a sub n is 3 n minus 1 since the value of n is up to 20 so the value of n is 20 3 times 20 minus 1 so 3 times 20 is 60 minus 1 equals 60 minus 1 is 59 this is the last turn now what about the value of n how many terms do we have since the value of n starts from 3 after 20 or go have 18 terms because 20-3 17 + 1 is 18 they got started from 3 up to 20 now let us use the formula S sub N equals n over 2 times a sub 1 plus a sub n then n is 18 over 2 times a sub 1 that is 8 plus a sub end of this 59 18 divided by 2 is 9 times 8 plus 59 that is 67 and then let's multiply 9 times 6 to 7 that is 603 so the sum is 603 what about number 3 summation of 3 n plus 4 when n starts from 2 up to 16 so let us determine the value of a sub 1 first it's a 1 the formula is 3 n plus 4 so since the value of n is 2 we have 3 times 2 plus 4 3 times 2 that is 6 plus 4 so a sub 1 is 10 then a sub n is 3n plus 4 since the value of n is up to 16 so 3 times 16 plus 4 3 times 16 is 48 plus 4 48 plus 4 that is 52 so this will be the value of a sub n now what about the value of n since the value of n starts from 2 after 16 how many terms do we have so there are 15 terms if you find any difficulty in finding the number of n you simply subtract 16 minus 2 that is 14 plus 1 that is 15 so the number of term is shifting now let us find the value or let us find the sum this summation number one we have S sub N equals n over two times a sub one plus a sub n so n is 15 over 2 a sub one is 10 plus a sub n is 52 and then let's simplify we have 15 over 2 times 10 plus 52 is 62 let's divide six two by two that is 15 times 30 divided by the state t1 now let us multiply 31 by 15 so 15 times 31 is 465 so the sum is 465 what about number for summation of n minus 5 where any starts from 2 up to 24 so let's find a sub 1 so a sub 1 is n minus 5 so the value of and start from 2 that is 2 minus 5 2 minus 5 is negative 3 that is a sub 1 then a sub n the formulas n minus 5 since n is up to 24 so 24 minus 5 that is 19 so this should be a sub n what about for the value of n so since we started from two up to 24 24 minus two is 22 plus one that is 23 so there are 23 terms number four so let us just the formula S sub N equals n over two times a sub one plus a sub n so we have 23 over 2 times a sub 1 is negative 3 plus 19 so we have 23 over 2 negative 3 plus 19 that this 16 let's divide 16 by 2 that is 8 times 23 now let us multiply 23 times 8 23 times 8 that is 184 so the sum is 184