Thursday, September 29, 2022

# [TAGALOG] Grade 10 Math Lesson: SOLVING ARITHMETIC SEQUENCE (Part III) – FINDING COMMON DIFFERENCE

Finding the common difference of an arithmetic sequence of what's up guys welcome to my youtube channel this is humor ph .

And willing to give but before we go to our lesson please don't forget to subscribe and click the notification bell and without further ado let's get to our you more cohen welcome to our lesson for today .

Iii negan is a part to an arithmetic sequence this is a challenge question at all find the common difference of an arithmetic sequence with first term value of 11 and the sixth term is 56. so don't support common difference how are you going to .

Do the solution for that so let me show you now the complete solution for this challenge question and find the common difference of an arithmetic sequence with first term value of 11 and the sixth term is fifty six so obviously um .

Difference so same parent same process so first you have to write the formula and formula a sub n is equal to a sub 1 plus the quantity of n minus 1 times the common difference again a sub n is the nth term c a sub 1 uh first terms and lagging so a sub 1 attention will be with first term value of 11 so on a sub 19 i .

11. next common difference i answer seems uncommon difference of pinapple had not of course question mark munayan because we still don't know that is uh the one being required for us to find then n is the term position turn position last term nothing number fifty six no and on turn position i .

Six so n is equal to six n is equal to six therefore an a sub n is of a sub six that is equal to 56 and sixth term is 56 so a sub 6 nothing i 56 and so from platinum given and we have the formula we can now get our solution so for the solution .

Gonna elect isolat monk formula muna para madali time substitute for our next step after you write the formula a sub n is equal to a sub one plus the quantity of n minus one times d you have to substitute all the given atoms 56 is equal to 11 plus the quantity of .

Six minus one times the again be careful suppose substitute kasiyan ang kadalasana 56 is equal to 11 plus five times d and multiply minus five and d so that is five d capacity multiply five times d that is finally in parenthesis then after that angle not in beto same as the part in .

Glasgow angola is one so written a month cd but in this case additional property of equality transposition transposition a term for a shortcut there is no such thing as transposition parallel negative sahabila so by doing that at .

The human area 56 minus 11 is equal to 5d again in the lipid case 110 the gingka bishop 56 and c5 dna in one dot and solve bundle minus 11. so 56 minus 11 that is 45 is equal to 5d now multiplication property of equality we're in we're going to multiply one .

Fifth so both side of equation so again multiplied on one fifth or young shortcut divided five sa both side of the equation so divide by five that's a five d and divide by five then sub forty five capacitive dividing on five magnitude .

One so one d and one d is equal to d and 45 over five is nine so therefore nine is equal to d params t is equal to nine so this is now the common difference so if it's a b uncommon difference in sequence 56 and this is now our complete solution okay parama's mind in .

The hand so let's have another example what is the common difference of an arithmetic sequence with 1st and 17th term values of negative 1 and 36 respectively ayan so pinapahana policy common difference and the uncommon difference and given saturn's first and 17th term .

Values the negative one and 63 respectively so etosi guru napan singh nyung were the respectively most samantha problems sama so happened encountered your word respectively seventeenth term i 63. be careful with that .

Now let's do the solution for this formula a sub n is equal to a sub 1 plus the quantity of n minus 1 times d and given let's find out what are you given for this ace of one again since my respective disc wouldn't be first and first a negative .

One all right next d so d is common difference shayang pinapahana i mean n n is the term position and the term position there is since 17th term so seven i am so these are the given array negative one question mark 17 and .

63 and we'll go to our solution so it's a solution copy him now let's say formula a sub n is equal to a sub 1 plus the quantity of n minus one times d and after that substitute all the given d to suffer melania .

And be careful again so be careful substitute see a sub n a sixty three say a sub one double plugging 63 is equal to negative 1 plus 17 minus 1 times d and 63 is equal to negative 1 plus 16 that is 17 minus 1 multiplied by d after that atom multiply .

Nothing so 63 is equal to negative 1 plus 16 d and then since i'm going not in my a1 cd 1cb addition property of equality is equal to 16 d and then a plus one and after 63 plus one that .

Will give us 64 is equal to 16d and multiplication property of the quality and that indito divide cannon 16s of both side of the equations so 16d 16 technical mpe is is that number or i mean variable young division so both .

Because 16 over 16 is equal to one so one being london or d then 64 divide 16 that is four so our common difference now is four is equal to d or paramas d is equal to four same language difference so i hope that gets more companion and confident let's go to our .

You merge it please go back to the video and panorama proceed to your jail i am so let's have our humor so i'll give you one question what is the common difference of an arithmetic sequence with first and fifteenth term values of 58 and 30 respectively .

All right so i'll give you time to answer this and conquer time peace feel free to pause the video and after that i'll give you the answer i am so your timer starts now all right time's up so let us now answer this one .

What is the common difference of an arithmetic sequence with first and fifteenth term values 58 and 30 respectively so my respectively oled so first 58 15 15 30 so formula so a is equal to a sub 1 plus the .

Quantity of n minus 1 times t and let's find the given a sub 1 so first term unknown get first term beta so 58th and first term not n so that is a 1 is equal to 58 then common difference d that's required to find so indeep nothing else question mark .

And n depends a sub n so 15th terminal 15. so on position i punch 15. and a sub n is the a sub 15 nathan kasingham and a15 let's go for solution literal formula a sub n is equal to a sub 1 plus the quantity of n minus one times d and then substitute all the given the answer formula nothing be careful .

Again so a sub 15 that b to a thirty a sub one that a fifty eights end up at a fifteen c d is three t is equal to 58 plus 15 minus one times d and and then so 30 is equal to 58 plus 14 because 15 .

Minus 1 times d and multiply 14 times d angle in parenthesis so that will give us 3 is equal 58 plus 14 d and thirty minus twenty minus fifty eight so the answer there will be negative .

Negative integers so that's it so negative 28 is equal to 14 d and apply the multiplication property of equality for enzymes and what ability 28 divide 14 that will give us negative 2 i am so negative divide positive that's negative so that's part .

Of integers again i'm mastering that in new integers and this is just equal to d is equal to negative 2. so i prefer this and this is now the final answer therefore the common difference that we're looking for here is negative .

And i want to see you again in our next video bye-bye thank you so much for listening thank you so much for watching guys i hope you enjoyed it learned something today and if you did learn from this please don't forget to subscribe and click notification bell and please .

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