Intro, Lecture 3

Sum of First N terms of A.P Type I, II, and III

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 Sum of First N terms of an A.P Type I, Type II, and Type III questions

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If the sum of 20 terms of an A.P is 780 and the first term is 20, find the 10th term.

 

By using the formula, we learned to drive in the last post. Click here

 

We can solve this question of A.P.
Firstly we have to find out the common difference because that is not given here.

 

How to find the common difference and other details of A.P click here

 

{ S }_{ n }=\frac { n }{ 2 } [2a+(n-1)c.d]
780=\frac { 20[2\times 20+(20-1)c.d] }{ 2 }
780=\frac { 20 }{ 2 } [40+19c.d]
780=10[40+19c.d]
780=10\times 40+19c.d
780=400+190c.d
780-400=190c.d
380=190c.d
\frac { 380 }{ 190 } =c.d
2= c.d
Now we have to find the 10th term which is the real question to solve here
{ S }_{ n }={ a }_{ 1 }+(n-1)c.d
We had learned to derive the above formula in the previous post. Click here
{ S }_{ 10 }=20+(10-1)2
{ S }_{ 10 }=20+9\times 2
{ S }_{ 10 }=20+18
{ S }_{ 10 }=38
Hence the 10th term is 38

Sum of first n terms of an A.P: Type II question

How many terms of an A.P of the following form should be taken to give the sum of 780?
By using the formula, we learned to drive in the last post. Click here
\frac { n[2a+(n-1)c.d] }{ 2 }
\frac { n }{ 2 } [2\times 20+(n-1)2]
\frac { n }{ 2 } [40+2n-2]
780\times 2=n(38+2n)
1560=38n+{ 2n }^{ 2 }
0=38n+{ 2n }^{ 2 }-1560
Or
{ 2n }^{ 2 }+38n-1560=0
All the numbers above are divisible by 2 i.e. 2, 38, and 1560, so we can divide by 2 to make the calculation easier.
\frac { { 2n }^{ 2 } }{ 2 } +\frac { 38n }{ 2 } -\frac { 1560 }{ 2 } =0
{ n }^{ 2 }+19n-780=0

 

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{ n }^{ 2 }+39n-20n-780=0

 

={ n }(n+39)-20(n+39)=0

 

(n-20)  and   (n+39)
This means n=20   and   n= -39

 

Since the number of terms in an A.P can’t be negative, the answer is 20 terms which are there in this A.P

Sum of first n terms of an A.P: Type III question

 

What is the sum of first 20 terms of an A.P whose nth term can be given by
 
{ a }_{ n }=17+3n
Solution:
Here we have not been given the first term and the common difference of this A.P which are essentially required to solve the questions given in the A.P.
But we have been given this form { a }_{ n }=17+3n where n is unknown.
So what we can do that we can put n=1 on our own side to solve this incomplete question.
{ a }_{ 1 }=17+3\times 1
{ a }_{ 1 }=20

 

n=2

 

{ a }_{ 2 }=17+3\times 2
20=17+6
{ a }_{ 2 }=23

 

n=3
{ a }_{ 3 }=17+3\times 3
20=17+9
{ a }_{ 3}=26

 

So, now we have found the three terms of this A.P which are:

 

 20, 23, 26… up to 20 terms
Now we have to find the common difference using this formula
{ a }_{ 2 }-{ a }_{ 1 }={ a }_{ 3 }-{ a }_{ 2 }=c.d for more details click here
Common difference (c.d) = 23 – 20 = 26 – 23 = 3
Now we have the first term { a }_{ 1 }=20 and the common/constant difference =3.

 

And, the number of first terms of this A.P for which the sum is to be found =20.
Now we can use the formula for finding the sum of first n terms of an A.P:
{ S }_{ n }=\frac { n }{ 2 } [2a+(n-1)c.d]
 
{ S }_{ 20 }=\frac { 20 }{ 2 } [2\times 20+(20-1)3]
 
{ S }_{ 20 }=\frac { 20 }{ 2 } [2\times 20+19\times 3]
 
{ S }_{ 20 }=\frac { 20 }{ 2 } [40+57]
 
{ S }_{ 20 }=\frac { 20 }{ 2 } \times 97
 
{ S }_{ 20 }=10\times 97
 
{ S }_{ 20 }=970
Hence the answer is 970 which is the sum of first 20 terms of this A.P.
 

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