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

- March 26, 2018
- Posted by: allexamshelps
- Category: Mathematics

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1,469 total views, 1 views today

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

^{th}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 10^{th} 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 10**^{th} term is 38

^{th}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

##### The mathematics of Quadratic equation will be uploaded very soon.

##### { 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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