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If the Sum of the First N Terms of an Ap is 4n − N2, What is the First Term (That Is S1)? What is the Sum of First Two Terms? - Mathematics

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Sum

If the sum of the first n terms of an AP is 4n − n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly find the 3rd, the10th and the nth terms.

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Solution 1

Given that,

Sn = 4n − n2

First term, a = S1 = 4(1) − (1)2 = 4 − 1 = 3

Sum of first two terms = S2

= 4(2) − (2)2 = 8 − 4 = 4

Second term, a2 = S2 − S1 = 4 − 3 = 1

d = a2 − a = 1 − 3 = −2

an = a + (n − 1)d

= 3 + (n − 1) (−2)

= 3 − 2n + 2

= 5 − 2n

Therefore, a3 = 5 − 2(3) = 5 − 6 = −1

a10 = 5 − 2(10) = 5 − 20 = −15

Hence, the sum of first two terms is 4. The second term is 1. 3rd, 10th, and nth terms are −1, −15, and 5 − 2n respectively.

Solution 2

In the given problem, the sum of n terms of an A.P. is given by the expression,

Sn = 4n -n2

So here, we can find the first term by substituting n = 1 ,

Sn = 4n -n2

S1 = 4(1) - (1)2

    = 4 - 1 

    = 3

Similarly, the sum of first two terms can be given by,

S2 = 4(2) - (2)2

      = 8 - 4 

      = 4

Now, as we know,

an = Sn - Sn-1

So,

a2 = S2 - S1

      = 4 - 3 

     = 1

Now, using the same method we have to find the third, tenth and nth term of the A.P.

So, for the third term,

a3 = S3 - S2

   `=[4(3)-(3)^2]-[4(2)-(2)^2]`

   ` = (12-9)-(8-4)`

     = 3 - 4

     = - 1

Also, for the tenth term,

`a_10 = A_10 - S_9`

      `=[4(10)-(10)^2]-[4(9)-(9)^2]`

       = (40 - 100 ) - ( 36 - 81 )

       = - 60 + 45

       = - 15 

So, for the nth term,

`a_n = S_n - S_(n-1)`

     `=[4(n)-(n)^2]-[4(n-1)-(n-1)^2]` 

     `=(4n -n^2)-(4n-4-n^2 - 1 + 2n)`

     `=4n - n^2 - 4n + 4 + n^2 + 1 -2n`

       = 5 - 2n

Therefore, `a = 3 , S_2 = 4 , a_2 = 1 , a_3 = -1 , a_10 = -15`.

Concept: Sum of First n Terms of an AP
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APPEARS IN

NCERT Class 10 Maths
Chapter 5 Arithmetic Progressions
Exercise 5.3 | Q 11 | Page 113
RD Sharma Class 10 Maths
Chapter 5 Arithmetic Progression
Exercise 5.6 | Q 45 | Page 53
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