Math, asked by pateldharmistha17, 11 months ago

If sn=2n^2+3n then d=

Answers

Answered by Anonymous
36

Question:

If S(n) = 2n² + 3n then, find the common difference (d) of the AP.

Answer:

d = 4

Note:

• A sequence in which, the difference between the consecutive terms are same is called AP (Arithmetic Progression).

• Any AP is given as ; a , (a + d) , (a + 2d) , .....

• The nth term of an AP is given by ;

T(n) = a + (n - 1)d , where a is the first term and d is the common difference of the AP .

• The common difference of an AP is given by ;

d = T(n) - T(n-1) .

• The sum of first n terms of an AP is given by ;

S(n) = (n/2)[2a + (n-1)d] .

• The nth term of an AP is given by ;

T(n) = S(n) - S(n-1) .

Solution:

It is given that ;

S(n) = 2n² + 3n

Thus;

S(n-1) = 2(n-1)² + 3(n-1)

Also;

=> T(n) = S(n) - S(n-1)

=> T(n) = [2n² + 3n] - [2(n-1)² + 3(n-1)]

=> T(n) = 2n² + 3n - 2(n-1)² - 3(n-1)

=> T(n) = 2[n² - (n-1)²] + 3[n - (n-1)]

=> T(n) = 2[n+(n-1)]•[n-(n-1)] + 3[n - (n-1)]

=> T(n) = 2(n+n-1)(n-n+1) + 3(n-n+1)

=> T(n) = 2(2n-1) + 3

=> T(n) = 4n - 4 + 3

=> T(n) = 4n - 1

Thus,

T(n-1) = 4(n-1) - 1

Now;

=> d = T(n) - T(n-1)

=> d = [4n - 1] - [4(n-1) - 1]

=> d = 4n - 1 - 4(n-1) + 1

=> d = 4n - 1 - 4n + 4 + 1

=> d = 4

Hence,

The common difference of the AP is 4 , ie ;

d = 4 .

Answered by RvChaudharY50
181

Question :---

  • If sum of n terms of AP is denoted as 2n² + 3n, than find the common Difference d of AP.

Points To Remember :---

A sequence is said to be in AP (Arithmetic Progression), if the difference between its consecutive terms are equal.

• The nth term of an AP is given as ;

T(n) = a + (n-1)•d , where a is the first term and d is the common difference.

• The common difference of an AP is given as ;

d = T(n) - T(n-1)

• If the number of terms in an AP is n ( where n is odd ) ,then there will be a single middle term.

Also, [(n+1)/2]th term will be its middle term.

• If the number of terms in an AP is n ( where n is even ) ,then there will be two middle terms.

Also, (n/2)th and (n/2 + 1)th terms will be its middle terms.

• The sum up to nth terms of an AP is given as ;

S(n) = (n/2)•[2a + (n-1)•d] where a is the first term and d is the common difference.

• The nth term of an AP is also given as ;

T(n) = S(n) - S(n-1)

_____________________________

My Approach :----

→ Sn = 2n² + 3n

Taking n common

→ Sn = n (2n + 3)

Multiply and Divide by 2 ,

Sn = n/2 [ 4n + 6 ]

Adding and Subtracting 4 now,

Sn = n/2 [ 4n + 6 + 4 - 4 ]

→ Sn = n/2 [ 6+4 + (4n-4) ]

→ Sn = n/2 [ 10 + 4(n-1) ]

→ Sn = n/2 [ 2×5 + 4(n-1) ]

_____________________________

When we compare this with our Sum formula

(n/2)[2a + (n-1)d] = n/2 [ 2×5 + 4(n-1) ]

we get Easily , a = 5 and d = 4 ...

So, we can Say that, value of d is 4 .

Similar questions