If sn=2n^2+3n then d=
Answers
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 .
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)
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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) ]
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When we compare this with our Sum formula
→ (n/2)[2a + (n-1)d] = n/2 [ 2×5 + 4(n-1) ]