(a) In an arithmetic progression, the sum, Sn
, of the first n terms is given by Sn=2n^2 + 8n. Find the first term and the common difference of the progression.

ans ; a =64 d= 3/4

Question

(a) In an arithmetic progression, the sum, Sn
, of the first n terms is given by Sn=2n^2 + 8n. Find the first term and the common difference of the progression. 

ans ; a =64 d= 3/4

jhonyy9 · Answer

so for the first term n=1 what mean that Sn = 10

parthkohli · Answer

$$a_n = S_n - S_{n - 1}= 2n^2 + 8n - \left(2(n-1)^2 + 8(n-1)\right)$$

parthkohli · Answer

You know what... I don't think you even need to derive the formula for the $n$th term. But you can apply a similar method to find the common difference.$$S_1 = a = 2 + 8 = 10.$$$$S_2 = a + (a + d) = \color{#c00}{2a + d} = 2(2)^2 + 8(2) = \color{#c00}{24}$$The first equation tells us that $a = 10$, so we'll plug that into$$2a + d = 24\Rightarrow 20 + d = 24 \Rightarrow d = 4$$

anonymous · Answer

Assume n=1,2,3..
then s1=2.1^2+8.1=10
s2=2.2^2+8.2=24
and s3=2.3^2+8.3=42
1st term=10
2nd term=s2-s1=24-10=14
3rd term=s3-s2=42-24=18
a=10 and d=4

Check, Sn=n/2{2a+(n-1)d}
Sn=n/2{2.10+(n-1)4}
Sn=n/2(20+4n-4)
Sn=n(10+2n-2)
Sn=2n^2+8n
Done!!!