Question

If the sum Sn of n terms of a progression is a cubic polynomial in n, then Find the progression whose sum of n terms is Sn – Sn-1 .

Answer 1

Answer:

Sn=3n2+5n 

Tn=Sn−Sn−1

Tn=(3n2+5n)−(3(n−1)2+5(n−1))

Tn=(3n2+5n)−(3(n2−2n+1))−5n+5

Tn=3n2+5n−3n2+6n−3−5n+5

Tn=6n+2

Again T1=6×1+2=8,T12=82=64

T2=6×2+2=14,T22=142=196

T3

Answer 2

Answer:

the progression whose sum of n term of  $S_{n} -S_{n-1}$ is Tn which is nth term of A.P.

Step-by-step explanation:

Explanation:

Given ,

n is the sum of $S_{n}$ terms

Let assume the  sequence be A.P

So ,   $S_{n}$ = sum of n term of an A.P

               = $\frac{n}{2}[2a +(n-1)d]$  .........(i)

Therefore , $S_{n-1}$ = sum of (n-1 ) term of an A.P.

                          =$\frac{n-1}{2}[2a +((n-1)-1)d]$

Now open the brackets ,

$S_{n-1}$ = $\frac{n-1}{2}[2a +(n-2)d]$  ......(ii)

Step1:

As we know that the nth term is called $T_{n}$

Now subtract equation (i) from equation (ii)

∴ $S_{n} -S_{n-1} = \frac{n}{2}[2a +(n-1)d]-\frac{n-1}{2}[2a +(n-2)d]$

              = $an +\frac{n(n-1)}{2}d -a(n-1) -\frac{(n-1)(n-2)}{2} d$

            = $an +\frac{n(n-1)}{2}d -an +a -\frac{(n-1)(n-2)}{2} d$

         = $a +\frac{n(n-1)}{2}d -\frac{(n-1)(n-2)}{2} d$ (where 'an' and 'an' are cancle out )

        Take common $\frac{(n-1)}{2} d$

⇒$S_{n} -S_{n-1} = a + \frac{(n-1)d}{2}[n-(n-2)]$

                  = a + (n-1)d. = $T_{n}$

Final answer:

Hence , the progression whose sum of n term of  $S_{n} -S_{n-1}$ is Tn which is nth term of A.P.