A series whose nth term is n-x-y then sum of r terms will be

The correct option is A Given, t_n = n/x+y ⇒ t_1 = 1/x+y, t_2 = 2/x+y ⇒ d = t_2 t_1 = 1/x' Where d is common difference of given AP. ∴ S_r = r/2[2a_1 + (r 1) ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Differentiation to Solve Modified Sum of Binomial Coefficients

A series whos...

Question

A series whose nth term is $(\frac{n}{x}) + y$ then sum of r terms will be

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
Given, $t_{n} = \frac{n}{x} + y \Rightarrow t_{1} = \frac{1}{x} + y, t_{2} = \frac{2}{x} + y \Rightarrow d = t_{2} - t_{1} = \frac{1}{x^{'}}$
Where d is common difference of given AP.
$∴ S_{r} = \frac{r}{2} [2 a_{1} + (r - 1) d] = \frac{r (r + 1)}{2 x} + r y$

Suggest Corrections

Similar questions

Q. A series, whose

n t h

term is

(\frac{n}{x}) + y

, then sum of

r

terms will be

Q. A series whose

n^{t h}

term is

⟮ \frac{n}{x} ⟯ + y

, the sum of r terms will be

Q. Find the sum of

n

terms of the series whose

n t h

term is :

n (n + 3)

Q. Sum to n terms the series , whose nth term is 2n + 3n -1.

Q. A series, whose

n^{t h}

term is

(\frac{n}{x}) + y

, then sum of

r

terms will be :

Join BYJU'S Learning Program

A series whose nth term is (nx)+y then sum of r terms will be

The correct option is A Given, tn=nx+y⇒t1=1x+y, t2=2x+y⇒d=t2−t1=1x′ Where d is common difference of given AP. ∴Sr=r2[2a1+(r−1)d]=r(r+1)2x+ry

A series whose nth term is $(\frac{n}{x}) + y$ then sum of r terms will be

The correct option is A
Given, $t_{n} = \frac{n}{x} + y \Rightarrow t_{1} = \frac{1}{x} + y, t_{2} = \frac{2}{x} + y \Rightarrow d = t_{2} - t_{1} = \frac{1}{x^{'}}$
Where d is common difference of given AP.
$∴ S_{r} = \frac{r}{2} [2 a_{1} + (r - 1) d] = \frac{r (r + 1)}{2 x} + r y$