Question

5.
Find the sum of
Sum of 'p' terms
of the
Series whose nth term
is n/a+b​

Answer 1

Answer:

Let the first term of a series be x and common difference be d

The nth term of an Arithmetic progression series is x+(n−1)d=x−d+nd

Given that the nth term in the series is n/a+b

By comparing we get d=1/a and x=b+1/a

Sum of p terms of the series is p /2(2x+(p−1)d)

=p/2(2b+2/a+p-1/a)

=p/2(2b+p+1/a)

Hope it helps<3

Answer 2

Step-by-step explanation:

Let the first term of a series be x and common difference be d

The nth term of an Arithmetic progression series is

a+(n−1)d = a−d+nd

Given that the nth term in the series is

$\frac{n}{a} + b$

By comparing , we get,

$d = \frac{1}{a} \\ and \: x= b + \frac{1}{a}$

Now the sum of p terms of the series is

$\frac{p}{2} (2a + (p - 1)d) = \frac{p}{2} (2b + \frac{2}{a} + \frac{p - 1}{a} ) \\ = \frac{p}{2} (2b + \frac{p + 1}{a} )$