Question

find the sum of n term of the series 2a+(3a-b)+(4a-2b)+....

Answer 1

a=2a
d=(3a-b)-2a
=a-b
s=
$\frac{n}{2} (2a + (n - 1)d)$
$= \frac{n}{2} (2 \times 2a + (n - 1)(a - b))$
$\frac{n}{2} (4a + na + nb - a + b)$
$\frac{n}{2}(3a + n(a + b) + b)$

Answer 2

Answer:

Step-by-step explanation:

Sn=n/2[2a+(n-1)d]

Sn=n/2[2(2a)+(n-1)a-b]

Sn=4a+(n-1)(a-b)]

Sn=n/2[4a+na-nb-a+b]

Sn=n/2(3a+na-nb-a+b)

Answer......