Question

The sum of n terms of a+b, 2a, 3a-b, ..... is
(a) n(a-b)+2b
(b) n(a+b)​

Answer 1

Step-by-step explanation:

formula :- Tn = a+(n-1)d

a = a+b , d = 2a - ( a+b)

d = a-b

n संख्याओं का योग = (a+b) + (n-1)(a-b)

= (a+b)+n(a-b) -1(a-b)

= a+b+ na-nb-a+b

= 2b + n(a-b)

= n(a-b) + 2b

Answer 2

The correct answer is $\frac{n}{2}(a(1+n)+b(3-n))$.

Given: Sequence = a+b, 2a, 3a-b, ......

To Find: Sum of n terms of series.

Solution:

Sum of an arithmetic progression = $\frac{n}{2}(2a+(n-1)d)$.

where n is number if terms, a is first term, d is common difference.

Here,

First term = a+b

Number of terms = n

Common difference = difference between any two consecutive terms.

Common difference = second term - first term

= 2a - (a+b)

= 2a-a-b

Common difference = a-b

Now put values in formula of sum of terms.

Sum of an arithmetic progression = $\frac{n}{2}(2(a+b)+(n-1)(a-b))$

= $\frac{n}{2}(2a+2b+na-nb-a+b)$

= $\frac{n}{2}(a+3b+na-nb)$

= $\frac{n}{2}(a(1+n)+b(3-n))$

Hence, the sum of n terms of series is $\frac{n}{2}(a(1+n)+b(3-n))$.

#SPJ2