The sum of progression (a+b), a, (a-b) n term is

Correct Answer is : n2[2a+(3−n)]\; n/2[2 a+(3-n)]2n[2a+(3−n)]

Correct Answer is : n2[2a+(3−n)]\; n/2[2 a+(3-n)]2n[2a+(3−n)]

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Sequence and Series

Question : 9 of 86

Marks: +1, -0

(a+b), a, (a-b) \dots n

term is

Solution:

Concept:

The given progression is an arithmetic progression (AP) with first term

a+b

and common difference

-b

Explanation:

First term

A = a+b

Common difference

D = a - (a+b) = -b

Sum of

n

terms of AP:

S_n = \frac{n}{2}[2A + (n-1)D]

Substitute:

2A = 2a+2b

(n-1)D = -(n-1)b

2A + (n-1)D = 2a+2b - b(n-1) = 2a + 3b - bn = 2a + b(3-n)

Thus

S_n = \frac{n}{2}[2a + (3-n)b]

Answer:

Option B:

\frac{n}{2}[2a+(3-n)b]

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