SSC CGL 17 Aug 2017 Shift 2 Paper

Question : 64 of 100

Marks: +1, -0

What is the sum of the first 12 terms of an arithmetic progression if the 3rd term is -13 and the 6th term is - 4 ?

Solution:

Let a be the first term and d be the common difference of an A.P.
Then

A_{3}=a+(3-1) d

⇒

-13=a+2 d

...(i)
and

A_{6}=a+(6-1) d

⇒

-4=a+5 d

...(ii)
On solving Eqs. (i) and (ii),

a=-19

and

d=3

By formula,

S_{n}= {n}/{2}[2 a+(n-1) d]

S_{12}= {12}/{2}[2(-19)+(12-1) 3]

=6[-38+33]=6(-5)=-30

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