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Question

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?

A
67
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B
45
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C
30
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D
48
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Solution

The correct option is C 30
Let the first term be a and d be the common difference
3rdterm=a+2d and 6thterm=a+5d
(a+5d)=4 ......(1)
(a+2d)=13) ......(2)
Solving (1) &(2), we get
d=3,a=19
Sum of first 12 terms=n2[2a+(n1)×d]=122[2×(19)+(121)3]=30.

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