TS ICET 23 May 2019 Shift 1 Paper

Question : 128 of 200

Marks: +1, -0

If seventh and eleventh terms of an arithmetic progression are 31 and 47 respectively, then fifteenth term is

Solution:

let a and d be the first term, the common difference of the given progression

a+6 d=31

→ 1

a+10 d=47

→ 2
Solving 1 and 2; a = 7; d = 4

∴ t _{15}= a +14 d =7+56=63

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