Question

Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Answer 1

$\huge \bf{Solution.}$

Let the first term and the common difference of the AP be a and b respectively.

Then,

Third term=16

a+(3–1) d=16

a+2d=16

and, A.T.Q.; 7th term=5th term + 12

a+(7–1)d=a+(5–1)d+12

a+6d=a+4d+12

6d–4d=12

2d=12

d=12/2 (12 upon 2)=6

putting d=6 in (1), we get

a+2(6)=16

a+12=16

a=16–12=4

Hence the required AP is

4,4+6,4+6+6,4+6+6+6,.......

i.e, 4,10,16,22........

Answer 2

Answer:

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