Question

Question 16 Determine the A.P. whose third term is 16 and the 7 th term exceeds the 5 th term by 12.

Class 10 - Math - Arithmetic Progressions Page 106

Answer 1

let first term be a and the common difference be d.
T3 = a + (3-1)d = a + 2d
therefore 16 = a + 2d

and
5th term T5 = a+4d, 7th term = a +6d (bith using same logic as above)

According to question,
a +6d - 12 = a+4d
that implies   2d = 12 
d = 6....
now, 
a + 2d = 16
a +2x6 = 16
a = 16 - 12 = 4
 this means that the A.P. = 4, 10, 16, 22, 28.........

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Answer 2

a3 = 16

a + (3 − 1) d = 16
a + 2d = 16 ... (i)
a7 − a5 = 12
[a+ (7 − 1) d] − [a + (5 − 1) d]= 12
(a + 6d) − (a + 4d) = 12
2d = 12
d = 6
From equation (i), we get,
a + 2 (6) = 16
a + 12 = 16
a = 4
Therefore, A.P. will be
4, 10, 16, 22, …