um..anyone...question number 16.
Attachments:
Answers
Answered by
0
There is an AP, The 4th term is three times the first. So the relation is
3a = a + 3d, or
2a = 3d, or
a = (3/2)d …(1)
The 7th term exceeds twice the third term by 1. The relation is
a+6d = 2(a+2d)+1, or
a + 6d = 2a+4d+1, or
2a-a = 6d-4d-1, or
a = 2d-1 …(2). Equate (1) and (2)
a = (3/2)d = 2d-1, or
3d = 4d-2, or
4d-3d=2, or d = 2 and a = 3.
The first term is 3 and the common difference is 3.
Check: The series of the AP is 3,5,7,9,11,13,15.
The 4th term is 9, which is three times the first term. Correct.
The 7th term is 15, which is twice the third term plus 1. Correct.
kusum72:
i want question number 16 answer
Done
Check this one out
Answered by
0
Answer toQ16
Hzishwizjsjwjsb
Hzishwizjsjwjsb
Attachments:
its wrong...in question it is given 7th term exceeds twice the 3rd term by 1
oh....
yes
Similar questions