The 19th term of an A.P. is equal to three times its sixth term. If its 9th term is 19, find the A.P.
Answers
Answer:
The required AP is 3, 5 ,7, 9 ,....
Step-by-step explanation:
Given:
a19 = 3(a6) and a9 = 19
Let the first term of an A.P be 'a' & Common difference be 'd'.
By using the formula , nth term , an = a + (n -1)d
Case : 1
a19 = 3(a6)
a + (19 -1)d = 3 ( a + (6 - 1) d
a + 18d = 3(a + 5d)
a + 18d = 3a + 15d
a - 3a + 18d - 15d = 0
-2a + 3d = 0 ………..(1)
Case : 2
a9 = 19
a + (9 - 1)d = 19
a + 8d = 19
a = 19 - 8d…………(2)
On putting the value of 'a' in eq 1.
-2a + 3d = 0
-2(19 - 8d) + 3d = 0
- 38 + 16d + 3d = 0
- 38 + 19d = 0
19d = 0 + 38
19d = 38
d = 38/19
d = 2
On putting the value of 'd' in eq 2,
a = 19 - 8d
a = 19 - 8(2)
a = 19 - 16
a = 3
First term , a = 3
Second term ,a2 = a + d = 3 +2 = 5
Third term , a3 = a2 + d = 5 + 2 = 7
Fourth term, a4 = a3 + d = 7 + 2 =9
Hence, the required AP is 3, 5 ,7, 9 ,....
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