The sum of the 2nd and 7th term of an A.p.is 30.if its 15th term is 1 less than twice its 8th term, find the A.P.
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1
First term = a
Difference = d
2nd term = a + d(2 - 1) = a + d
7th term = a + d(7 - 1) = a + 6d
a + d + a + 6d = 30
2a + 7d = 30 --------------> (1)
8th term = a + d(8 - 1) = a + 7d
15th term = a + d(15 - 1) = a + 14d
a + 14d - 1 = 2(a + 7d)
a + 14d - 1 = 2a + 14d
So a = 1
But from (1),
2a + 7d = 30
2 + 7d = 30
7d = 28
d = 28/7
d = 4
Therefore the AP is 1, 4, 7, .......
Hope this helps! :)
AccioNerd:
So sorry, i did a calculation mistake, a = 1 and so in equation 2a + 7d = 30, d will be equal to 4 and the AP will 1, 5 7. i'm so sorry for the mistake
So please again solve this question
Okay hang on
thanks
Here you go
Check from a = 1
Yes
I accidentally took it as a = -1, sorry abt that
Yes my a I also -1
Yup
Answered by
7
Here is your solution
Given:-
t2+t7=30
=>a+d+a+6d=30
=>2a+7d=30..................... (1)
A/q
15th term=2×8th term-1
=>a+14d=2×(a+7d)-1
=>a+14d=2a+14d-1
=>14d-14d+1=2a-a
=>1=a
Putting the value of a in equation (1)
2a×7d=30
2×1+7d=30
2+7d=30
7d=30-2
7d=28
d=4
A.p are a,a+d,a+2d,a+3d,a+4d
=>A.p are 1,5,9,13.........
HOPE IT HELPS YOU
Given:-
t2+t7=30
=>a+d+a+6d=30
=>2a+7d=30..................... (1)
A/q
15th term=2×8th term-1
=>a+14d=2×(a+7d)-1
=>a+14d=2a+14d-1
=>14d-14d+1=2a-a
=>1=a
Putting the value of a in equation (1)
2a×7d=30
2×1+7d=30
2+7d=30
7d=30-2
7d=28
d=4
A.p are a,a+d,a+2d,a+3d,a+4d
=>A.p are 1,5,9,13.........
HOPE IT HELPS YOU
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