SOLVE 16 one plzzzzzz
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In this first term is 5 and 10 is the common difference
Therefore 31 term is
a+29d=5+290=295
Now 130 more then 31 term so
a+nd=295+130
n is the required term
5+n(10)=425
10n=420
n=42
So answer is 42 term
And hope it helps
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HERE IS YOUR ANSWER..⬇⬇
➡
Given,
10th term, a10 = 52
=> a + +(10-1)d = 52
=> a + 9d = 52 ...........(1)
Also,
a17 = 20+a13
=> a+(17-1)d = 20 + {a+ (13-1)d}
=> a + 16d = 20+(a+12d)
=> a + 16d = 20 + a + 12d
=> a - a + 16d - 12d = 20
=> 16d-12d = 20
=> 4d = 20
=> d = 20/4
=> d = 5
Putting the value of "d" in eq. (1),
we get,
a + 9×(5) = 52
=> a +45 = 52
=> a = 52 - 45
=> a = 7
Hence, the required AP is 7,12,17,22,27,.........
_________________________________
✝✝…HOPE…IT…HELPS…YOU…✝✝
➡
Given,
10th term, a10 = 52
=> a + +(10-1)d = 52
=> a + 9d = 52 ...........(1)
Also,
a17 = 20+a13
=> a+(17-1)d = 20 + {a+ (13-1)d}
=> a + 16d = 20+(a+12d)
=> a + 16d = 20 + a + 12d
=> a - a + 16d - 12d = 20
=> 16d-12d = 20
=> 4d = 20
=> d = 20/4
=> d = 5
Putting the value of "d" in eq. (1),
we get,
a + 9×(5) = 52
=> a +45 = 52
=> a = 52 - 45
=> a = 7
Hence, the required AP is 7,12,17,22,27,.........
_________________________________
✝✝…HOPE…IT…HELPS…YOU…✝✝
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