consider an AS whose 6th term is 40 and 9th term is 58. find 25th term of the sequence . find the sum of first n terms of the sequence
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(i)
6th term is 40.
a + 5d = 40
(ii)
9th term is 58.
a + 8d = 58
On solving (i) & (ii), we get
a + 5d = 40
a + 8d = 58
-----------------
3d = 18
d = 6
Substitute d = 6 in (i), we get
a + 5d = 40
a + 30 = 40
a = 10
25th term of the sequence:
aₙ = a + (n - 1) * d
a₂₅ = 10 + (25 - 1) * 6
= 10 + 24 * 6
= 154
Sum of first n terms of the sequence:
S₂₅ = (25/2)[2a + (n - 1) * d]
= (25/2)[20 + (24 * 6)]
= (25/2)[164]
= 2050
Hope this helps!
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Answer:
yeerydzarshf
Step-by-step explanation:
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