Find the 31st term of the AP ,whose fifth term is 32 and 8th term is 41.
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0
Answer:
31st term of the AP is 110.
Solution:-
Let:-
The first term of AP be a.
The common difference be d.
According to the question, we have
=>>Tn = a + (n - 1)d
=>> T5 = a + (5 - 1)d
=>> 32 = a + 4d _______eq1
And:-
=>> T8 = a + 7d
=>> 41 = a + 7d __________eq2
Subtracting eq1 from eq2 we get,
=>> (a + 7d) - (a + 4d) = 41 - 32
=>> a + 7d -a -4d = 9
=>> 3d = 9
=>> d = 3
Now,using the value of d in eq1 we get
=>> 32 = a + 4 × 3
=> 32 - 12 = a = 20
Therefore:-
=>> T31 = a + (n - 1)d
=>> T31 = 20 + 30×3
=>> T31 = 20 + 90
=>> T31 = 110
Hence:-
31st term of the AP is 110.
Answered by
0
Answer:
110
Step-by-step explanation:
see in the attachment above
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