Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
correct ans dena plz.........✌️
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Answer:
Let a be the First term, a
3
be the third term, a
5
be the 5th term and a
7
be the 7th term
We know, a
n
=a+(n−1)d
⇒ a
3
=16 [ Given ]
⇒ a+(3−1)d=16
⇒ a+2d=16 ....... (1)
⇒ Now, a
7
−a
5
=12 [ Given ]
⇒ [a+(7−1)d]−[a+(5−1)d]=12
⇒ 2d=12
∴ d=6
From equation (1), we get
a+2(6)=16
a+12=16
∴ a=4
So first term is 4
We can find AP by adding d continuously.
∴ Required AP is 4,10,16,22,28,34,40,.
hope it helps you ...... bro Aap konse class me ho
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