Example-7. There are 50 terms in an arithmetic progression whose third term is 12 and
the last term is 106. Find the 29th term of this arithmetic progression.
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2
ANSWER
Given, a3=12, a50=106
an=a+(n−1)d
a3=a+(3−1)d
12=a+2d ... (i)
a50=a+(50−1)d
106=a+49d ... (ii)
On subtracting (i) from (ii), we get
94=47d
∴d=2
From equation (i), we get
12=a+2(2)
⇒a=12−4=8
Now a29=a+(29−1)d
=8+(28)2
=8+56
=64
Therefore, 29th term is 64.
Answered by
0
Answer:
Given, a
3
=12, a
50
=106
a
n
=a+(n−1)d
a
3
=a+(3−1)d
12=a+2d ... (i)
a
50
=a+(50−1)d
106=a+49d ... (ii)
On subtracting (i) from (ii), we get
94=47d
∴d=2
From equation (i), we get
12=a+2(2)
⇒a=12−4=8
Now a
29
=a+(29−1)d
=8+(28)2
=8+56
=64
Therefore, 29
th
term is 64.
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