An Al consists of 50 terms of which
3rd term 12 and the last tim
is 106. find 30th term.
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1
Answer:
The 30th term is 66.
Step-by-step explanation:
Let,
the first term and the common difference of the AP be a and d respectively .
Given,
No. of terms = 50
We have,
t3 = 12
=> a + 2d = 12 -----(1)
Also,
t50 = 106
=> a + 49 d = 106 ------(2)
Subtracting (2) from (1) , we get
a + 49d - a - 2d = 106 - 12
=> 47d = 94
=> d = 94/47
•°• d = 2
Now substituting the value of d in (1)
a + 2d = 12
=> a + 2*2 = 12
=> a + 4 = 12
=> a = 12 - 4
•°• a = 8
•°• t30 = a + 29 d
= 8 + 29*2
= 8 + 58
= 66
•°• The 30th term is 66.
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