maths ch 5 ex 5.2 question 8 class 10
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Given that,
3rd term, a3 = 12
50th term, a50 = 106
We know that,
an = a+(n−1)d
a3 = a+(3−1)d
12 = a+2d ……………………………. (i)
In the same way,
a50 = a+(50−1)d
106 = a+49d …………………………. (ii)
On subtracting equation (i) from (ii), we get
94 = 47d
d = 2 = common difference
From equation (i), we can write now,
12 = a+2(2)
a = 12−4 = 8
a29 = a+(29−1) d
a29 = 8+(28)2
a29 = 8+56 = 64
Therefore, 29th term is 64
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